我有第一個(gè) PHP 腳本，它允許我在遠(yuǎn)程服務(wù)器上顯示圖像。這個(gè)有效，我想設(shè)定一個(gè)條件，以便當(dāng)它在 $file 變量中找不到圖像時(shí)，它會(huì)在 $newfile 變量中顯示圖像。但我收到錯(cuò)誤消息“警告：file_get_contents()：C:\wamp64\www 中的文件名不能為空...”我的錯(cuò)誤在哪里？<!-- Old Script --><?php$file = '//Alcyons/it/PhotoShoot/Photos_Outil/A1111_0070_1.jpg';$type = pathinfo($file, PATHINFO_EXTENSION);?></td><td valign=top align=center>  <img src="<?php echo "data:image/$type;base64,",     base64_encode(file_get_contents($file)) ?>" border=0 width="180"></a></td><td width=10></td><!-- New Script --><?php$file = '//Alcyons/it/PhotoShoot/retail/Photos/SS20,FW1920/A1127G_00_1.jpg';$newfile = '//Alcyons/it/PhotoShoot/retail/Photos/SS20,FW1920/A1119_4023_1.jpg';$type = pathinfo($file, PATHINFO_EXTENSION);?></td><td valign=top align=center>  <img src="<?php   if ($file = NULL){    echo "data:image/$type;base64,", base64_encode(file_get_contents($newfile));}else {    echo "data:image/$type;base64,", base64_encode(file_get_contents($file));}?>" border=0 width="180"></a></td><td width=10></td>