zip_read()
(PHP 4 >= 4.1.0, PHP 5 >= 5.2.0, PHP 7, PECL zip >= 1.0.0)
读取ZIP存档文件中下一项
说明
zip_read(resource $zip): resource
读取ZIP存档文件中下一项。
参数
- $zip
一个ZIP压缩文件,该ZIP归档文件之前应由函数zip_open()打开。
返回值
成功的时候返回该当前实体资源供zip_entry_...系列函数后续使用;如果没有更多的读取项,则会返回FALSE
如果遇到错误则会返回相应的错误码。
参见
zip_open()
打开ZIP存档文件zip_close()
关闭一个ZIP档案文件zip_entry_open()
打开用于读取的目录实体zip_entry_read()
读取一个打开了的压缩目录实体
*Here is a simple example* <?php $zp = zip_open('file.zip'); while ($file = zip_read($zp)) { echo zip_entry_name($file).PHP_EOL; } ?> The output will be something similar to: myfile.txt mydir/
Note: Only the first 65535 entries will be returned, even if your archive contains more entries. See https://bugs.php.net/bug.php?id=59118 for details.
If you get an error like this: Warning: zip_read() expects parameter 1 to be resource, integer given in xxxxxx on line x It's because zip_open() failed to open the file and returned an error code instead of a resource. It took me a while to figure out why it failed to open the file, until I tried to use the FULL path to the file. <?php // Even if the file exists, zip_open() will return an error code. $file = 'file.zip'; $zip = zip_open($file); // The workaround: $file = getcwd() . '/file.zip'; // Or: $file = 'C:\\path\\to\\file.zip'; ?> This worked for me on Windows at least. I'm not sure about other platforms.