Using sed to find first and replace the first occurrence on a line rather than the last [duplicate]

Question

I have the following sed command: sed 's/\('\''\).*div><div>/,/'

which take everything between a single quote ' and div><div> and replaces it with a comma ,. Which works near enough perfectly to how I want it. However there are some lines that contains two div><div> and my command is taking the second one to be its stoppping point, where I'm trying to cut it off at the first.

To try and provide more clarity, heres the line in the file that I am trying to extract data from:

'>Person A</a></div><div>Teaching A</div></div></td><td width='50%'><div style='height: 50px; margin-bottom: 6px;'><div style='font-weight:bold'>Unknown or external</div><div>Teaching B<

I am trying to replace everything up until Teaching A so my output should look should like ,Teaching A. However the output I am getting is ,Teaching B.

How could I manipulate my sed command to pick up on the first instance of div><div> instead of the last?

Answer 1

@AdminBee: I was also going to suggest non-greedy matching
@Dr Little: What was your solution? Please tell us.

If I understand correctly, this should also work: sed 's/\('\''\).*<.a><div><div>/,/'.

Please understand that it is not recommended to parse HTML files with regex. For example, one time I was parsing 10s of thousands of HTML files with vim+regex, it was a time-sensitive task and I regret doing it that way. Why? Because the task probably would have been completed way faster if I used an actual XML/HTML parser to parse the text files / extract lines and data.