Shell – How to output file & ignoring lines that start with “?”

shell-scriptsubversiontext processing

I do svn status --show-updates and then I want to either

Q1:
ignore (not to display) lines that start with ?
Q2:
display only lines that start with * Note that there are few spaces before * occurs. That means that * is not the first character on the line.

How can I do that?

Best Answer

You can express those conditions using regular expressions and use grep to filter the results based on those.

The first one is ^?. The carat is a special character that represents the beginning of a line; so that expression matches the beginning of the line immediately followed by a ?.

The second one is ^ *\*. The * is a special character that qualifies the preceding character - it means the preceding character may appear zero or more times. Since * is a special character, the one you're looking for needs to be escaped, hence, \*. So that expression will match the beginning of a line followed by zero or more spaces, followed by an asterisk.

For your first condition, use the -v option for grep to negate the results.

So finally,

svn status --show-updates | grep -v '^?'

svn status --show-updates | grep '^ *\*'

Regular expressions are very powerful, so many Unix tools can use them. They are very much worth learning. There is a great tutorial at regular-expressions.info.

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Text Processing – Command to Display First and Last Lines of a File

You can use sed or awk to make it with one command. However you'll loose at speed, cause sed and awk will need to run through the whole file anyway. From a speed point of view it's much better to make a function or every time to combination of tail + head. This does have the downside of not working if the input is a pipe, however you can use proccess substitution, in case your shell supports it (look at example below).

first_last () {
    head -n 10 -- "$1"
    tail -n 10 -- "$1"
}

and just launch it as

first_last "/path/to/file_to_process"

to proceed with process substitution (bash, zsh, ksh like shells only):

first_last <( command )

ps. you can even add a grep to check if your "global conditions" exist.

How to print all the lines not containing a string along with adjacent lines

I'm not sure how structured your data is but if it is just as you show, how about the following:

grep -B2 'Auto-Installed: [^1]'

That assumes that every stanza includes an Auto-Installed line, which might not be correct.

Here's an awk program which I think does exactly as you asked.

awk 'BEGIN{deleted=3}
     !deleted{printf "%s",l[NR%3]}
     deleted {--deleted}
     {l[NR%3]=/./?$0"\n":$0}
     /Auto-Installed: 1/{deleted=3}
     END{for(i=NR+deleted;i<NR+3;++i)printf "%s",l[i%3]}'

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