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Shell Programming and Scripting

Print out specific pattern column data

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# 1  
Old 12-29-2009
Print out specific pattern column data
Input file:
Code:
adc_0123
haa_1000
bcc_520
adc_0150
bcc_290
adc_0112
haa_8000
adc_0139
haa_7000

Output file:
Code:
adc_0123 adc_0123
              haa_1000
              bcc_520
adc_0150 adc_0150
              bcc_290    
adc_0112 adc_0112
              haa_8000
adc_0139 adc_0139
              haa_7000

I would like to print out again those data that beginning with "adc_" at column 1. It should be able to solve by awk command.
Thanks for any suggestion.
# 2  
Old 12-29-2009
Yes, awk would be a good choice. What did you try?
# 3  
Old 12-29-2009
Try:
Code:
awk '{ str=$0; if(/^adc_/) print str str; else print "\t"str; }' file

# 4  
Old 12-29-2009
Hi Aia,
This is what I have try before:
Code:
awk '$0~/adc_/{print "";t=$1}{for (i=1;1<NF;i++){if ($i~/adc_/){print t,$_} else if ($i ~! /adc_/){print $_}}}' input_file

But it can't get my desired output result.
The dennis's awk code work nice for my problem.
But it got some syntax error

---------- Post updated at 01:05 AM ---------- Previous update was at 01:04 AM ----------

Thanks dennis,
Your awk code work perfectly for my problem.
Thanks again Smilie
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