2011-04-30 80 views
0
<?php 
function saveFile($url, $filename) { 
    $data = file_get_contents($url); 
    $fh = fopen($filename, 'w') or die("can't open file"); 
    fwrite($fh, $data); 
    fclose($fh); 
} 
$myFile = "images.txt"; 
$fh = fopen($myFile, 'r'); 
$theData = fread($fh, filesize($myFile)); 
$data = explode("\n", $theData); 

$count = count($data); 

for($i=0;$i<=$count;$i++) { 

    saveFile($data[$i], basename($data[$i])); 
} 

?> 

这是我的代码我每次更换打印错误PHP下载文件,读取问题

Warning: fopen(10017226314.jpg) [function.fopen]: failed to open stream: Invalid argument in C:\xampp\htdocs\download\download.php on line 4 
can't open file 

应该是什么原因一个动态的文件名?请帮忙。由于

+2

这需要基本调试。你看过'$ filename'吗?我想你会发现它不包含有效的文件名。 – 2011-04-30 08:58:20

回答

1

请更换您的fopen函数的参数从wr+

function saveFile($url, $filename) { 
    $data = file_get_contents($url); 
    $fh = **fopen($filename, 'r+')** or die("can't open file"); 
    fwrite($fh, $data); 
    fclose($fh); 
}