I am trying to build a small sh script that does a basic find / replace but with a small twist, that I wonder if I am going about it the wrong way.

here is the old string


here is what i need


now the easy part is:

find . -type f -name "*.php" -exec sed -i'' -e 's/mysql_fetch_array(/->fetch_array(/g' {} +

But I don't know how to pull $foo from the parantheses, and then shove it up in front ... Do I do this in the same sed regex .. Do I store it in a variable and reconstruct it? What's the most efficient way to go about this?

I have tried multiple things, but I am having trouble with escaping the $ etc etc ... My mind is thoroughly exhausted!


It should be noted that I was needing to match mysql_fetch_array specifically as well -- my bad for not specifying it in the OP

find . -type f -name "*.php" -exec sed -i -e 's/^.*mysql_fetch_array(\$\([^)]*\))/$\1->fetch_array()/' {} \;
  • Ahh! I see! Use regex first to isolate and then reconstruct it ... clever! – Zak Feb 17 '17 at 21:45
  • Yeah. Matching the regex meta characters - '$' and '()' here - is so often a pain. If the replacement is simple enough, just ignoring them in the match and including them 'manually' in the replacement is easier. – Eli Heady Feb 17 '17 at 21:50
  • question ... Just tested with blah_blah_blah($something) and it replaced it --- Is there a way to get it to look specifically for mysql_fetch_array? There are other like queries that I don't want it to touch. – Zak Feb 17 '17 at 21:56
  • Changed in edit. Add the specific string you want to match before the first paren. – Eli Heady Feb 17 '17 at 21:57
  • Great, I edited my OP as well to denote that I was slightly unclear -- Thanks for you great solution and explanation! – Zak Feb 17 '17 at 22:03

Use this:

find . -type f -name "*.php" -exec sed -i'' -e 's/mysql_fetch_array(\$foo)/\$foo->fetch_array\(\)/g' {} +

Use \ to escape $ character, and in new value for replacement, parenthesis must be escaped too.

So, you need to use \ for ( and )

  • I think 'variable_here' was a placeholder – Eli Heady Feb 17 '17 at 21:44
  • yes '$variable_here' is a placeholder hence the term "reusable string" in the title – Zak Feb 17 '17 at 21:47
  • @EliHeady hmm, I think it's a specific name! If it be a placaholder, your answer is better than me – MrRolling Feb 17 '17 at 21:48
  • @Zak so please use foo or bar for placaholders in your questions :) – MrRolling Feb 17 '17 at 21:51

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