用PHP下載服務器上的幾個文件的內容

Question

我有2個文件 - 1.csv and 2.csv。我可以通過已知的方法下載1.csv用PHP：

$file = '1.csv'; 
header('Content-Description: File Transfer'); 
header('Content-Type: application/jpeg'); 
header('Content-Disposition: attachment; filename='.basename($file)); 
header('Content-Transfer-Encoding: binary'); 
header('Expires: 0'); 
header('Cache-Control: public'); 
header('Pragma: public'); 
header('Content-Length: ' . filesize($file)); 
ob_clean(); 
flush(); 
readfile($file); 

但我的目的是要在這兩個文件的下載的文件內容 - 1.csv和2.csv。有沒有人有一些想法如何實現這一點？ 預先感謝您。

Answer 1

串聯如果我理解正確，你這個應該這樣做。

$file1 = '1.csv'; 
$file2 = '2.csv'; 
header('Content-Description: File Transfer'); 
header('Content-Type: text/comma-separated-values'); 
header('Content-Disposition: attachment; filename='.basename($file1, '.csv') . '-' . basename($file2, '.csv') . '.csv'); 
header('Content-Transfer-Encoding: binary'); 
header('Expires: 0'); 
header('Cache-Control: public'); 
header('Pragma: public'); 
header('Content-Length: ' . (filesize($file1) + filesize($file2))); 
ob_clean(); 
flush(); 
readfile($file1); 
readfile($file2); 

考慮使用多個文件的循環和心靈的Content-Type頭的CSV文件的變化。

Answer 2

$file1 = '1.csv'; 
$file2 = '2.csv'; 
header('Content-Description: File Transfer'); 
header('Content-Type: application/jpeg'); 
header('Content-Disposition: attachment; filename="1-2.csv"'); 
header('Content-Transfer-Encoding: binary'); 
header('Expires: 0'); 
header('Cache-Control: public'); 
header('Pragma: public'); 
header('Content-Length: ' . (filesize($file1)+filesize($file2))); 
ob_clean(); 
flush(); 
echo file_get_contents($file1).file_get_contents($file2); 

的下載2項文件內容