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Need help using % or like option in shell script

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# 1  
Old 06-14-2010
Need help using % or like option in shell script
Hi All,
I need help from you guys.
In the shell script I have, I have a variable that has a value as follows
Code:
a=abc_20100614_12346

I want to do the following logic,
Code:
if a like '%20100614%' then 
true
else
false 
fi

Guys please help me out on this.
I tried the follwoing option
Code:
if [[ $a = 20100614* ]]; then
true
else false 
fi.

This didnt work for me.It's throwing an error message as follows
Code:
[[: not found

More info on [[ from unix system
Code:
efmsdapp:/export/home/hut$ type [[
[[ is a reserved shell keyword
efmsdapp:/export/home/hut$ help [[
ERROR: Key '[[' not found (he1)
efmsdapp:/export/home/hut$

Moderator's Comments:
Mod Comment Use code tags please, ty.
Last edited by zaxxon; 06-14-2010 at 12:07 PM.. Reason: code tags
# 2  
Old 06-14-2010
Code:
if [ -n $( echo $a | grep '20100614' ) ]

Code:
if [[ "$a" =~ "20100614" ]]

Last edited by dr.house; 06-14-2010 at 10:49 AM.. Reason: 2nd version added
This User Gave Thanks to dr.house For This Post:
srini02
# 3  
Old 06-14-2010
A couple of inexpensive, posix-compliant approaches:
Code:
#!/bin/sh

a=abc_20100614_12346
date=20100614



# One way

case $a in
    *$date*) echo MATCH;;
    *) echo NOT A MATCH;;
esac



# Another way

if [ "$a" != "${a#*$date}" ]; then
    echo MATCH
else
    echo NOT A MATCH
fi

Regards,
Alister
This User Gave Thanks to alister For This Post:
srini02
# 4  
Old 06-14-2010
Quote:
Originally Posted by dr.house
Code:
if [ -n $( echo $a | grep '20100614' ) ]

Code:
if [[ "$a" =~ "20100614" ]]

Hi,Thanks for the speed reply.
When i tried " if [ -n $( echo $a | grep '20100614' ) ] " this one
its saying syntax error.
here is the code I used,
Code:
a=21345zef
b=21345
case $a in
    *$date*) d=MATCH;;
    *)  d=NOT A MATCH;;
esac
echo $d
if [ -n $( echo $a | grep '20100614' ) ] then 
echo 'eureka'
else
echo 'try again'
fi



---------- Post updated at 10:50 AM ---------- Previous update was at 10:47 AM ----------
Quote:
Originally Posted by alister
A couple of inexpensive, posix-compliant approaches:
Code:
#!/bin/sh
 
a=abc_20100614_12346
date=20100614
 
 
 
# One way
 
case $a in
    *$date*) echo MATCH;;
    *) echo NOT A MATCH;;
esac
 
 
 
# Another way
 
if [ "$a" != "${a#*$date}" ]; then
    echo MATCH
else
    echo NOT A MATCH
fi

Regards,
Alister
Hi Alister,
Thanks for the speed reply.
The first code worked like a charm for me.
But I tried the second one also but its saying bad substituion.
Here is the code I used
Code:
a=21345zef
b=21345
if [ "$a" != "${a#*$date}" ]; then
    echo MATCH
else
    echo NOT A MATCH
fi

Moderator's Comments:
Mod Comment Use code tags please, ty.
Last edited by zaxxon; 06-14-2010 at 12:09 PM.. Reason: code tags
# 5  
Old 06-14-2010
Your shell may not support the "${var#pattern}" form of parameter substitution.

Regards,
Alister
This User Gave Thanks to alister For This Post:
srini02
# 6  
Old 06-14-2010
Quote:
Originally Posted by srini02
Code:
if [ -n $( echo $a | grep '20100614' ) ] then


[/COLOR]
Code:
if [ -n $( echo $a | grep '20100614' ) ]
then


Code:
if [ -n $( echo $a | grep '20100614' ) ] ; then





This User Gave Thanks to dr.house For This Post:
srini02
# 7  
Old 06-14-2010
Quote:
Originally Posted by dr.house
Code:
if [ -n $( echo $a | grep '20100614' ) ]; then

A couple of observations regarding this code:

1. The -n operator takes only one operand. If $a contains any IFS whitespace, the unquoted command substitution will result in more than one word and a syntax error. Since we don't know for certain anything about the range of possible values for $a, this unquoted version might be buggy. It also wouldn't hurt to double quote "$a" within the command substitution.

2. You could simplify your approach to use grep's exit status directly:
Code:
if echo "$a" | grep -q '20100614'; then

Regards,
Alister
This User Gave Thanks to alister For This Post:
srini02
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