regex - Find and replace first double-quote before special character via grep -

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I have a whole bunch of files that have all the blocks of text that look like this: / P> 

  "The opening of each file is different text between double quote and closing right-quote (or whatever is said)" "

Maybe not relevant, but in the past i have grep used to search and change it like this:

grep -rl 'search './path/to/files/ | xargs sed - I's / is there any way to do something like that, but use a rezux to replace the plain old double-quote opening with a left quote ( "),

Code>) The only reliable way to change the right-two-coat characters is to search on the correct quote, then according to the previous bi-quote she is backwards. I think I'm sure it's possible or what to do. this.

I could only do this with PHP scripts, but then I will not find if it is possible from the command line.

You can use sed: 

  sed - I.bak 's / "\ ([^" *] "\" / "\ 1 /" file cat file "in the opening and closing of the double quote in each file, right-quote (or whatever is said) There is a separate text between "

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