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10-05-2006
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Search for a non zero character in file
Hi,
I have a log file which has entries as Staged 0 records from fn.dat (0 failed) 01/01 01:01:01 I 0 Error Transactions I want to find out any line that has an entry like "(1 failed)" or "(2 failed)" or any number in general ( >0 ) similarly it should search for string like "1 Error" or "2 Error" ie anything greater than 0 Thanx in advance |
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10-05-2006
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Alternative in Python, assuming (...) is always at the end Code:
for lines in open("input.dat"):
lines = lines.strip()
if lines.find("(") != -1:
print lines
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10-09-2006
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Code:
[failed|Error] This is not going to do what you expect. It will match *any single* character in that regex character class. For example: Code:
echo "Staged 10 records from fn.dat (2 accepted)" | sed -n '/[1-9][0-9]* [failed|Error]/p' will match because we have a numeric followed by space and the letter "a" which is in the character class. Not only that, it will also match 10 records To understand the logic of the regex do this: Code:
echo "Staged 10 records from fn.dat (2 accepted)" | grep --color '[1-9][0-9]* [failed|Error]' If your grep's GREP_COLOR environment variable is configured you should get this: ---> $ Staged 10 records from fn.dat (2 accepted) To cut a long story short, the correct solution could be any of the following: Code:
awk '$0 ~ /[1-9][0-9]* (failed|Error)/ {print}' file
sed -n '/[1-9][0-9]* \(failed\|Error\)/p' file # yucky !
sed -nr '/[1-9][0-9]* (failed|Error)/p' file # more readable with option -r
grep '[1-9][0-9]* \(failed\|Error\)' file # yucky again!
grep -E '[1-9][0-9]* (failed|Error)' file
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