Say I have a folder called Foo located in /home/user/ (my /home/user also being represented by ~).
I want to have a variable
a="~/Foo" and then do
cd $a
I get
-bash: cd: ~/Foo: No such file or directory
However if I just do cd ~/Foo it works fine. Any clue on how to get this to work?
You can do (without quotes during variable assignment):
a=~/Foo
cd "$a"
But in this case the variable $a will not store ~/Foo but the expanded form /home/user/Foo. Or you could use eval:
a="~/Foo"
eval cd "$a"
You can use $HOME instead of the tilde (the tilde is expanded by the shell to the contents of $HOME).
Example:
dir="$HOME/Foo";
cd "$dir";
Although this question is merely asking for a workaround, this is listed as the duplicate of many questions that are asking why this happens, so I think it's worth giving an explanation. According to https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06:
The order of word expansion shall be as follows:
Tilde expansion, parameter expansion, command substitution, and arithmetic expansion shall be performed, beginning to end.
When the shell evaluates the string cd $a, it first performs tilde expansion (which is a no-op, since $a does not contain a tilde), then it expands $a to the string ~/Foo, which is the string that is finally passed as the argument to cd.
A much more robust solution would be to use something like sed or even better, bash parameter expansion:
somedir="~/Foo/test~/ing";
cd "${somedir/#\~/$HOME}"
or if you must use sed,
cd $(echo "$somedir" | sed "s#^~#$HOME#")
If you use double quotes the ~ will be kept as that character in $a.
cd $a will not expand the ~ since variable values are not expanded by the shell.
The solution is:
eval "cd $a"
Related
I'm sure this has been asked before but I can't find anything. We have inscrutable login names on a shared machine and want to use shell variables to substitute the hard-to-remember login names for people's real names.
For example, let's say Omar's login name is xyz123. I can do this:
$ omar=xyz123
$ echo ~$omar
and output looks fine:
~xyz123
but if I type this:
$ ls ~$omar
there is an error:
ls: cannot access ~xyz123: No such file or directory
I think it's because tilde expansion happens before variable expansion but can't figure out how to get around this.
Perhaps this answer is related although I'm not sure:
How to manually expand a special variable (ex: ~ tilde) in bash
bash expands the tilde before the variable. See https://www.gnu.org/software/bash/manual/bash.html#Shell-Expansions
The shell will see if the literal characters $ o m a r are a login name. As they are not, the tilde is not expanded. The shell eventually sees $omar as a variable and substitutes that. It then hands the expanded word ~xyz123 to echo which just prints it.
Similarly, it hands the word ~xyz123 to ls. Since ls does not do its own tilde expansion, it is looking for a file in your current directory named ~xyz123 with a literal tilde. Since such a file does not exist you get that error.
If you want ls ~$var to list files, you need eval ls ~$var. Or, since eval is considered very unsafe to use casually, you could do this instead:
ls "$(getent passwd "$omar" | cut -d: -f6)"
I would go with checking if "$omar" is a valid user with id and then using eval to force double expansion. So protect against evil eval and then do it.
if ! id "$omar" >/dev/null 2>&1;
echo "Error: user with the name $omar does not exist!" >&2
exit 1
fi
eval echo "\"~$omar\""
Say I have a folder called Foo located in /home/user/ (my /home/user also being represented by ~).
I want to have a variable
a="~/Foo" and then do
cd $a
I get
-bash: cd: ~/Foo: No such file or directory
However if I just do cd ~/Foo it works fine. Any clue on how to get this to work?
You can do (without quotes during variable assignment):
a=~/Foo
cd "$a"
But in this case the variable $a will not store ~/Foo but the expanded form /home/user/Foo. Or you could use eval:
a="~/Foo"
eval cd "$a"
You can use $HOME instead of the tilde (the tilde is expanded by the shell to the contents of $HOME).
Example:
dir="$HOME/Foo";
cd "$dir";
Although this question is merely asking for a workaround, this is listed as the duplicate of many questions that are asking why this happens, so I think it's worth giving an explanation. According to https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06:
The order of word expansion shall be as follows:
Tilde expansion, parameter expansion, command substitution, and arithmetic expansion shall be performed, beginning to end.
When the shell evaluates the string cd $a, it first performs tilde expansion (which is a no-op, since $a does not contain a tilde), then it expands $a to the string ~/Foo, which is the string that is finally passed as the argument to cd.
A much more robust solution would be to use something like sed or even better, bash parameter expansion:
somedir="~/Foo/test~/ing";
cd "${somedir/#\~/$HOME}"
or if you must use sed,
cd $(echo "$somedir" | sed "s#^~#$HOME#")
If you use double quotes the ~ will be kept as that character in $a.
cd $a will not expand the ~ since variable values are not expanded by the shell.
The solution is:
eval "cd $a"
Say I have a folder called Foo located in /home/user/ (my /home/user also being represented by ~).
I want to have a variable
a="~/Foo" and then do
cd $a
I get
-bash: cd: ~/Foo: No such file or directory
However if I just do cd ~/Foo it works fine. Any clue on how to get this to work?
You can do (without quotes during variable assignment):
a=~/Foo
cd "$a"
But in this case the variable $a will not store ~/Foo but the expanded form /home/user/Foo. Or you could use eval:
a="~/Foo"
eval cd "$a"
You can use $HOME instead of the tilde (the tilde is expanded by the shell to the contents of $HOME).
Example:
dir="$HOME/Foo";
cd "$dir";
Although this question is merely asking for a workaround, this is listed as the duplicate of many questions that are asking why this happens, so I think it's worth giving an explanation. According to https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06:
The order of word expansion shall be as follows:
Tilde expansion, parameter expansion, command substitution, and arithmetic expansion shall be performed, beginning to end.
When the shell evaluates the string cd $a, it first performs tilde expansion (which is a no-op, since $a does not contain a tilde), then it expands $a to the string ~/Foo, which is the string that is finally passed as the argument to cd.
A much more robust solution would be to use something like sed or even better, bash parameter expansion:
somedir="~/Foo/test~/ing";
cd "${somedir/#\~/$HOME}"
or if you must use sed,
cd $(echo "$somedir" | sed "s#^~#$HOME#")
If you use double quotes the ~ will be kept as that character in $a.
cd $a will not expand the ~ since variable values are not expanded by the shell.
The solution is:
eval "cd $a"
I found a weird behavior that I don't know how to workaround.
$ var1=*
$ echo $var1
Audiobooks Downloads Desktop (etc.)
$ ls $var1
Audiobooks:
Downloads:
(etc)
All seems OK. At declaration, the variable gets expanded and everything else works. But see this:
$ var2=~/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm
$ echo $var2
/home/yajo/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm
$ ls $var2
ls: no se puede acceder a /home/yajo/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm: No existe el fichero o el directorio
$ ls /home/yajo/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm
/home/yajo/rpmbuild/RPMS/noarch/enki-12.10.3-1.fc18.noarch.rpm /home/yajo/rpmbuild/SRPMS/enki-12.10.3-1.fc18.src.rpm
/home/yajo/rpmbuild/RPMS/noarch/enki-12.10.3-1.fc19.noarch.rpm /home/yajo/rpmbuild/SRPMS/enki-12.10.3-1.fc19.src.rpm
This time, at declaration only ~ gets expanded, which produces that I cannot pass it as an argument to ls. However, passing the same string literally produces the expected results.
Questions are:
Why sometimes expand and sometimes not?
How to mimic the behavior of $var1 with $var2?
Thanks.
Extra notes:
I tried the same with double and single quotes, but with the same bad results.
The order in which the shell parses various aspects of the command line is not obvious, and it matters for things like this.
First, the wildcards aren't expanded at declaration, they're expanded after the variable value is substituted (note: in these examples I'll pretend I have your filesystem):
$ var1=*
$ echo "$var1" # double-quotes prevent additional parsing of the variable's value
*
$ echo $var1 # without double-quotes, variable value undergoes wildcard expansion and word splitting
Audiobooks:
Downloads:
(etc)
BTW, the ~ is expanded at declaration, confusing things even further:
$ var2=~
$ echo "$var2" # again, double-quotes let me see what's actually in the variable
/home/yajo
The problem with ~/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm1 is that while the shell does wildcard expansion (*) on the value after substitution, it doesn't do brace expansion ({SRPMS,RPMS/*}), so it's actually looking for directory names with braces and commas in the name... and not finding any.
The best way to handle this is generally to store the file list as an array; if you do this right, everything gets expanded at declaration:
$ var2=(~/rpmbuild/{SRPMS,RPMS/*}/enki-*.rpm)
$ echo "${var2[#]}" # This is the proper way to expand an array into a word list
/home/yajo/rpmbuild/RPMS/noarch/enki-12.10.3-1.fc18.noarch.rpm etc...
Note that arrays are a bash extension, and will not work in plain POSIX shells. So be sure to start your script with #!/bin/bash, not #!/bin/sh.
My question is very similar to How to copy multiple files from a different directory using cp?
I don't want to use an explicit loop. Here is what I do:
$ FILES_TOOLS="fastboot,fastboot-HW.sh"
$ cp $HOME/tools/{$FILES_TOOLS} $TOP_DIR/removeme
cp: cannot stat `/home/johndoe/tools/{fastboot,fastboot-HW.sh}': No such file or directory
The files are present and destination is valid, because:
$ cp $HOME/tools/{fastboot,fastboot-HW.sh} $TOP_DIR/removeme
$ echo $?
0
I tried to remove the double quote from FILES_TOOLS, no luck.
I tried to quote and double quote {...}, no luck
I tried to backslash the brackets, no luck
I guess this is a problem of when the shell expansion actually occurs.
This answer is limited to the bash.
Prepend an echo to see what your cp command turns into:
echo cp $HOME/tools/{$FILES_TOOLS} $TOP_DIR/removeme
You have to insert an eval inside a sub-shell to make it work:
cp $( eval echo $HOME/tools/{$FILES_TOOLS} ) $TOP_DIR/removeme
I guess this is a problem of when the shell expansion actually occurs.
Yes. Different shells have different rules about brace expansion in relation to variable expansion. Your way works in ksh, but not in zsh or bash. {1..$n} works in ksh and zsh but not in bash. In bash, variable expansion always happens after brace expansion.
The closest you'll get to this in bash is with eval.
As long as the contents of the braces are literals, you can use brace expansion to populate an array with the full path names of the files to copy, then expand the contents of the array in your cp command.
$ FILES_TOOLS=( $HOME/tools/{fastboot,fastboot-HW.sh} )
$ cp "${FILES_TOOLS[#]}" $TOP_DIR/removeme
Update: I realized you might have a reason for having the base names alone in the variable. Here's another array-based solution that lets you prefix each element of the array with a path, again without an explicit loop:
$ FILES_TOOLS=( fastboot fastboot-HW.sh )
$ cp "${FILES_TOOLS[#]/#/$HOME/tools/}" $TOP_DIR/removeme
In this case, you use the pattern substitution operator to replace the empty string at the beginning of each array element with the directory name.