Viewing Thread:
"How would I..."

I'm building a CMS at the moment, and it's my first real attempt at using the image library functions. I'm going to use a source image, resize and resample it, and output in custom dimensions. Simple enough. But I'm also building into the system a small BBoard-style text formatting system (like the square bracket one on here), and I want the user to be able to put [IMG=x] - 'x' being the image ID, then the script scans the string, pulls out that text, and lets me use x as a variable to find the corresponding image. I'm probably being a right dunce, and there's probably a simple answer, so any help would be appreciated, cheers!


EDIT: Hmmm, found the function "stristr(haystack, needle)", which might do what I'm after.

The function you're probably after is eregi_replace(). The PHP manual is quite good on that function - it tells you how to do the regular pattern you need to use.

I'm crap with pattern matching, so I can't give you a specific code for it, but the jist is you say "The tag starts [IMG= and ends with a ], so extract the bit in the middle".

Explain. Please. And not in the form of a riddle, Dungeon Master.

Regular expressions are the way to go with problems like this.

$image_id = substr($content, strrpos($content, '[IMG=') + 1);
$image_id = str_replace("]", "", $image_id);

That'll work, right?

I'm building a CMS at the moment, and it's my first real attempt at using the image library functions. I'm going to use a source image, resize and resample it, and output in custom dimensions. Simple enough. But I'm also building into the system a small BBoard-style text formatting system (like the square bracket one on here), and I want the user to be able to put [IMG=x] - 'x' being the image ID, then the script scans the string, pulls out that text, and lets me use x as a variable to find the corresponding image. I'm probably being a right dunce, and there's probably a simple answer, so any help would be appreciated, cheers!


EDIT: Hmmm, found the function "stristr(haystack, needle)", which might do what I'm after.