well, if the problem is on line 48, the most sinister actors would be the 2 arguments '$dir1' AND '$dir/thumbs/$photo1'
- try outputting these strings before you call the function, and see what they contain.
e.g. echo "++$dir1++$dir/thumbs/$photo1++";
wilch
Junior Poster in Training
84 posts since Aug 2007
Reputation Points: 25
Solved Threads: 17
well, if the problem is on line 48, the most sinister actors would be the 2 arguments '$dir1' AND '$dir/thumbs/$photo1'
- try outputting these strings before you call the function, and see what they contain.
e.g. echo "++$dir1++$dir/thumbs/$photo1++";
wilch
Junior Poster in Training
84 posts since Aug 2007
Reputation Points: 25
Solved Threads: 17
Your issue is your using single quotes ' instead of double quotes " as a string with a variable in it.
on line 48 you have:
createthumb('$dir1','$dir/thumbs/$photo1',100,100);//Not working(problem here only)
Try:
createthumb("$dir1","$dir/thumbs/$photo1",100,100);//Not working(problem here only)
the problem is, that when you use single quotes '$myvar' will actually be the string $myvar instead of the variable myvar. PHP does not parse inside single quotes, only double quotes.
hope that helps.
kylegetson
Junior Poster in Training
89 posts since Sep 2009
Reputation Points: 26
Solved Threads: 12
function createthumb() expect the first parameter to be image file name and i guess $filename is the directory path.
I think there is problem with you passing as $dir1, it seems to be made from $dir and some dir structure.Also what comes in $dirid is not been clear
network18
Practically a Master Poster
619 posts since Sep 2009
Reputation Points: 29
Solved Threads: 76
did you noticed that you checking with image type against jpeg/jpg and png, check which format image you passing, when blank thumbnail gets created.
network18
Practically a Master Poster
619 posts since Sep 2009
Reputation Points: 29
Solved Threads: 76