如何下载临时文件
2022-08-30 16:45:36
我正在尝试创建一个简短的PHP脚本,该脚本采用JSON字符串,将其转换为CSV格式(using),并使该CSV可用作下载的.csv文件。我的想法是用来不用担心cronjobs或磁盘空间不足,但我似乎无法让奇迹发生。fputcsv
tmpfile()
这是我的尝试,这是从readfile
的PHP文档中转换的示例:
$tmp = tmpfile();
$jsonArray = json_decode( $_POST['json'] );
fputcsv($tmp, $jsonArray);
header('Content-Description: File Transfer');
header('Content-Type: text/csv');
header('Content-Disposition: attachment; filename='.basename($tmp));
header('Content-Transfer-Encoding: binary');
header('Expires: 0');
header('Cache-Control: must-revalidate');
header('Pragma: public');
header('Content-Length: ' . filesize($tmp));
ob_clean();
flush();
readfile($tmp);
我觉得 和 标题是错误的,但我不确定如何修复它们。例如,似乎没有返回任何内容,我不确定传输为二进制编码。同样,也是空白的。有没有办法做我在这里尝试的事情?Content-Disposition
Content-Transfer-Encoding
basename($tmp)
text/csv
echo filesize($tmp)
[编辑]
**以下是我根据接受的答案编写的工作代码:*
$jsonArray = json_decode( $_POST['json'], true );
$tmpName = tempnam(sys_get_temp_dir(), 'data');
$file = fopen($tmpName, 'w');
fputcsv($file, $jsonArray);
fclose($file);
header('Content-Description: File Transfer');
header('Content-Type: text/csv');
header('Content-Disposition: attachment; filename=data.csv');
header('Content-Transfer-Encoding: binary');
header('Expires: 0');
header('Cache-Control: must-revalidate');
header('Pragma: public');
header('Content-Length: ' . filesize($tmpName));
ob_clean();
flush();
readfile($tmpName);
unlink($tmpName);