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#1
02-04-2011
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Sed/Awk: Finding first instance of a string and printing the following characters
Hi all, I'm new to scripting, so sorry if this is a simple question! I want to find the first occurrence of the string "00" in a file, and then print the first 4 characters that follow that string. So for example, if the file begins: Code:
ELT190 +0019879753004367 BDGF... I want it to print "1987" only I can use sed/awk and I think grep as well. Thanks a lot
Last edited by zaxxon; 02-04-2011 at 11:30 AM.. Reason: code tags |
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#2
02-04-2011
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Code:
$ echo "ELT190 +0019879753004367 BDGF..."| sed 's/.*+00\(....\).*/\1/g' 1987 |
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#3
02-04-2011
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Thanks for that,
It's almost right, but not quite. That line prints the first 4 characters after the LAST occurrence of +00, I want the first 4 characters after the FIRST occurrence of "00" Thanks |
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#5
02-04-2011
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Code:
$ echo "ELT190 +0019879753004367 00notthissss BDGF..."| awk -F "00" '{print substr($2,0,4)}'
1987 |
| The Following User Says Thank You to zaxxon For This Useful Post: | ||
jhunter87 (02-04-2011) | ||
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#6
02-04-2011
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That's working now. Thanks a lot!
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| Tags |
| awk, first occurrence, print, sed |
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