Preg Help with Lookahead

MaxDeMarzi

Hello,

I am trying to parse a text file and insert into mysql, however some of the fields allow a New Line character (\n) in the text and it is messing things up.

I cannot simple replace them all because that would screw things up even more.

What I need is a preg_replace expression that finds all \n that do NOT have 6 numbers right afterwards and deletes them. Deleting just them, using a look forward or "Assertions".

So "blah blah.\nBlah Blah" would be found but
"blah blah.\n123456" would not be found

I've been racking my brains for hours, and so far no luck. I need more brains, maybe you guys can help me with this?

Thank you!
Max

LordShryku

It's ereg, but should do the trick.
ereg_replace("(\n)([^0-9]{6})", "\2", $text);

MaxDeMarzi

Thanks, but no luck.

It does take care of most of the problem, but it misses some things:

/n1742 sq ft
/n2 Apts

It is almost as if the {6} does not work right.

Any other ideas?

Thanks,
Max

MaxDeMarzi

It is ugly, but I finally found a way that works:

$content = ereg_replace("(\n)([1-9])([0-9])([0-9])([0-9])([0-9])([0-9])|", "NewLineHere\2\3\4\5\6\7|", $content);
$content = ereg_replace("(\n)", "", $content);
$content = ereg_replace("(NewLineHere)", "\n", $content);

The delimiter is a pipe, and thankfully all the 6 digits I have to save were followed by the delimeter. So I ended up searching for those, "Saving" them using the NewLineHere text, nuking the new lines in the rest of the file, and then putting the correct new lines back.

I hope this helps anyone who runs into a similar problem. 😃

Regards,
Max

Weedpacket

/\n(?!\d{6})/

Replace that with an empty string, of course.

Since preg_replace() is binary-safe, if the file is using an MS-DOS newline convention (a carriage return/line feed pair), then that will have to be dealt with (\r\n instead of just \n). Otherwise the stray \r characters that remain can cause weirdness.