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I want to process PHP files on the content but i'm only interested in the parts between < ?php ... ?>. The rest should be removed and saved to a new file. Below I've already part of the code based on the code from this page multi line sed search. But it only gives the first part in .

start="<?php"
end="?>"
temp=$(sed -n '1h;1!H;${;g;s/.*'"$start"'//pI;d;}' <<< echo $filename)
output=$(sed -n '1h;1!H;${;g;s/'"$end"'.*//p;d;}' <<< "$temp")

Somebody has a solution? awk, grep is also oké. THanks

jjacobs
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1 Answers1

2

this one-liner may work for you:

 awk '/<?php/{f=1} /?>/&&f{print;f=0} f'  file

if you don't want the <?php and ?> in output:

 awk '/<?php/{f=1;next;} /?>/&&f{f=0} f' file

test: (I hope the output is what you are looking for. to save to new file, you could redirect output to a new file)

kent$  cat test.txt 
<?php
all
text

here
should
be
saved

even if 

empty

lines

?>


foo

bar

should be removed

kent$  awk '/<?php/{f=1} /?>/&&f{print;f=0} f' test.txt                                                                                                                     
<?php
all
text

here
should
be
saved

even if 

empty

lines

?>
Kent
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  • You don't need an explicit print or next, just put the conditions in the right order : `awk '//{f=0}'` and `awk 'f&&/?>/{f=0} f; / – Ed Morton Dec 19 '12 at 12:53