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JAKZ AIZZAT
7,813 PointsWhy I can't output the image in rotate?
<?php
$img = imagecreatefromjpeg('simpletext.jpg');
$imgRotated = imagerotate($img, 45,-1);
imagejpeg($imgRotated,'newsimpletext.jpg',100);
?>
<img src="simpletext.jpg" />
<hr>
<img sec="newsimpletext.jpg" />
I want to output the image in rotate version but it has error
Warning: imagejpeg() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\developphp\index.php on line 6
2 Answers
Aimee Ault
29,193 Pointsimagerotate actually will return to you a boolean (false) if it fails, so the problem is likely either there or in setting the initial $img. Are you sure it's loading an image into that variable?
JAKZ AIZZAT
7,813 PointsI have tried with other image and its still doesn't work. :(
This is its error: Warning: imagejpeg() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\developphp\index.php
JAKZ AIZZAT
7,813 PointsJAKZ AIZZAT
7,813 PointsYa. Im sure. Im trying to echo the original image at bottom of line code and its work.
Actually i'm following this tutorial : http://www.developphp.com/view.php?tid=1138
Aimee Ault
29,193 PointsAimee Ault
29,193 PointsHave you tried it with a different jpeg image? I just tried your code out and it works for me. Do you see any errors in your error_log for the call to imagerotate?