我正在创建一个包含2个文件但在打开时显示错误的zip文件。但是当我用代码编辑器打开它时会显示一些错误。
<?php
$files = array(
'localhost/apache_pb2_ani.gif',
'localhost/apache_pb2.png');
$zipname = 'file.zip';
$zip = new ZipArchive;
$zip->open($zipname, ZipArchive::CREATE);
foreach ($files as $file) {
$zip->addFile($file);
}
$zip->close();
//if true, good; if false, zip creation failed
header('Content-Type: application/zip');
header('Content-disposition: attachment; filename='.$zipname);
header('Content-Length: ' . filesize($zipname));
readfile($zipname);
?>
错误是:
Warning:filesize(): stat failed for file.zip in C:\xampp\htdocs\download.php on line 19
Warning: readfile(file.zip): failed to open stream: No such file or directory in <b>C:\xampp\htdocs\download.php on line 20
什么似乎是错误?
答案 0 :(得分:0)
如果我在$files
数组中使用的文件路径不正确,我可以重现警告。
您确定图像的路径(带有“localhost /”)是否正确?
$files = array(
'localhost/apache_pb2_ani.gif',
'localhost/apache_pb2.png'
);