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Top Forums Shell Programming and Scripting Loop problem with one more problem Post 302239009 by cfajohnson on Monday 22nd of September 2008 05:05:42 PM
Old 09-22-2008
Quote:
Originally Posted by aliahsan81
Hi ALL
I am greping some thing from a file,some thing is "/var/www/html" /var/www"/example" and so on its showing like this when u use echo.

Please use English in this forum. Write "you", not "u".
Quote:
i want to pass each one of them in a loop for processing.Like start with /var/www/html then 2nd and so one
But that not working any iead how i give one by one value.

What does "not working" mean? What does happen? What do you want to happen?
Quote:
My code

Please put code inside [CODE] tags.
Quote:
Code:
docroot=$(grep -H DocumentRoot /etc/httpd/conf.d/*.conf |
 awk -F' ' '{ print $3 }'| sort -u | uniq)


You don't need uniq; you have already removed duplicates with sort -u.

In fact, you don't need sort, either. It can be done with awk alone:

Code:
docroot=$(grep -H DocumentRoot /etc/httpd/conf.d/*.conf |
 awk -F' ' '!x[$3]++ { ++n } END { print n }'

Quote:
Code:
doccount=$(grep -H DocumentRoot /etc/httpd/conf.d/*.conf |
 awk -F' ' '{ print $3 }'| sort -u | uniq | wc -l)

for (( i=0; i <= $doccount; ++i ))


It is safer to stick to standard syntax:

Code:
i=0
while [ $i -le $doccount ]
do
   : whatever
   i=$(( $i + 1 ))
done

Quote:
Code:
 do
        confd="$docroot\n"

Code:
        confd="$docroot"

Quote:
Code:
echo $confd
        find $confd -type d -perm /o=w | while read DIR
do
do some thing 

done ## while done
done ##for done

 

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