Al wrote:
On 5/22/2010 1:02 PM, Robert Cummings wrote:
tedd wrote:
At 4:27 PM +0200 5/21/10, Anton Heuschen wrote:
So in the file it would look like (from the original file the user
uploads
that is)
1
2
3
4
5
6
but when the file is saved to the server it must look like
1
2
3
4
5
6
If that is all (i.e., removing double linefeeds), then this will do it:
$text_array = array();
$text_array = explode("\n\n", $input_text);
$output_text = implode("\n",$text_array);
Sorry tedd, this is broken. It doesn't solve problems with runs of
greater than 2 newlines which is even in the example :) I would use the
following instead which is also line break agnostic with final output in
the style for your system:
<?php
$data = preg_replace( "#[\r\n]+#", PHP_EOL, $input );
?>
Cheers,
Rob.
Rob: Your solution doesn't remove the blank lines [\r\n]+ use instead [\r\n]{2,}
So 2 or more becomes only 1.
In general, problem is trickier when the following are considered. # means any
number. 0, 1.....
some textEOL
#spacesEOL
more text
some text#spacesEOL
#spacesEOL
.... any number of these
#spacesEOL
more text
some textEOL
EOL
...any number of these
EOL
some text
The white space before the EOLs can also be tabs
Look at the solution I posted earlier. The trim() removes all the white spaces
My solution worked well where spaces were not an issue. Your solution
breaks my more general solution. Although I did realize I should have
trimmed the final output since any empty lead lines will not be removed.
Please review and see why you're comment to use [\r\n]{2,} does not work
properly. Correcting for lead blank lines and handling spaces in a blank
line is also quite simple without having to use the heavy solution of
foreach:
<?php
$data = preg_replace( "#[\r\n]+[[:space:]]+[\r\n]+#", "\n", $input );
$data = preg_replace( "#[\r\n]+#", PHP_EOL, $input );
$data = trim( $input );
?>
Without benchmarking, I'm willing to bet this is faster and less memory
intensive than your foreach solution :)
Cheers,
Rob.
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