Shell Programming and Scripting

This will take the value in the variable ( ${VAR} ) and find the longest pattern that matches the pattern after ## and remove it.

This value in ${VAR} was /abc1/abc2/abc3 and the pattern given to match was "*/". This matched "/abc1/abc2/" as the longest matching pattern from the start of the variable value and removed it leaving only "abc3".

You can get the shortest match from the start by using only 1 # as in ( ${VAR#*/} )

Code:

[user#host ~]$ VAR=/abc1/abc2/abc3
[user#host ~]$ echo ${VAR##*/}
abc3

[user#host ~]$ echo ${VAR#*/}
abc1/abc2/abc3

The shortest match was "/" so the result was that the beginning of the variable now starts with "abc1" and not "/abc1".

You can match from the end of the variable by using '%' and '%%' in the same manner.

Code:

[user#host ~]$ VAR=/abc1/abc2/abc3
[user#host ~]$ echo ${VAR%/*}
/abc1/abc2

[user#host ~]$ echo ${VAR%%/abc2*}
/abc1

Taken from man page for bash:

${parameter##word}
The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches
the beginning of the value of parameter, then the result of the expansion is the expanded value of
parameter with the shortest matching pattern (the ‘‘#'' case) or the longest matching pattern (the
‘‘##'' case) deleted. If parameter is @ or *, the pattern removal operation is applied to each
positional parameter in turn, and the expansion is the resultant list. If parameter is an array
variable subscripted with @ or *, the pattern removal operation is applied to each member of the
array in turn, and the expansion is the resultant list.

I'm trying to create a script that places you in any directory that you input the name.(No need for the whole path)

I've finished doing it but I'm now experiencing a problem because I'm trying to echo: "Directory doesn't exist" using an if condition but it's not working.

When I write tftftft as a folder name the find command gives an explicit message that the directory doesn't exist but my if statement(That contains an echo) isn't popping out.

Here's the test part:

Code:

...
if [ ! -e $variableWithPathInside ]
then 
echo "Directory isn't existent"
else
cd $variableWithPathInside
...
fi

So why isn't my test working?

Actually I'm an idiot.

But how could I express that if the find doesn't find anything it shows up a message?

read -p "Name of directory" $rep
if [ -d find -name $rep ] .....?

i'm going to guess that since the directory doesn't exist, $variableWithPathInside is empty. so you're test to see if <file> exists is testing nothing and returning a success. bash (and other shell scripting languages) is different in that you don't need to declare variables. calling to a previously undeclared/undefined variable just returns "", just like it's doing in your script. it's going to the else block and executing a cd command with no parameters, which just changes directories to the users home: defined in env as $HOME.

instead, try something like this:

Code:

# $variableWithPathInside is already defined by user
 
dirLoc=`find <startLocation> -type d -name $variableWithPathInside`
 
if [ -z $dirLoc ] #tests to see if variable has zero length.  can use -n for nonzero length
then
    echo "Directory isn't existent"
else
    cd $dirLoc
fi

give that a go

If I understand you correctly:

Code:

DIR=""
while [[ -z "${DIR}" || ! -d "${DIR}" ]]
do
        read -p "Enter dir: " DIR
done

The double-brackets are necessary for the || construct inside. -d is preferred to -e when checking for directories, since -d will only succeed for directories, while -e will accept any kind of file at all.

Try this instead:

Code:

read -p "Enter dir: " DIR
if [ ! -d "$DIR" ]
then 
     echo "Directory $DIR dosn't exist"
else
     cd $DIR
     ...
fi

that's not going to work. he said he doesn't want the user to have to enter the entire path. he just wants to enter a directory and the use the find command to find it...when using the find command with the -type d argument, it's already testing for it to be a directory...no need to test for it again.

sorry missed that