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错误:imagecopymerge()期望参数2是资源(Error: imagecopymerge() expects parameter 2 to be resource)

尝试在PHP中将图像设置为变量时出错:

$barcode= 'barcode_test.png';
$image = ImageCreate($width, $height);
ImageCopyMerge($image,$barcode,0,0,0,0,400,100,0);

PHP警告:imagecopymerge()期望参数2是资源,给定字符串

如何让$barcode成为资源,而不是字符串?


Getting an error trying to set an image as a variable in PHP:

$barcode= 'barcode_test.png';
$image = ImageCreate($width, $height);
ImageCopyMerge($image,$barcode,0,0,0,0,400,100,0);

PHP Warning: imagecopymerge() expects parameter 2 to be resource, string given

How do I make $barcode a resource, not a string?

更新时间:2023-02-03 19:02

最满意答案

在您的情况下,您必须使用imagecreatefrompng来获取文件的资源。 请尝试以下方法:

$barcode = imagecreatefrompng('barcode_test.png');
$image = ImageCreate($width, $height);
ImageCopyMerge($image,$barcode,0,0,0,0,400,100,0);

如果您正在使用其他文件类型,则必须根据扩展名选择以下函数之一: imagecreatefromgd2imagecreatefromgd2partimagecreatefromgdimagecreatefromgifimagecreatefromjpegimagecreatefrompngimagecreatefromstringimagecreatefromwbmpimagecreatefromwebpimagecreatefromxbmimagecreatefromxpm


In your case you have to use imagecreatefrompng to get a resource of the file. Try the following:

$barcode = imagecreatefrompng('barcode_test.png');
$image = ImageCreate($width, $height);
ImageCopyMerge($image,$barcode,0,0,0,0,400,100,0);

If you are using another file type you have to choose one of the following function depending on the extension: imagecreatefromgd2, imagecreatefromgd2part, imagecreatefromgd, imagecreatefromgif, imagecreatefromjpeg, imagecreatefrompng, imagecreatefromstring, imagecreatefromwbmp, imagecreatefromwebp, imagecreatefromxbm, imagecreatefromxpm

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