bash脚本awk，带有传递变量和if语句

Question

I am writing a bash script using awk with variable passing in AIX powerpc machine. The code written works fine after I read some questions & answers in this site. :)

Now, I'd like to add if statement in my bash script and I got awk syntax errors. My requirement (see my script below): - if awk pattern matching is true, print $0 - else print "transaction: p value not found."

Content of trans.txt

bash-4.2$ cat trans.txt

10291413
8537353
8619033
8619065
8625705

Could someone help me please. Thank you.

content of getDetail.sh

#!/usr/bin/bash

sysDir="/var/syslog-ng/TEST/STS"
syslogFile="DPSTSLog2014-10-22T09.log"
pattern="Latency"

filename=$1
while read f; do
  concatPattern="$f.*$pattern"
  awk -v p="$concatPattern" '$0 ~ p {print $0}' $sysDir/$syslogFile
done < $filename

run command execution below

./getDetails.sh trans.txt

result

2014-10-22T09:15:53+11:00,10.16.198.50,info,latency,[info] xmlfirewall(ws-TrustValidate): trans(10291413): Latency:   0   0   0   0  14  14   0  14   0   0   0  14   0   0   0   0
2014-10-22T09:15:38+11:00,10.16.198.50,info,latency,[info] xmlfirewall(ws-TrustValidate): trans(8619033): Latency:   0   0   0   0  73  73   0  73   0   0   0  73   0   0   0   0
2014-10-22T09:18:04+11:00,10.16.198.50,info,latency,[info] xmlfirewall(ws-TrustValidate): trans(8625705): Latency:   0   0   0   0  13  13   0  13   0   0   0  13   0   0   0   0

Answer 1

awk -v p="$concatPattern" '
    $0 ~ p {print $0; found=1}
    END   {if (! found) print "transaction: p value not found"}
' $sysDir/$syslogFile