Bash script to find the path of an executable - linux

Question

EDIT: I KNOW THIS IS REDUNDANT, IT IS HOMEWORK, I HAVE WRITTEN MY OWN CODE AND NEED HELP TROUBLESHOOTING>
As stated, I must write a BASH script to determine whether an executable file is in the users path.
such that if you type
./findcmd ping it returns /bin/ping
I have some code written, But it does not properly work and I hope someone can help me troubleshoot. When I type ./findcmd ping it just returns my file does not exist.(with any other file I try as well that I know exists.)
#!/bin/bash
#
# Invoke as ./findcmd command
#
# Check for argument
if [[ $# -ne 1 ]]
then
echo 'useage: ./findcmd command'
exit 1
fi
#
# Check for one argument
if [[ $# -eq 1 ]]
then
pathlist=`echo $PATH | tr ':' ' '`
for d in $pathlist;
do
if [[ ! -d $d || ! -x $d || ! -r $d ]]
then
echo 'You do not have read end execute
permissions!'
exit 2
fi
if [[ $(find $d -name $1 -print | wc -l) -ne 0 ]]
then
echo 'The file does not exist in the PATH!'
exit 0
fi
done
fi
exit 0
#
#

Answer 1

No need to use a bash array, tr'ing the ':' with ' ' will work just fine in a for loop.
#!/bin/bash
#
# Invoke as ./findcmd command
#
# Check for argument
if [[ $# -ne 1 ]]
then
echo 'usage: ./findcmd command'
exit 1
fi
f=$1
# No need to check the $# again, there's at least one arg and other will be ignored..
# Otherwise you can wrap this in a loop and keep shift'ing args and checking one by one
pathlist=`echo $PATH | tr ':' '\n'`
for d in $pathlist;
do
#echo command is your friend
#echo "Checking for $f in $d"
path="$d/$f"
if [[ -f "$path" && -x "$path" ]]; then
# PATH is not recursive, therefore no need to use find command
# Simply checking that the file exists and is executable should be enough
echo "Found $f at '$path'"
# Note the same filename may be present farther down the PATH
# Once the first executable is found, exit
exit 0
fi
done
# Getting here means file was not found
echo "$f could not be found"
exit 1
Here are the results:
rbanikaz#lightsaber:~$ ./which.sh grep
Found grep at '/usr/bin/grep'
rbanikaz#lightsaber:~$ ./which.sh foo
foo could not be found
rbanikaz#lightsaber:~$

Answer 2

The which command already does this...
Techinically this is a solution...
#!/bin/bash
which $1
I probably wouldn't submit it for as assignment though...
Update
Messing around a bit and I think the following will code will get your past your current bug:
#!/bin/bash
#
# Invoke as ./findcmd command
#
# Check for argument
if [[ $# -ne 1 ]]
then
echo 'useage: ./findcmd command'
exit 1
fi
#
# Check for one argument
if [[ $# -eq 1 ]]
then
d=$1
pathlist=($(echo $PATH | tr ':' ' '))
echo $pathlist
i=0
while read line; do
a4[i++]=$line
done < <(echo "$PATH" | tr ':' '\n')
n=${#a4[#]}
for ((i=0; i < n; i++)); do
if [[ ! -d $d || ! -x $d || ! -r $d ]]
then
echo 'You do not have read end execute
permissions!'
exit 2
fi
if [[ $(find $d -name $1 -print | wc -l) -ne 0 ]]
then
echo 'The file does not exist in the PATH!'
exit 0
fi
done
fi
exit 0
#
#
Pretty much, it uses a solution in this SO question to split the $PATH variable into an array and then loops through it, applying the logic you had inside your while statement.