I have a file and I need to filter lines that have (or don\'t have) N occurrences of a pattern. I.e., if my pattern is the letter o and I what to match lines where the letter o occurs exactly 4 times,
This is an extension of my previous question recommended by the answerer. Basically, I need sed to print text between delimiters. The delimiters could span multiple lines like:
I am trying to repla开发者_如何学Cce a[ \'xxx\' ] by a[ xxx ] using sed: sed -e \'s/a[ \'\\(.*\\)\' ]/a[ \\1 ]/\' ./Test
Suppose I have op(abc)asdfasdf and I need sed to print abc between the brackets. What would work for me? (Note: I only want the text between first pair of delimiters on a line, and nothing if a partic
I\'ve tried this on multiple small files, everything works fine. But when testing sed on multiple files with size 37kB each, only one file completely processed while other files become horrible.
I want开发者_StackOverflow社区 to search a configuration file for this expression: \"central.database\".
the data i have looks something like this a,b,c,d a,b1 b2,c,d A,B,C,D What is happening is that in fie开发者_运维问答ld 2 there is occasionally a new line character in the second field so the line
I am very new to sed and so even with looking at examples I am totally at a loss as how to go about writing the correct code for my need (this one is close but it seems not for multi-line replacement.
Can anyone explain this sed one-liner in English (the more detail, the better)? @sed \'s/\\($*\\)\\.o[ :]*/\\1.o $@ : /g\' < $*.d > $@; \\
According to开发者_高级运维 http://linux.about.com/od/commands/l/blcmdl1_sed.htm suppress automatic printing of pattern
加载中,请稍侯......