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Or:
ab=$(echo $dirk | awk '/ab/' RS=" ")

Not worried about the case of running this script before 4am?

awk -F'[()]' '/CURDEPTH/ && $2>10{exit 1}' test1.txt || mailx -s ...

Phew!



@anurag.singh
When using "grep" to search for a string containing space characters, remember to put double quotes round the string.

Small variation:
awk '{sub(/ *$/,x)} NR==FNR{A[$1]=$0;m++;next} NR-FNR==m{B[$1]=$0;next} {print A[$1],B[$2]}' FS=\| OFS=\~ fromfile tofile FS=\~ keyfile

You can leave out \2:
sed 's/^\(ASX1[89]......\)07/\100/;s/^\(ASX1[89]......\)12/\100/' infile


----
GNU sed has alternation as well as back reference:
sed -r...

I presumed 5:221 to 5:299 should not be printed...
awk 'v!=$1{v=$1; a=$2; b=$3} $2>a{a=$2} $3>b{b=$3} {for(i=a;i<=b;i++)print v":"i; a=i}' infile

Hi, this particular expansion cuts the longest match of space followed by any character off of the end of the string. It is described here with examples: 2.6.2 Parameter Expansion...

Couple of issues here.
1) Environment Variable names are case-sensitive. $DIR and $dir are not the same variable.
2) Your messages appear to be the wrong way round. When we find no files (and the...

Try:
awk '!/date_conditions/{print $2}' RS= infile

Maybe something like this?
grep -c '^[^$]*DOMAINSOLVER[ \t]*ACMS' file

@anurag:That should be sort -rnk2, no? Otherwise number > 9 would not get represented properly..
Also there is no need for temp files and an extra sort step if you sort like this:
sort -k1,1...

There's no need to use external commands:
#!/bin/ksh

while read line
do
id=${line%,*}
name=${line#*,}
echo $id
echo $name
done < temp.txt