mistake in for loop

Question

When I run my script I get this error:

234.sh: line 3: syntax error near unexpected token `do
234.sh: line 3: `for folder in $array ; do

I don't see the mistake. Help?

#!/bin/bash
base=$(pwd)
array=`find * -type d`
 for folder in $array ; do
  cd $folder ;
  grep -n $1 * | while read line ;
   do    name=$(echo "$line" | cut -f1 -d:) ;
        if [ "$name" == "1234.sh" ]; then
        continue ;
        else
        string=$(echo "$line" | cut -f2 -d:) ;
        a=$(expr $string - 10)
        if [ $a -lt 1 ] ; then 
        a=1 ;
        fi ;
        b=$(expr $string + 10) ;   
        echo "-----------------------"
        echo $name:$a
        sed -n $a,${b}p $name;
        fi ;
    done
   cd $base ;
done

That shouldn't be necessary; it's legal to terminate the statement with a semi-colon. — chepner
– chepner, Commented Jul 31, 2012 at 13:09
Your variable named "array" is not an array, it is a string. — jordanm
– jordanm, Commented Jul 31, 2012 at 13:12

Henk Langeveld · Accepted Answer · 2012-08-01 11:33:30Z

3

A few suggestions:

Make array a proper array, not just a string. (This is the only suggestion that actually addresses your syntax error.)
Quote parameters
Use IFS to allow read to split your line into its two components
Use a subshell to eliminate the need to cd $base.
Most of your semicolons are unnecessary.

#!/bin/bash
array=( `find * -type d` )
for folder in "${array[@]}" ; do
  ( cd $folder
    grep -n "$1" * | while IFS=: read fname count match; do
      [ "$fname" == "1234.sh" ] && continue

      a=$(expr $count - 10); [ $a -lt 1 ] && a=1
      b=$(expr $count + 10) 
      echo "-----------------------"
      echo $fname:$a
      sed -n $a,${b}p $fname
    done
  )
done

edited Aug 1, 2012 at 11:33

Henk Langeveld

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answered Jul 31, 2012 at 13:36

chepner

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Comments

Igor Chubin · Accepted Answer · 2012-07-31 13:04:30Z

1

#!/bin/bash

base=$(pwd)
array=`find . -type d`
for folder in $array
do
  cd $folder
  grep -n $1 * | while read line
  do    
      name=$(echo "$line" | cut -f1 -d:)
      if [ "$name" == "1234.sh" ]
      then
        continue
      else
        string=$(echo "$line" | cut -f2 -d:)
        a=$(expr $string - 10)
        if [ $a -lt 1 ]
        then 
          a=1
        fi
        b=$(expr $string + 10)
        echo -----------------------
        echo $name:$a
        sed -n $a,${b}p $name
      fi
  done
  cd $base
done

answered Jul 31, 2012 at 13:04

Igor Chubin

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Comments

Henk Langeveld · Accepted Answer · 2012-08-01 11:24:11Z

1

What you're trying to accomplish looks like a context grep for a specified pattern on all files in a directory tree.

I suggest you use Gnu grep Context Line Control

#!/bin/bash
base=$(pwd)
spread=10
pattern=$1

find . -type d | while read dir; do
    (cd $dir && egrep -A $spread -B $spread $pattern *)
done

This is the simple version, without dealing with 1234.sh or empty directories

answered Aug 1, 2012 at 11:24

Henk Langeveld

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Comments

Henk Langeveld · Accepted Answer · 2012-08-01 11:29:32Z

1

This solution is even less complex, and in addition deals with the exempt file name. It depends on the xargs and Gnu grep Context Line Control as well.

#!/bin/bash
spread=10
pattern=$1

find . -type f ! -name "1234.sh" |
    xargs egrep -A $spread -B $spread $pattern

answered Aug 1, 2012 at 11:29

Henk Langeveld

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1 Comment

Henk Langeveld Over a year ago

If presentation is very important, more work is needed for this solution.

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