What\'s the simplest way to print all matches (either one line per match or one line per line of input) to a regular expression on a unix command line? Note that there may be 0 or more than 1 match pe
The following is the output of \"grep\" command. grep -R \'table=\"transaction\"\' /home/shantanu/* /hom开发者_如何学Pythone/shantanu/conf/transaction/Transaction.hbm.xml:<class name=\"com.common.
I am trying to grep a file for the first 2 matches of a string (there will only ever be a maximum 2 matches) including some context (grep -B 1 -A 5), split each set of 7 lines into two separate variab
I am using the following to search a directory recursively for specific string and replace it with another:
I want to write a bash command that greps all *.txt file with a pattern in current folder to another folder. Should I use find or for loop? I tried using find but it seems to complicate things.
I want to count the number of matches there is on one single line (or all lines as there always will be only one line).
awk \'$0 ~ str{print b}{b=$0}\' str=\"findme\" path_to_file with th开发者_高级运维is I can get the line before the found string\'s line.
If I run grep -C 1 match over the following file: a b match1 c d e match2 f match3 g I get the following o开发者_JS百科utput:
I’m trying to count a particular word occurrence in a whole directory. Is this possible? Say for example there is a directory with 100 files all of whose files may have the word “aaa” in them. How
I had two wordpress installs where a hacker tried to inject some php code into the existing php files
加载中,请稍侯......