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The function of substr clause in awk command

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# 1  
Old 10-31-2009
Lightbulb The function of substr clause in awk command
Hello all,

Please help me in letting me know the function of *substr* function in awk...

actually i am new with this function i can play with awk but for this function needs your help in making me understand the correct way of using it.

I am writting a code please advice regarding this prespective.

Code:
 
#!/bin/sh
YESTERDAY=`head -1 /tmp/prv_date` \\ prv_date contains (e.g. 2009-10-30)
cd /tmp
cat ~scu/temp/man*.old ~scu/temp/man*.run | awk -F":" -v today=$YESTERDAY 'BEGIN{c=0;dc=0;} 
FNR{dc=c;
if(c==0 && substr($0,12,16)=="Date: "today) {c++;}; 
if(dc==1 && substr($0,12,6)=="Date: " && substr($0,18,10)!=today) {c=0;dc=0;};
if(c==1 && (substr($0,26,13)=="Platform CAPS" || substr($0,26,13)=="Over max CAPS")) {print $0}; }' > output$YESTERDAY.txt

Quote:
please explain line by line if possible, especially (FNR and substr) function
# 2  
Old 10-31-2009
You can find the answer and many more here:

The AWK Manual - Table of Contents
The GNU Awk User's Guide

Regards
# 3  
Old 10-31-2009
Thanks Franklin52,

But for a descripton regarding my query is still unsloved as i cannot find any details regarding substr function.

Dear All,

Please help
# 4  
Old 10-31-2009
Here you are:

The AWK Manual - String Functions
String Functions - The GNU Awk User's Guide
# 5  
Old 10-31-2009
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