Home Backend Development PHP Tutorial 为什么小弟我的var_dump($file_suffix=="php")总是false?明明值一样的啊

为什么小弟我的var_dump($file_suffix=="php")总是false?明明值一样的啊

Jun 13, 2016 am 10:52 AM
entry file name quot zip

为什么我的var_dump($file_suffix=="php")总是false?明明值一样的啊.
我是想读取压缩包里面的php文件内容,结果遇到这个怪问题给困惑啊
var_dump($file_suffix=="php");明明值一样,怎么是false啊...

PHP code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->//获取后缀函数function suffix($filename){    $string= strrpos($filename,'.');    $suffix = substr($filename,$string+1);    return $suffix;}$zip = zip_open("test.zip"); //打开 ZIP 文件if ($zip) {    while ($zip_entry = zip_read($zip)) {  //读取 ZIP 文件中的下一个项目        echo "Name: " . $file_name=zip_entry_name($zip_entry) . "<br>";   //返回 ZIP 文件中的一个项目的名称        echo "Actual Filesize: " . zip_entry_filesize($zip_entry) . "<br>";  //返回 ZIP 文件中的一个项目的实际文件尺寸        echo "Compressed Size: " . zip_entry_compressedsize($zip_entry) . "<br>";  //返回 ZIP 文件中的一个项目的被压缩尺寸        echo "Compression Method: " . zip_entry_compressionmethod($zip_entry) . "<br>";  //返回 ZIP 文件中的一个项目的压缩方法        echo $file_suffix=suffix($file_name);        var_dump($file_suffix=="php");        if($file_suffix=='php'){           zip_entry_open($zip, $zip_entry, "r"); //打开 ZIP 文件中的一个项目以供读取            echo "File Contents:" . "<br>";            $buf = zip_entry_read($zip_entry, zip_entry_filesize($zip_entry)); //读取 ZIP 文件中的一个打开的项目            echo "$buf";            zip_entry_close($zip_entry);  //关闭 ZIP 文件中的一个项目        }                   echo "<br>";    }    zip_close($zip);  //关闭 ZIP 文件}
Copy after login


------解决方案--------------------
能一样吗?
echo "Name: " . $file_name=zip_entry_name($zip_entry) . "
"; //返回 ZIP 文件中的一个项目的名称
中 $file_name=zip_entry_name($zip_entry) . "
" 这不是后面还有
吗?
在 echo 中赋值的话,应写作
echo "Name: " .( $file_name=zip_entry_name($zip_entry) ). "
";

echo "Name: " , $file_name=zip_entry_name($zip_entry) , "
";
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