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Check if parameter passes in contains certain string

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# 8  
Old 02-15-2005
Quote:
Originally Posted by kash
if [ `echo $1 |grep "WA"` ]; then
echo "Found"
else
echo "Not Found"
fi
Whether that will work depends on the text being tested. Try it with something like:
./kash "/x/y/z/WA = x"
# 9  
Old 08-15-2008
Quote:
Originally Posted by jim mcnamara
The shell also supports different kinds of string operators. The most nearly universal one is expr - in this case expr index

Try man expr.
Code:
let is_there=0
check ()
{
  if [ $is_there -gt 0 ] ; then
     echo "$str has a WA in it"
  else
     echo "$str has no WA in it"
  fi         
}

str="/someplace/WA/01/"
is_there=`expr index $str "WA"`
check;

Test whether expr return zero or a non-zero result.
For the record, this doesn't work, since expr index finds the first occurrence of ANY CHARACTER from the second arg.

Code:
$ expr index quickbrownfox zzzzzzkzzzzzzzzzz
5

case ... esac solved my problem.
# 10  
Old 08-15-2008
While case...esac is fine, if you have your heart set on using 'expr' you can use expr this way:

expr $parameter : ".*$search" >/dev/null && echo Found!!!

In this form, expr returns the # of characters matched from the beginning of $parameter. That's why I had to put a ".*" in the beginning of the search string. Also, expr writes the number of characters matched to stdout (so I directed to /dev/null). Of course, you can just use this value in a test like this:

if [ expr $parameter : ".*$search" -gt 0 ]
then
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