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one simple shell script problem

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# 1  
Old 04-18-2009
CPU & Memory one simple shell script problem
Hi everyone, I am facing to one shell script problem, which is as following


Write a shell script that:
  • Takes a number of arguments.
  • For each argument, print out all files in the current directory that contain this substring in their name.
I know I need to use grep for the second part, but I am confused on how to take arguments and print out all files?

anyone can help? Thanks very much...~
# 2  
Old 04-18-2009
can you provide sample input and output of your requirement and what have you done so far ?
# 3  
Old 04-18-2009
sorry there is no sample input and output..and since I was stucked on the first part, so I haven't done anything yet...Smilie
# 4  
Old 04-19-2009
Code:
echo *first* *second* *third* ...

If you really hate to have to type the *, and want to have a vertical list, and want to avoid yo display "*first*" when no matching file exist, then:

Code:
colemar@deb:~$ cat shaloovia
#!/bin/ksh
for arg in $*; do
  for file in *$arg*; do
    [[ -f $file ]] && print -- $file
  done
done

# 5  
Old 04-20-2009
hope below help you some

Code:
(
   read aa
   set $aa
   while [ $# -gt 0 ];do
    echo "-----$1-----"
    ls | grep $1
    echo "-----$1-----"
   	shift
   done
)

# 6  
Old 04-20-2009
thx
million thanks to all.. they are very helpful..~Smilie
# 7  
Old 04-21-2009
Quote:
Originally Posted by colemar
Code:
echo *first* *second* *third* ...

If you really hate to have to type the *, and want to have a vertical list, and want to avoid yo display "*first*" when no matching file exist, then:

Code:
colemar@deb:~$ cat shaloovia
#!/bin/ksh
for arg in $*; do
  for file in *$arg*; do
    [[ -f $file ]] && print -- $file
  done
done

I am supposed to write bash shell script..so i think i should write
#!/bin/sh instead of #!/bin/ksh

is that rite?Smilie
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