[英]Why won't PHP read this file?
我正在尝试使其检索服务器上的图像文件,但是如果图像文件名中有空格,则该行将不起作用。例如,即使我逃脱了,死角和空气之间也有空格添加%20之后,该函数将返回一个空字符串..但如果该文件的名称中没有空格,例如“ http://www.m.trialsite.com/images/thumb/Espresso.jpg ”; 会工作的! ..我要去哪里错了?
$filename = 'http://www.m.trialsite.com/images/thumb/dead air.jpg';
function readfile_chunked($filename,$retbytes=true) {
$chunksize = 1*(1024*1024); // how many bytes per chunk
$buffer = '';
$cnt =0;
// $handle = fopen($filename, 'rb');
$filename = str_replace(' ','%20',$filename);
$handle = fopen($filename, 'rb');
if ($handle === false) {
return false;
}
$filename = str_replace(' ','%20',$filename);
while (!feof($handle)) {
$buffer = fread($handle, $chunksize);
echo $buffer; var_dump($buffer); exit;
ob_flush();
flush();
if ($retbytes) {
$cnt += strlen($buffer);
}
}
$status = fclose($handle);
if ($retbytes && $status) {
return $cnt; // return num. bytes delivered like readfile() does.
}
return $status;
}
使用preg_replace("/\\s+/","_",$nome);
重命名文件,然后恢复它会起作用
$directory = '/public_html/testfolder/';//example
if ($handle = opendir($directory)) {
while (false !== ($fileName = readdir($handle))) {
$newName = preg_replace("/\s+/","_",$fileName);
rename($directory . $fileName, $directory . $newName);
}
closedir($handle);
}
如果您这样做是怎么办的:
$filename = str_replace(' ','%20', 'http://www.m.trialsite.com/images/thumb/dead air.jpg');
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