this is my first post here (maybe more later ^^)
So, here we are :
I'm using Gitlab-CI, docker, makefile, firebase.
So in my Makefile :
firebase-deploy: ## Deploy the application
$(FIREBASE) deploy -m "$(comment)"
in my CI :
variables:
DOCKER_DRIVER: overlay2
MESSAGE: $CI_COMMIT_MESSAGE
deploy-firebase:
stage: deploy
script:
- make install
- make deploy comment="$MESSAGE"
The problem is actually, i have
docker-compose exec -T js node_modules/firebase-tools/lib/bin/firebase.js deploy -m "test ci ?
/bin/sh: syntax error: unterminated quoted string
So, as you can see, the double quote at the end is ... not here.
I tested a few things
\"Comment\"
\'Comment\'
"Comment\"
And more, but none of them work.
One thing interesting is that if i commit FROM gitlab.com it work as expected, but from my computer on push : nop.
Do you have any advice, or something to look for ?
Thanks!
If your commit message contains end of lines you cannot pass it to your script as it is. Try, maybe, to substitute end of lines by another character, e.g. -. Of course the way to do it depends on the shell that executes your CI scripts. If it is bash, for instance, the following should work:
make deploy comment="${MESSAGE//$'\n'/-}"
If you must also escape other characters you can chain substitutions. Example if you want to substitute end of lines and double quotes by - and ':
c="${MESSAGE//$'\n'/-}"; c="${c//\"/\'}"; make deploy comment="$c"
To learn more about these bash substitutions just type man bash and read section EXPANSION, sub-section Parameter Expansion. The parameter expansion I used is:
${parameter/pattern/string}
that replaces pattern by string in the value of variable parameter (in bash parlance parameter is another name for variable). When the pattern begins with a / all matches of pattern are replaced with string, else only the first match is replaced. So the first parameter expansion:
c="${MESSAGE//$'\n'/-}"
substitutes all end-of-line characters in the value of variable MESSAGE by - (more on $'\n' later) and assigns the result to variable c. Note the enclosing double quotes, they are needed because you can have spaces in the message. And the second:
c="${c//\"/\'}"
substitutes all " in the value of variable c by ' and assigns again the result to variable c.
In this case the most complicated is to put a end-of-line character in the first pattern, a double quote in the second and a single quotes in the second string. The $'string' construct is explained in the QUOTING section of the bash manual:
Words of the form $'string' are treated specially. The word expands to
string, with backslash-escaped characters replaced as specified by the
ANSI C standard.
So $'\n' is replaced by the end-of-line character. In the second pattern and string you just need to escape the double and single quotes with a backslash to have them interpreted literally.
Finally the 3 commands of this little shell script are chained by ;, the bash sequence operator. They are executed one after the other in the same bash invocation.
Related
In the first part of my question I will provide some background info as a
service to the community. The second part contains the actual question.
Part I
Assume I've created the following alias:
alias ls='ls -r'
I know how to temporarily unalias (i.e., override this alias) in the following
ways, using:
1) the full pathname of the command: /bin/ls
2) command substitution: $(which ls)
3) the command builtin: command ls
4) double quotation marks: "ls"
5) single quotation marks: 'ls'
6) a backslash character: \ls
Case 1 is obvious and case 2 is simply a variation. The command builtin in case 3 was designed to ignore shell functions, but apparently it also works for circumventing aliases. Finally, cases 4 and 5 are consistent with both the POSIX standard (2.3.1):
"a resulting word that is identified
to be the command name word of a
simple command shall be examined to
determine whether it is an unquoted,
valid alias name."
and the Bash Reference Manual (6.6):
"The first word of each simple
command, if unquoted, is checked to
see if it has an alias."
Part II
Here's the question: why is case 6 (overriding the alias by saying \ls)
considered quoting the word? In keeping with the style of this question, I am looking for references to the "official" documentation.
The documentation says that a backslash only escapes the following
character, as opposed to single and double quotation marks, which quote a
sequence of characters. POSIX standard (2.2.1):
"A backslash that is not quoted shall
preserve the literal value of the
following character, with the
exception of a < newline >"
Bash Reference Manual (3.1.2.1):
"A non-quoted backslash ‘\’ is the
Bash escape character. It preserves
the literal value of the next
character that follows, with the
exception of newline."
(BTW, isn't "the next character that follows" a bit of overkill?)
A possible answer might be that this situation isn't that special: it is
similar to a few cases in ANSI-C quoting, e.g. \nnn. However, that is still
escaping a single character (the eight-bit character whose value is the octal
value nnn), not a sequence of characters.
Historically, and maintained by POSIX, quoting any part of the word causes the entire word to be considered quoted for the purposes of functions and alias expansion. It also applies to quoting the end token for a here document:
cat << \EOF
this $text is fully quoted
EOF
Just for completion, here's yet another way to suppress alias & function lookups (by clearing the entire shell environment for a single command):
# cf. http://bashcurescancer.com/temporarily-clearing-environment-variables.html
env -i ls
\ls only quotes the first character rather than the whole word. It's equivalent to writing 'l's.
You can verify it like this:
$ touch \?l
$ \??
bash: ?l: command not found
If \?? quoted the whole word it would say ?? not found rather than ?l not found.
I.e. it has the same effect as:
$ '?'?
bash: ?l: command not found
rather than:
$ '??'
bash: ??: command not found
There's a bash file with something like this:
FOO=${BAR:-"/some/path/with/$VAR/in/it"}
Are those double quotes necessary? Based on the following test, I'd say no, and that no quote at all is needed in the above assignment. In fact, it's the user of that variable that needs to expand it within double quotes to avoid wrong splitting.
touch 'some file' # create a file
VAR='some file' # create a variable for that file name
FOO=${BAR:-$VAR} # use it with the syntax above, but no quotes
ls -l "$FOO" # the file does exist (here we do need double quotes)
ls -l $FOO # without quotes it fails searching for files `some` and `file`
rm 'some file' # remove temporary file
Am I correct? Or there's something more?
Are those double quotes necessary?
Not in this case, no.
Am I correct?
Yes. And it's always the user of the variable that has to quote it - field splitting is run when expanding the variable, so when using it it has to be quoted.
There are exceptions, like case $var in and somevar1=$somevar2 - contexts which do not run field splitting, so like do not require quoting. But anyway, quotes do not hurt in such cases and can be used anyway.
Or there's something more?
From POSIX shell:
2.6.2 Parameter Expansion
In addition, a parameter expansion can be modified by using one of the following formats. In each case that a value of word is needed (based on the state of parameter, as described below), word shall be subjected to tilde expansion, parameter expansion, command substitution, and arithmetic expansion.
${parameter:-word}
Because field splitting expansion is not run over word inside ${parameter:-word}, indeed, quoting doesn't do much.
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$variable'
In this format, $ is escaped and the variable is not expanded.
I tried the following variations but they were rejected:
repo forall -c '...."$variable" '
repo forall -c " '....$variable' "
" repo forall -c '....$variable' "
repo forall -c "'" ....$variable "'"
If I substitute the value in place of the variable the command is executed just fine.
Please tell me where am I going wrong.
Inside single quotes everything is preserved literally, without exception.
That means you have to close the quotes, insert something, and then re-enter again.
'before'"$variable"'after'
'before'"'"'after'
'before'\''after'
Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?
Do not concatenate strings interpreted by a shell
You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).
Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.
For example, the following is very unsafe. DON'T DO THIS
script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"
If the contents of $myvar is untrusted, here is an exploit:
myvar='foo"; echo "you were hacked'
Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:
script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"
Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.
The repo command can't care what kind of quotes it gets. If you need parameter expansion, use double quotes. If that means you wind up having to backslash a lot of stuff, use single quotes for most of it, and then break out of them and go into doubles for the part where you need the expansion to happen.
repo forall -c 'literal stuff goes here; '"stuff with $parameters here"' more literal stuff'
Explanation follows, if you're interested.
When you run a command from the shell, what that command receives as arguments is an array of null-terminated strings. Those strings may contain absolutely any non-null character.
But when the shell is building that array of strings from a command line, it interprets some characters specially; this is designed to make commands easier (indeed, possible) to type. For instance, spaces normally indicate the boundary between strings in the array; for that reason, the individual arguments are sometimes called "words". But an argument may nonetheless have spaces in it; you just need some way to tell the shell that's what you want.
You can use a backslash in front of any character (including space, or another backslash) to tell the shell to treat that character literally. But while you can do something like this:
reply=\”That\'ll\ be\ \$4.96,\ please,\"\ said\ the\ cashier
...it can get tiresome. So the shell offers an alternative: quotation marks. These come in two main varieties.
Double-quotation marks are called "grouping quotes". They prevent wildcards and aliases from being expanded, but mostly they're for including spaces in a word. Other things like parameter and command expansion (the sorts of thing signaled by a $) still happen. And of course if you want a literal double-quote inside double-quotes, you have to backslash it:
reply="\"That'll be \$4.96, please,\" said the cashier"
Single-quotation marks are more draconian. Everything between them is taken completely literally, including backslashes. There is absolutely no way to get a literal single quote inside single quotes.
Fortunately, quotation marks in the shell are not word delimiters; by themselves, they don't terminate a word. You can go in and out of quotes, including between different types of quotes, within the same word to get the desired result:
reply='"That'\''ll be $4.96, please," said the cashier'
So that's easier - a lot fewer backslashes, although the close-single-quote, backslashed-literal-single-quote, open-single-quote sequence takes some getting used to.
Modern shells have added another quoting style not specified by the POSIX standard, in which the leading single quotation mark is prefixed with a dollar sign. Strings so quoted follow similar conventions to string literals in the ANSI standard version of the C programming language, and are therefore sometimes called "ANSI strings" and the $'...' pair "ANSI quotes". Within such strings, the above advice about backslashes being taken literally no longer applies. Instead, they become special again - not only can you include a literal single quotation mark or backslash by prepending a backslash to it, but the shell also expands the ANSI C character escapes (like \n for a newline, \t for tab, and \xHH for the character with hexadecimal code HH). Otherwise, however, they behave as single-quoted strings: no parameter or command substitution takes place:
reply=$'"That\'ll be $4.96, please," said the cashier'
The important thing to note is that the single string that gets stored in the reply variable is exactly the same in all of these examples. Similarly, after the shell is done parsing a command line, there is no way for the command being run to tell exactly how each argument string was actually typed – or even if it was typed, rather than being created programmatically somehow.
Below is what worked for me -
QUOTE="'"
hive -e "alter table TBL_NAME set location $QUOTE$TBL_HDFS_DIR_PATH$QUOTE"
EDIT: (As per the comments in question:)
I've been looking into this since then. I was lucky enough that I had repo laying around. Still it's not clear to me whether you need to enclose your commands between single quotes by force. I looked into the repo syntax and I don't think you need to. You could used double quotes around your command, and then use whatever single and double quotes you need inside provided you escape double ones.
just use printf
instead of
repo forall -c '....$variable'
use printf to replace the variable token with the expanded variable.
For example:
template='.... %s'
repo forall -c $(printf "${template}" "${variable}")
Variables can contain single quotes.
myvar=\'....$variable\'
repo forall -c $myvar
I was wondering why I could never get my awk statement to print from an ssh session so I found this forum. Nothing here helped me directly but if anyone is having an issue similar to below, then give me an up vote. It seems any sort of single or double quotes were just not helping, but then I didn't try everything.
check_var="df -h / | awk 'FNR==2{print $3}'"
getckvar=$(ssh user#host "$check_var")
echo $getckvar
What do you get? A load of nothing.
Fix: escape \$3 in your print function.
Does this work for you?
eval repo forall -c '....$variable'
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$variable'
In this format, $ is escaped and the variable is not expanded.
I tried the following variations but they were rejected:
repo forall -c '...."$variable" '
repo forall -c " '....$variable' "
" repo forall -c '....$variable' "
repo forall -c "'" ....$variable "'"
If I substitute the value in place of the variable the command is executed just fine.
Please tell me where am I going wrong.
Inside single quotes everything is preserved literally, without exception.
That means you have to close the quotes, insert something, and then re-enter again.
'before'"$variable"'after'
'before'"'"'after'
'before'\''after'
Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?
Do not concatenate strings interpreted by a shell
You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).
Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.
For example, the following is very unsafe. DON'T DO THIS
script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"
If the contents of $myvar is untrusted, here is an exploit:
myvar='foo"; echo "you were hacked'
Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:
script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"
Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.
The repo command can't care what kind of quotes it gets. If you need parameter expansion, use double quotes. If that means you wind up having to backslash a lot of stuff, use single quotes for most of it, and then break out of them and go into doubles for the part where you need the expansion to happen.
repo forall -c 'literal stuff goes here; '"stuff with $parameters here"' more literal stuff'
Explanation follows, if you're interested.
When you run a command from the shell, what that command receives as arguments is an array of null-terminated strings. Those strings may contain absolutely any non-null character.
But when the shell is building that array of strings from a command line, it interprets some characters specially; this is designed to make commands easier (indeed, possible) to type. For instance, spaces normally indicate the boundary between strings in the array; for that reason, the individual arguments are sometimes called "words". But an argument may nonetheless have spaces in it; you just need some way to tell the shell that's what you want.
You can use a backslash in front of any character (including space, or another backslash) to tell the shell to treat that character literally. But while you can do something like this:
reply=\”That\'ll\ be\ \$4.96,\ please,\"\ said\ the\ cashier
...it can get tiresome. So the shell offers an alternative: quotation marks. These come in two main varieties.
Double-quotation marks are called "grouping quotes". They prevent wildcards and aliases from being expanded, but mostly they're for including spaces in a word. Other things like parameter and command expansion (the sorts of thing signaled by a $) still happen. And of course if you want a literal double-quote inside double-quotes, you have to backslash it:
reply="\"That'll be \$4.96, please,\" said the cashier"
Single-quotation marks are more draconian. Everything between them is taken completely literally, including backslashes. There is absolutely no way to get a literal single quote inside single quotes.
Fortunately, quotation marks in the shell are not word delimiters; by themselves, they don't terminate a word. You can go in and out of quotes, including between different types of quotes, within the same word to get the desired result:
reply='"That'\''ll be $4.96, please," said the cashier'
So that's easier - a lot fewer backslashes, although the close-single-quote, backslashed-literal-single-quote, open-single-quote sequence takes some getting used to.
Modern shells have added another quoting style not specified by the POSIX standard, in which the leading single quotation mark is prefixed with a dollar sign. Strings so quoted follow similar conventions to string literals in the ANSI standard version of the C programming language, and are therefore sometimes called "ANSI strings" and the $'...' pair "ANSI quotes". Within such strings, the above advice about backslashes being taken literally no longer applies. Instead, they become special again - not only can you include a literal single quotation mark or backslash by prepending a backslash to it, but the shell also expands the ANSI C character escapes (like \n for a newline, \t for tab, and \xHH for the character with hexadecimal code HH). Otherwise, however, they behave as single-quoted strings: no parameter or command substitution takes place:
reply=$'"That\'ll be $4.96, please," said the cashier'
The important thing to note is that the single string that gets stored in the reply variable is exactly the same in all of these examples. Similarly, after the shell is done parsing a command line, there is no way for the command being run to tell exactly how each argument string was actually typed – or even if it was typed, rather than being created programmatically somehow.
Below is what worked for me -
QUOTE="'"
hive -e "alter table TBL_NAME set location $QUOTE$TBL_HDFS_DIR_PATH$QUOTE"
EDIT: (As per the comments in question:)
I've been looking into this since then. I was lucky enough that I had repo laying around. Still it's not clear to me whether you need to enclose your commands between single quotes by force. I looked into the repo syntax and I don't think you need to. You could used double quotes around your command, and then use whatever single and double quotes you need inside provided you escape double ones.
just use printf
instead of
repo forall -c '....$variable'
use printf to replace the variable token with the expanded variable.
For example:
template='.... %s'
repo forall -c $(printf "${template}" "${variable}")
Variables can contain single quotes.
myvar=\'....$variable\'
repo forall -c $myvar
I was wondering why I could never get my awk statement to print from an ssh session so I found this forum. Nothing here helped me directly but if anyone is having an issue similar to below, then give me an up vote. It seems any sort of single or double quotes were just not helping, but then I didn't try everything.
check_var="df -h / | awk 'FNR==2{print $3}'"
getckvar=$(ssh user#host "$check_var")
echo $getckvar
What do you get? A load of nothing.
Fix: escape \$3 in your print function.
Does this work for you?
eval repo forall -c '....$variable'
In the first part of my question I will provide some background info as a
service to the community. The second part contains the actual question.
Part I
Assume I've created the following alias:
alias ls='ls -r'
I know how to temporarily unalias (i.e., override this alias) in the following
ways, using:
1) the full pathname of the command: /bin/ls
2) command substitution: $(which ls)
3) the command builtin: command ls
4) double quotation marks: "ls"
5) single quotation marks: 'ls'
6) a backslash character: \ls
Case 1 is obvious and case 2 is simply a variation. The command builtin in case 3 was designed to ignore shell functions, but apparently it also works for circumventing aliases. Finally, cases 4 and 5 are consistent with both the POSIX standard (2.3.1):
"a resulting word that is identified
to be the command name word of a
simple command shall be examined to
determine whether it is an unquoted,
valid alias name."
and the Bash Reference Manual (6.6):
"The first word of each simple
command, if unquoted, is checked to
see if it has an alias."
Part II
Here's the question: why is case 6 (overriding the alias by saying \ls)
considered quoting the word? In keeping with the style of this question, I am looking for references to the "official" documentation.
The documentation says that a backslash only escapes the following
character, as opposed to single and double quotation marks, which quote a
sequence of characters. POSIX standard (2.2.1):
"A backslash that is not quoted shall
preserve the literal value of the
following character, with the
exception of a < newline >"
Bash Reference Manual (3.1.2.1):
"A non-quoted backslash ‘\’ is the
Bash escape character. It preserves
the literal value of the next
character that follows, with the
exception of newline."
(BTW, isn't "the next character that follows" a bit of overkill?)
A possible answer might be that this situation isn't that special: it is
similar to a few cases in ANSI-C quoting, e.g. \nnn. However, that is still
escaping a single character (the eight-bit character whose value is the octal
value nnn), not a sequence of characters.
Historically, and maintained by POSIX, quoting any part of the word causes the entire word to be considered quoted for the purposes of functions and alias expansion. It also applies to quoting the end token for a here document:
cat << \EOF
this $text is fully quoted
EOF
Just for completion, here's yet another way to suppress alias & function lookups (by clearing the entire shell environment for a single command):
# cf. http://bashcurescancer.com/temporarily-clearing-environment-variables.html
env -i ls
\ls only quotes the first character rather than the whole word. It's equivalent to writing 'l's.
You can verify it like this:
$ touch \?l
$ \??
bash: ?l: command not found
If \?? quoted the whole word it would say ?? not found rather than ?l not found.
I.e. it has the same effect as:
$ '?'?
bash: ?l: command not found
rather than:
$ '??'
bash: ??: command not found