PHP读取文件并创建下拉菜单 [英] php read file and create drop down menu

问题描述

我有以下php代码，它们创建下拉菜单来读取包含名称的文本文件

I have following php code which create Drop-Down menu reading text file which contain name

files.txt

files.txt

JAMES
MARK
TONY

drop-down.php

drop-down.php

<?php
$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>';
?>

这是代码的POST部分，它是一个循环

Here is the POST section of code, it is a loop

foreach ($_POST["bname"] AS $id => $value)
{
...
...   
USERNAME: "'.$_POST["file"][$id].'"

一切正常，但是当我以其他形式提交数据时，我只会得到用户的第一个字母，就像我选择 JAMES 并提交数据时，我只会得到<$ c 。$ _ POST 中的$ c> J 字母。我想要全名 JAMES ，我的代码有什么问题？

Everything working but when i submit data to other form i am getting only first letter of users like if i select JAMES and submit data i am only getting J letter in .$_POST. I want Full name JAMES, what is wrong in my code?

推荐答案

您选择的名称属性错误，必须为：

The name attribute of your select is wrong, must be:

echo '<select name="file[]">';

或

echo '<select name="file[someNumber]">';

这篇关于PHP读取文件并创建下拉菜单的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！