foochuck Posted October 6, 2009 Share Posted October 6, 2009 I'm using this code to count images in a folder on my website: function countImgs() { global $imgCount, $folder; $dir = "./$folder/thumbnails/"; $imgCount = count(glob($dir . '*.jpg')); } Instead of counting all of the jpg's in the folder I want to count specific image sets in that folder. For example if my images are named: foo-01.jpg, foo-02.jpg, foo-03.jpg - I want to count all images that begin with the string 'foo'. Is this the correct way to check that? -> function countImgs() { global $imgCount, $folder; $dir = "./$folder/thumbnails/"; $imgCount = count(glob($dir . 'foo*')); } Thanks Link to comment https://forums.phpfreaks.com/topic/176724-counting-images-in-a-folder/ Share on other sites More sharing options...
.josh Posted October 6, 2009 Share Posted October 6, 2009 that glob would return everything starting with "foo", regardless of whether it is an image, some other file, or a subdirectory. You can change it to ...glob($dir . 'foo*.jpg') to get all .jpg files starting with foo. Link to comment https://forums.phpfreaks.com/topic/176724-counting-images-in-a-folder/#findComment-931753 Share on other sites More sharing options...
foochuck Posted October 6, 2009 Author Share Posted October 6, 2009 What if I want to replace 'foo' with a variable? How would I put $foo into this line of code to replace the string foo: $imgCount = count(glob($dir . 'foo*')); Link to comment https://forums.phpfreaks.com/topic/176724-counting-images-in-a-folder/#findComment-931763 Share on other sites More sharing options...
.josh Posted October 6, 2009 Share Posted October 6, 2009 well one way to do it is the same way you have $dir in there Link to comment https://forums.phpfreaks.com/topic/176724-counting-images-in-a-folder/#findComment-931774 Share on other sites More sharing options...
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