从文件中读取行并输出作为下拉列表的选项

发布于 2024-11-02 03:57:45 字数 425 浏览 1 评论 0原文

我有 $data 包含类似这样的内容并且经常更改：

backup 2011-04-15 16:39:18.559965.zip
backup 2011-04-15 16:39:56.289656.zip
backup 2011-04-15 16:41:34.463754.zip
backup 2011-04-15 16:41:54.089134.zip
backup 2011-04-15 16:42:18.742616.zip
backup 2011-04-16 13:12:33.083622.zip
backup 2011-04-16 13:14:53.387308.zip
backup 2011-04-17 00:30:32.591461.zip

我怎样才能将其转换为下拉列表，以便人们可以从中进行选择，并且我可以将结果用作 $_POST['file']？

原文

I have $data which contains something like this and changes often:

backup 2011-04-15 16:39:18.559965.zip
backup 2011-04-15 16:39:56.289656.zip
backup 2011-04-15 16:41:34.463754.zip
backup 2011-04-15 16:41:54.089134.zip
backup 2011-04-15 16:42:18.742616.zip
backup 2011-04-16 13:12:33.083622.zip
backup 2011-04-16 13:14:53.387308.zip
backup 2011-04-17 00:30:32.591461.zip

How can i make that it will transform it into a Dropdown list, so people can choose from and i can use result as $_POST['file']?

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离笑几人歌 2024-11-09 03:57:45

使用 fopen() 读取文本
使用 explode() 将文本拆分为数组
使用 foreach() 遍历数组
>
选项标签之间的 echo 元素。

编辑

我没有提供任何代码，因为您应该能够按照上述步骤自己编写代码。只需使用 Google 即可了解这些功能的工作原理。你只能通过自己实现一些东西来学习 PHP。不要指望人们为你编写完整的脚本。

无论如何，这应该可以工作:)

$path = "files.txt";
$file = fopen($path, 'r');
$data = fread($file, filesize($path));
fclose($file);

$lines =  explode(PHP_EOL,$data);
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="'. urlencode($line).'">'.$line.'</option>';
}
echo '</select>';

您也可以使用 file_get_contents() 而不是 fopen()。

$data = file_get_contents("files.txt");

编辑

添加了file()，如Wiseguy推荐的

$lines = file('files.txt');
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="'. urlencode($line).'">'.$line.'</option>';
}
echo '</select>';

Read text using fopen()
Split text into array using explode()
Loopt trough array using foreach()
echo elements between option tags.

EDIT

I did not provide any code because you should be able to write this yourself following the steps above. Simply use Google to find out how these functions work. You can only learn PHP by implementing things yourself. Don't expect people to write complete scripts for you.

Anyways, this should be working :)

$path = "files.txt";
$file = fopen($path, 'r');
$data = fread($file, filesize($path));
fclose($file);

$lines =  explode(PHP_EOL,$data);
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="'. urlencode($line).'">'.$line.'</option>';
}
echo '</select>';

You can also use file_get_contents() instead of fopen().

$data = file_get_contents("files.txt");

EDIT

Added file() like recommended by Wiseguy

$lines = file('files.txt');
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="'. urlencode($line).'">'.$line.'</option>';
}
echo '</select>';

回复收藏 0 原文

你怎么敢 2024-11-09 03:57:45

$lines = explode($data);
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="', $line, '">', $line, '</option>';
}
echo '</select>';

$lines = explode($data);
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="', $line, '">', $line, '</option>';
}
echo '</select>';

回复收藏 0 原文

~没有更多了~