Bash is not expanding the ~ character in the argument --home_dir=~. For example:
$ echo --home_dir=~
--home_dir=~
Bash does expand ~ when I leave out the hyphens:
$ echo home_dir=~
home_dir=/home/reedwm
Why does Bash have this behavior? This is irritating, as paths with ~ are not expanded when I specify that path as an argument to a command.
bash is somewhat mistakenly treating home_dir=~ as an assignment. As such, the ~ is eligible for expansion:
Each variable assignment is checked for unquoted tilde-prefixes immediately following a : or the first =. In these cases, tilde expansion is
also performed.
Since --home_dir is not a valid identifier, that string is not mistaken for an assignment.
Arguably, you have uncovered a bug in bash. (I say arguably, because if you use set -k, then home_dir=~ is an assignment, even though it is after, not before, the command name.)
However, when in doubt, quote a string that is meant to be treated literally whether or not it is subject to any sort of shell processing.
echo '--home_dir=~'
Update: This is intentional, according to the maintainer, to allow assignment-like argument for commands like make to take advantage of tilde-expansion. (And commands like export, which for some reason I was thinking were special because they are builtins, but tilde expansion would have to occur before the actual command is necessarily known.)
Like chepner says in their answer, according to the documentation, it shouldn't expand it even in echo home_dir=~. But for some reason it does expand it in any word that even looks like an assignment, and has done so at least as far back as in 3.2.
Most other shells also don't expand the tilde except in cases where it really is at the start of the word, so depending on it working might not be such a good idea.
Use "$HOME" instead if you want it to expand, and "~" if you want a literal tilde. E.g.
$ echo "~" --foo="$HOME"
~ --foo=/home/itvirta
(The more complex cases are harder to do manually, but most of the time it's the running user's own home directory one wants.)
Well, that's because in echo --home_dir=~, the '~' does not begin the word and the output of echo is not considered a variable assignment. Specifically, man bash "Tilde Expansion" provides expansion if
If a word begins with an unquoted tilde character (~); or
variable assignment is checked for unquoted tilde-prefixes immediately following a : or the first =.
You case doesn't qualify as either.
Related
I recently converted a shell script from bash to zsh and got a strange error. I had a command like
HOST="User#1.1.1.1"
scp "$BASE_DIR/path/to/file" $HOST:some\\path
This worked fine in bash, but zsh failed with a bad substitution. I fixed this by change $HOST to ${HOST}, but I'm curious as to why this was necessary. Also, strangely, I had a few such scp commands, and all of them "worked" except the first one. However, I ended up with a file called User#1.1.1.1 on my filesystem which was really unexpected. Why did this subtle change make such a big difference?
Two possible problems (1) Extra '$' at the beginning of the assignment, and (2) embedded spaces.
The first potential problem is the assignment in the style $var=foo. In zsh like in other sh-like engines (ksh, bash, ...), the assignment operation is VAR=value - no $.
The second potential problem are the spaces. No spaces are allowed between the variables name, the '=' and the value. Spaces in the value must be escaped (with quotes, or backslash)
Potential correction:
HOST=User#1.1.1.1
scp "$BASE_DIR/path/to/file" $HOST:some\\path
As chepner mentioned in the commments, zsh has modifiers that are added via :. So $HOST:some was interpreted as $HOST:s by zsh.
A list of modifiers can be found here: https://web.cs.elte.hu/local/texinfo/zsh/zsh_23.html
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$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'
I've written this script:
#!/bin/bash
file="~/Desktop/test.txt"
echo "TESTING" > $file
The script doesn't work; it gives me this error:
./tester.sh: line 4: ~/Desktop/test.txt: No such file or directory
What am I doing wrong?
Try replacing ~ with $HOME. Tilde expansion only happens when the tilde is unquoted. See info "(bash) Tilde Expansion".
You could also do file=~/Desktop without quoting it, but if you ever replace part of this with something with a field separator in it, then it will break. Quoting the values of variables is probably a good thing to get into the habit of anyway. Quoting variable file=~/"Desktop" will also work but I think that is rather ugly.
Another reason to prefer $HOME, when possible: tilde expansion only happens at the beginnings of words. So command --option=~/foo will only work if command does tilde expansion itself, which will vary by command, while command --option="$HOME/foo" will always work.
FYI, you can also use eval:
eval "echo "TESTING" > $file"
The eval takes the command as an argument and it causes the shell to do the Tilde expansion.
Nearly everywhere I've read, including Google's bash scripting style guide mention to necessity of quoting command substitutions (except when specifically desired of course).
I understand the when/where/why of quoting command substitutions during general use. For example: echo "$(cat <<< "* useless string *")" rather than echo $(...)
However for variable assignments specifically, I have seen so many examples as such:
variable="$(command)"
Yet I have found no instances where variable=$(command) is not equivalent.
variable="$(echo "*")" and variable=$(echo "*") both set the value to '*'.
Can anyone give any situations where leaving the substitution unquoted during variable assigment would actually cause a problem?
The shell does not perform word splitting for variable assignments (it is standardized that way by POSIX and you can rely on it). Thus you do not need double quotes (but you can use them without making the result different) in
variable=$(command) # same as variable="$(command)"
However, word-splitting is performed before executing commands, so in
echo $(command)
echo "$(command)"
the result may be different. The latter keeps all multi-space sequences, while the former makes each word a different argument to echo. It is up to you to decide which is the desired behavior.
Interesting shell quirk: there is one more place where quoting a substitution or not makes no difference, namely the expression in a case expr in construct.
case $FOO in
(frob) ...;;
esac
is indistinguishable from
case "$FOO" in
(frob) ...;;
esac
When using BASH, those two lines are 100% equivalent:
variable="$(command)"
variable=$(command)
while these two aren't:
echo $(command)
echo "$(command)"
Alas, the human brain isn't a computer. It's especially not as reliable when it comes to repeat a job.
So chances are that if you mix styles (i.e. you quote when you use command arguments but you don't quote when you assign a variable), that once in a while, you'll get it wrong.
Worse, once in a while you will want the command result to be expanded into individual words. And the next guy who reads your code will wonder if he's looking at a bug.
Conculsion: Since the first two lines are the same, it's easier on the average brain to always quote $() (even when it's not necessary) to make sure you always quote when you have to.
Testing with so many different commands, it is the same. Can't find a difference.