我有一個包含在表單操作中的按鈕，單擊該按鈕會調用下面的 download.php 并下載已經存在的文件。在 results.php 中形成動作<form method="post" action="download.php"> <button type="submit" class="btn btn-dark btn-block" data-toggle="dwn-one" title="Click to download"> Download </button> </form>下載.php<?php header("Content-disposition: attachment; filename=file1.txt"); header("Content-type: application/txt"); readfile("file1.txt"); header("results.php");?>上述功能運行良好，但現在我有很多文件要下載，所以我想避免多個下載按鈕，而是提供一個選擇功能，用戶可以選擇文件并單擊下載按鈕。到目前為止，我正在執行如下操作，但無法弄清楚如何調用 download.php。選擇代碼<form method = "post" action=""> <select class="form-control" name="downloaditem"> <option value="1">Select a file to download</option> <option value="2">File one</option> <option value="3">File two</option> <option value="4">File three</option> <option value="5">File four</option> </select> <button type="submit" class="btn btn-info" name="files">Download</button></form>isset 函數if(isset($_POST['files'])){ $filesel = $_POST['downloaditem']; if($filesel == "2") { //call to download.php }}請幫忙。
1 回答
慕的地6264312
TA貢獻1817條經驗 獲得超6個贊
只需將表單發送到 download.php 腳本。
<form method = "post" action="<url-to>download.php">
<select class="form-control" name="downloaditem">
<!-- It's not necessary to make the first option available for select -->
<option value="0" disabled>Select a file to download</option>
<option value="1">File one</option>
<option value="2">File two</option>
<option value="3">File three</option>
<option value="4">File four</option>
</select>
<button type="submit" class="btn btn-info" name="files">Download</button>
</form>
下載.php
// Get the file name and path to the file based on the given id
// or return null
function getFile($id) {
switch((int)$_POST['downloaditem']) {
case 1:
return ['file1', 'path/to/file1.txt'];
case 2:
return ['file1', 'path/to/file2.txt'];
case 3:
return ['file1', 'path/to/file3.txt'];
default:
return null;
}
}
if(isset($_POST['files'])) {
$file = getFile($_POST['downloaditem']);
if (null !== $file} {
header('Content-Disposition: attachment; filename=' . $file[0]. '.txt');
header('Content-Type: application/txt');
readfile($file[2]);
header('results.php');
} else {
// If function returns null show a HTTP 404 error
header('Content-Type: text/plain');
header('HTTP/1.1 404 Not Found');
}
}
- 1 回答
- 0 關注
- 139 瀏覽
添加回答
舉報
0/150
提交
取消