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@NogDogDec 14.2010 — #You cannot output the image such as with imagejpeg() within a file that is outputting anything else, including HTML <img> tags. If you are going to output it via imagejpeg(), then that is [i]all[/i] that file should output (with the applicable content-type header). Therefore, your separate main script would have an <img> tag whose src attribute calls the image-generating script.
Alternatively, if you are saving the file anyway, then just create an <img> tag that points to that saved file instead of outputting it with imagejpeg().@NogDogDec 14.2010 — #One thought is to make sure that your image file was not saved as UTF8 with a BOM (byte order mark), as the BOM will corrupt the image output. @NogDogDec 14.2010 — #Ugh...I've seen lots of people complaining about getting that function to work. I've not used it myself (or at least not in quite some time, I think), so all I can suggest is to do what I'd so: search through all the user comments on that function's manual page -- pending someone coming along here who has more experience with it. ?
[RESOLVED] Calling php file from another php file + GD library
I was trying to print on screen an image within a div by doing this in my layout php file:
[code=php]
<?php header(“Content-type: image/jpeg”); ?>
…
<img src=”index.php?var=<?php echo $value; ?>” />
but something isn’t going as expected.
the invoked code does this at the end:
[code=php]
$success = imagejpeg($image, ‘images/postcards/postcard.jpg’);
$success = imagejpeg($image);
imagedestroy($image);
As you can see, I’m trying to do two different things with the image:
1- To dump it and save it in a jpeg file, which is working right
2- To directly output the data as a jpg image on the browser. This is not working right. Despite the second line returns true as well as the first line does, the image is not displayed.
Why could the first choice be wrong and not display the image?
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@NogDogDec 14.2010 — #You cannot output the image such as with imagejpeg() within a file that is outputting anything else, including HTML <img> tags. If you are going to output it via imagejpeg(), then that is [i]all[/i] that file should output (with the applicable content-type header). Therefore, your separate main script would have an <img> tag whose src attribute calls the image-generating script.Alternatively, if you are saving the file anyway, then just create an <img> tag that points to that saved file instead of outputting it with imagejpeg().
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@NogDogDec 14.2010 — #0
@NogDogDec 14.2010 — #Ugh...I've seen lots of people complaining about getting that function to work. I've not used it myself (or at least not in quite some time, I think), so all I can suggest is to do what I'd so: search through all the user comments on that function's manual page -- pending someone coming along here who has more experience with it. ?