I have a script which can overwrite values in a configuration file using options, for example, option --password can overwrite the setting in the configuration file (please note, this is not a discussion about security). However a password can contain contain characters, that are by bash, recognized as special characters, and those characters needs to be escaped or placed within " ".
Now I understand this. Although I can't say whom will be using this script in the future, so I would like to save he or she the trouble of having an incorrect password simply because, he or she forgot to place the password within " " or escape the special character.
What is the best way of dealing with such an input?
Thanks in advance.
Hm.. Double quotes are not enough. Must use single quotes, because the rare situation, for example
mycommand --password "AAA$PWD" #is worng for any exported environment varname
mycommand --password 'AAA$PWD' #ok
Here is no way avoid this, because your users using a sort of shell, what have variable expansions and metachar-globbing. Your command getting already expanded args, so here is no way catch this in your script.
The only way - as #Bohemian told above - reading the password from a tty. You can write a simple wrapper around your current script, what will read the password from a tty and after will execute your script with properly escaped --pasword argument.
There is a simple, but not very intuitive solution.
Encapsulate the character that is causing problem with '"CARACTER"'. And put all the password string between single quotes.
In my case the character that was causing the problem was the ' (single quote).
So if I have a command like that:
mycommand --password 'AAA'PWD'
Replace by that:
mycommand --password 'AAA'"'"'PWD'
Now my password is a concatenation of three strings. Explanation:
'AAA' + "'" + 'PWD' # plus sign is just to make clear the contatenation.
That's works.
Related
I'm building a shell script (trying to be POSIX compliant) and I'm stuck in an issue.
The script is supposed to receive an URL and do some things with it's content.
myscript www.pudim.com.br/?&args=ok
The thing is, the ampersand symbol is interpreted as a command additive, and giving to my script only the www.pudim.com.br/? part as an argument.
I know that the right workaround would be to surround the URL with quotes but, because I need to use this script several times in a row, I wanted to paste the URL's without having to embrace it with quotes every time.
Is there some way to get the full URL argument, somehow bypassing the ampersand?
Quotes for full URL
Wrapping the URL in quotes will be your only chance. See popular shell utility curl, as it states for its core argument URL:
When using [] or {} sequences when invoked from a command line prompt,
you probably have to put the full URL within double quotes to avoid
the shell from interfering with it. This also goes for other
characters treated special, like for example '&', '?' and '*'.
See also this question and that.
Extra argument(s) for specifying query parameters
You can also pass query parameter (key-value pair) as separate argument. So you can bypass & as separator. See curl's -F option:
-F, --form <name=content>
Read URL from STDIN
If your script allows user interaction you could read the unescaped URL (including metachars as &) from an uninterpreted input-source. See this tutorial.
You can escape just the ampersand; quotes effectively escape every character between them.
myscript www.pudim.com.br/\?\&args=ok # The ? should be escaped as well
There is no solution that lets you avoid all quoting, as & is a shell metacharacter whose unquoted meaning cannot be disabled. The & terminates the preceding command, causing it to be run in a background process; adding some redundant whitespace, you attempt is the same as
myscript www.pudim.com.br/? &
args=ok
Unescaped, the ? will cause the URL to be treated as a pattern to expand. However, it's unlikely the pattern will match any existing file, and bash's default behavior is to treat an unmatched pattern literally. (The failglob option will treat it as an error, and the nullglob option will make the URL disappear completely from the command line, but neither option is enabled by default.)
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 am trying to give an argument to my python program through the terminal.
For this I am using the lines:
import sys
something = sys.argv[1]
I now try to put in a string like this through the bash terminal:
python my_script.py 2m+{N7HiwH3[>!"4y?t9*y#;/$Ar3wF9+k$[3hK/WA=aMzF°L0PaZTM]t*P|I_AKAqIb0O4# cm=sl)WWYwEg10DDv%k/"c{LrS)oVd§4>8bs:;9u$ *W_SGk3CXe7hZMm$nXyhAuHDi-q+ug5+%ioou.,IhC]-_O§V]^,2q:VBVyTTD6'aNw9:oan(s2SzV
This returns a bash error because some of the characters in the string are bash special characters.
How can I use the string exactly as it is?
You can put the raw string into a file, for example like this, with cat and a here document.
cat <<'EOF' > file.txt
2m+{N7HiwH3[>!"4y?t9*y#;/$Ar3wF9+k$[3hK/WA=aMzF°L0PaZTM]t*P|I_AKAqIb0O4# cm=sl)WWYwEg10DDv%k/"c{LrS)oVd§4>8bs:;9u$ *W_SGk3CXe7hZMm$nXyhAuHDi-q+ug5+%ioou.,IhC]-_O§V]^,2q:VBVyTTD6'aNw9:oan(s2SzV
EOF
and then run
python my_script.py "$(< file.txt)"
You can also use the text editor of your choice for the first step if you prefer that.
If this is a reoccurring task, which you have to perform from time to time, you can make your life easier with a little alias in your shell:
alias escape='read -r string ; printf "Copy this:\n%q\n" "${string}"'
It is using printf "%q" to escape your input string.
Run it like this:
escape
2m+{N7HiwH3[>!"4y?t9*y#;/$Ar3wF9+k$[3hK/WA=aMzF°L0PaZTM]t*P|I_AKAqIb0O4# cm=sl)WWYwEg10DDv%k/"c{LrS)oVd§4>8bs:;9u$ *W_SGk3CXe7hZMm$nXyhAuHDi-q+ug5+%ioou.,IhC]-_O§V]^,2q:VBVyTTD6'aNw9:oan(s2SzV
Copy this:
2m+\{N7HiwH3\[\>\!\"4y\?t9\*y#\;/\$Ar3wF9+k\$\[3hK/WA=aMzF°L0PaZTM\]t\*P\|I_AKAqIb0O4#\ cm=sl\)WWYwEg10DDv%k/\"c\{LrS\)oVd§4\>8bs:\;9u\$\ \*W_SGk3CXe7hZMm\$nXyhAuHDi-q+ug5+%ioou.\,IhC\]-_O§V\]\^\,2q:VBVyTTD6\'aNw9:oan\(s2SzV
You can use the escaped string directly in your shell, without additional quotes, like this:
python my_script.py 2m+\{N7HiwH3\[\>\!\"4y\?t9\*y#\;/\$Ar3wF9+k\$\[3hK/WA=aMzF°L0PaZTM\]t\*P\|I_AKAqIb0O4#\ cm=sl\)WWYwEg10DDv%k/\"c\{LrS\)oVd§4\>8bs:\;9u\$\ \*W_SGk3CXe7hZMm\$nXyhAuHDi-q+ug5+%ioou.\,IhC\]-_O§V\]\^\,2q:VBVyTTD6\'aNw9:oan\(s2SzV
In order to make life easier, shells like bash do a little bit of extra work to help users pass the correct arguments to the programs they instruct it to execute. This extra work usually results in predictable argument arrays getting passed to programs.
Oftentimes, though, this extra help results in unexpected arguments getting passed to programs; and sometimes results in the execution of undesired additional commands. In this case, though, it ended up causing Bash to emit an error.
In order to turn off this extra work, Bash allows users to indicate where arguments should begin and end by surrounding them by quotation marks. Bash supports both single quotes (') and double quotes (") to delimit arguments. As a last resort, if a string may contain single and double quotes (or double quotes are required but aren't aggressive enough), Bash allows you to indicate that a special- or whitespace-character should be part of the adjacent argument by preceding it with a backslash (\\).
If this method of escaping arguments is too cumbersome, it may be worth simplifying your program's interface by having it consume this data from a file instead of a command line argument. Another option is to create a program that loads the arguments from a more controlled location (like a file) and directly execs the target program with the desired argument array.
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've a bash script that simple has to add new user and sign a password that is passed when script is called:
./adduser_script username password
and the password is then used as a parameter in the script like this:
/usr/sbin/useradd ... -p `openssl passwd -1 "$2"` ...
the problem occurs of course when password contains special characters like $#, $* itd. So when i call the script:
/adduser_script username aa$#bbb
and after script ends password looks like: aabbb (so the special charakters are removed from original password). The question is how can I correctly pass the original password with special charakters to the script?
Thanks in advance,
Regards
have you tried strong qoutes ??
use 'aa$#bb' instead of weak qoutes i.e. "aa$#bb"
for example: check with echo command
echo "aa$#bb" will print aabb
while
echo 'aa$#bb' will print aa$#bb
In your script use
/usr/sbin/useradd ... -p `openssl passwd -1 '$2'` ...
now you need not to worry about qoutes while passing password as argument.
The problem is probably not in your script at all, but rather on how you call it. At least from the snippets you provide, it doesn't seem like the password field is being evaluated.
So, when you call the script, if an argument contains something like $a, bash will replace it with the value of the variable a, or an empty string if it is unset.
If $ needs to be in the password, then it needs to be in single quotes.
./adduser_script username 'password$#aaa'
You can also use double quotes with escape .
For example: set password "MyComplexP\#\$\$word"
I have been facing similar issue and here is my in take on it.
One thing we need to consider is the user of single and double quotes in our script file.
if we put a value, variable in single quote the value will be as it is and it will not be replaced with actual value we reference.
example,
name='java'
echo '$name' ---> prints $name to console , but
echo "$name" ---> prints java to console.
So, play around on which quote is best to used based on your circumstance.
/usr/sbin/useradd ... -p "$(openssl passwd -1 '$2')"