Ubuntu – Grep a line which start and end with a pre defined character

bashcommand linegrep

I am trying to fetch a line from a file file.txt which looks like this:

>This is line 1.</li>
>This is line 2.</li>
>This is line 3.</li>
>This is line 4.</li>

I need to fetch content which starts with > and ends in </li> so the output will be This is line 1. This is line 2. and so on. I have looked into this in forum but didnt found solution. This solution also didnt worked.

I ultimately have to fetch some lines from a webpage. So first I will curl webpage and then use grep command to grep that line which starts with > and ends in </li>.

Thanks.!

Best Answer

This should be enough:

grep '^>.*</li>$' input-file

The ^ and $ ensure that those parts are anchored at the start and end of the lines respectively.

You can also do:

grep -x '>.*</li>' input-file

-x looks for an exact match: the whole line should match the pattern (which implies ^ and $ is wrapped around the pattern).

Related Solutions

Ubuntu – Use bash to extract number from between square brackets

You can achieve this with just a single grep command. This is because GNU grep lets you use a Perl regular expression (-P), which supports zero-width lookaround assertions (\K and (?= ), in this case):

grep -oP '^\[\K\d+(?=\])' infile

As written, that will send the output to your terminal. To redirect it to a file, use:

grep -oP '^\[\K\d+(?=\])' infile > outfile

This method has the advantage of brevity and simplicity. It matches text that

is preceded by (\K)
- a [ character(\[) -- \ is needed as [ otherwise has a special meaning in regular expressions
- that appears at the beginning of a line (^);
consists of one or more (+) digits (\d);
is followed by ((?= ))
- a ] character (\]) -- like with [, \ forces ] to be matched literally.

Ubuntu – Repeat whole line matches with grep for multiple instances on the same line

grep doesn't have counter for matches, only -c counter for lines that have the match, but we can use awk to do that. As far as I understand , you want to print the line that matches x number of times based on the amount of matches. Well, here it is:

$ awk '{for(i=1;i<=NF;i++) if($i=="banana") counter++;for(j=1;j<=counter;j++) print NR,$0;counter=0 }' input.txt         
1 There is one banana here
2 There are two banana banana here
2 There are two banana banana here
3 There are three banana banana banana here
3 There are three banana banana banana here
3 There are three banana banana banana here
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have
4 Basically there is no limit to how many banana banana banana banana banana banana banana we can have

Basic idea here is that we loop over each word in a line, and count matches. If there's a match we increment counter and then use that counter to print the same line in a loop. Finally counter is reset and process repeats

Best Answer

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Ubuntu – Use bash to extract number from between square brackets

Ubuntu – Repeat whole line matches with grep for multiple instances on the same line

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