问题描述
在PHP中复制图像变量的最简单方法是什么.
What is the easiest way to duplicate an image variable in PHP.
通常您可以简单地执行$varnew = $varold.
Normally you can simply do $varnew = $varold.
但是如果使用GD库变量，则执行上述操作，然后编辑$ varnew，然后$ varold也将生效.
But with a GD library variable if I did the above and then edited $varnew then $varold would be effected too.
显然，一种方法是重新打开文件或创建新图像并将其复制到其中.有没有更简单的方法?
Obviously one way would be to reopen the file or to make a new image and copy it into it. Is there an easier way?
推荐答案
尝试失败的原因是因为该变量仅存储GD映像的句柄(内存指针). $varnew和$varold仍将存储相同的指针，从而指向内存中完全相同的图像.您必须使用图像复制，或者可能更糟的是，再次从文件中打开图像.
The reason your try didn't work is because the variable only stores a handle (a memory pointer) to the GD image. Both $varnew and $varold will still store the same pointer, thus pointing to the exact same image in memory. You have to use imagecopy or, maybe worse, open the image from file again.
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