File upload 通过simpleimage.php上载图像
我是php新手,我尝试使用Simpleimage.php,但他们从未打印过error image类型的错误File upload 通过simpleimage.php上载图像,file-upload,image-uploading,File Upload,Image Uploading,我是php新手,我尝试使用Simpleimage.php,但他们从未打印过error image类型的错误 $image = new SimpleImage(); $image->load('$_FILES[file][tmp_name]'); $image->resize(250,400); $image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.j
$image = new SimpleImage();
$image->load('$_FILES[file][tmp_name]');
$image->resize(250,400);
$image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.jpg');
邮政编码
伊玛吉
$image = new SimpleImage();
$image->load('$_FILES[file][tmp_name]');
$image->resize(250,400);
$image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.jpg');
我错在哪里了???我想您想在load方法中输入变量的值,而不是变量的名称,所以请尝试以下操作:
$image = new SimpleImage();
$image->load('$_FILES[file][tmp_name]');
$image->resize(250,400);
$image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.jpg');
$image->load($_FILES['file']['tmp_name']);
您应该在数组索引中使用'in
$image = new SimpleImage();
$image->load('$_FILES[file][tmp_name]');
$image->resize(250,400);
$image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.jpg');
启用显示错误:error\u reporting(E\u ALL)
并始终尝试修复它们
$image = new SimpleImage();
$image->load('$_FILES[file][tmp_name]');
$image->resize(250,400);
$image->save('http://localhost:8080/GuitarShop/images/chitarre/ciao.jpg');