I have a problem putting the content of pwd command into a shell variable that I'll use later.
Here is my shell code (the loop doesn't stop):
#!/bin/bash
pwd= `pwd`
until [ $pwd = "/" ]
do
echo $pwd
ls && cd .. && ls
$pwd= `pwd`
done
Could you spot my mistake, please?
Try:
pwd=`pwd`
or
pwd=$(pwd)
Notice no spaces after the equals sign.
Also as Mr. Weiss points out; you don't assign to $pwd, you assign to pwd.
$() mean in $(pwd)?
$() in bash just replaces itself with the output of the command executed with the braces; so $(pwd) would expand to /home/ubuntu for example, making the assignment the equivalent of: pwd='/home/ubuntu'
cd $pwd/newdir does this work if the newdir exists?
In shell you assign to a variable without the dollar-sign:
TEST=`pwd`
echo $TEST
that's better (and can be nested) but is not as portable as the backtics:
TEST=$(pwd)
echo $TEST
Always remember: the dollar-sign is only used when reading a variable.
In this specific case, note that bash has a variable called PWD that contains the current directory: $PWD is equivalent to `pwd`. (So do other shells, this is a standard feature.) So you can write your script like this:
#!/bin/bash
until [ "$PWD" = "/" ]; do
echo "$PWD"
ls && cd .. && ls
done
Note the use of double quotes around the variable references. They are necessary if the variable (here, the current directory) contains whitespace or wildcards (\[?*), because the shell splits the result of variable expansions into words and performs globbing on these words. Always double-quote variable expansions "$foo" and command substitutions "$(foo)" (unless you specifically know you have not to).
In the general case, as other answers have mentioned already:
var=value, not var = value$ means “take the value of this variable”, so you don't use it when assigning: var=value, not $var=value
$PWD is an internal part of bash. This gets confusing when you have a PWD=`pwd` in your script and then you pushd to another directory and expect $PWD is holding the dir before pushd and it doesn't!You can also do way more complex commands, just to round out the examples above. So, say I want to get the number of processes running on the system and store it in the ${NUM_PROCS} variable.
All you have to so is generate the command pipeline and stuff it's output (the process count) into the variable.
It looks something like this:
NUM_PROCS=$(ps -e | sed 1d | wc -l)
I hope that helps add some handy information to this discussion.
JOBMAX changes the number of background tasks that can be running at once; it's not hard to imagine a shell using NUM_PROCS for something similar.Here's your script...
DIR=$(pwd)
echo $DIR
while [ "$DIR" != "/" ]; do
cd ..
DIR=$(pwd)
echo $DIR
done
Note the spaces, use of quotes, and $ signs.
echo "$DIR", with the quotes; see BashPitfalls #14.
i=0oustide of the loop. then inside,i=$i + 1. And then also inside the loop, add something likeif [ $i > 25 ] then; break; endif;I'm not sure of the loop breaking in syntax in shell scripts, but it should be something like that.[ $i > 25 ], is exactly the same as[ $i ] >25; that is to say,25is treated as a filename. You need to use-gtfor numeric comparisons, or quote'>'for string comparisons. And adding one toiisi=$(( i + 1)); the syntax you have now tries to run the command+with the argument1andiset in its environment. And the keyword for ending anifin shell isfi.