动态填充的下拉菜单；$_POST 返回错误 [英] Dynamically populated drop-down; $_POST returning error

问题描述

<?php
mysql_connect("host", "user", "pw") or die(mysql_error());
mysql_select_db("db") or die(mysql_error());



$sql = "SELECT machine_id FROM machines";
$result = mysql_query($sql);

echo "<form action='results_send.php' method='post' >";
echo "<select name='machine_id'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['machine_id'] . "'>" . $row['machine_id'] . "</option>";
}
echo "</select></form>";
echo '<input type="submit" value="Submit">';



$machine = $_POST['machine_id'];

var_dump($machine);

?>

返回:

<form action='results_send.php' method='post' ><select name='machine_id'><option value='2011 Honda'>2011 Honda</option><option value='1999 Toyota'>1999 Toyota</option><option value='1999 Honda'>1999 Honda</option><option value='1999 Honda'>1999 Honda</option><option value='2013 Toyota'>2013 Toyota</option><option value='2012 Ford'>2012 Ford</option><option value='2012 Ford'>2012 Ford</option><option value='2012 Ford'>2012 Ford</option><option value='2012 Ford'>2012 Ford</option></select></form><input type="submit" value="Submit">NULL

所有这些都在同一个文档中:results_send.php

all of this is on the same document: results_send.php

我只试用了几天 php/mysql -- 非常感谢您的耐心.

I've been trying out php/mysql for only a couple days--the patience is appreciated.

推荐答案

我不明白您希望 $_POST 如何返回任何值，甚至希望我们了解您的问题，而您却不解释您是如何提交的你的表格.

I don't see how you want $_POST to return any value or even want us to understand your problem while you don't explain how you're submitting your form.

您必须为您的案例使用两种解决方案之一:

you have to use one of 2 solutions for your case:

• 使用 AJAX

• To use AJAX

使用具有不同名称的提交按钮的两种不同形式.然后使用 isset($_POST['name_of_button']) 检查发布时提交的表单.

To use two different forms with a submit buttons that have different names. Then use isset($_POST['name_of_button']) to check which form is submitted upon posting.

谢谢

这篇关于动态填充的下拉菜单；$_POST 返回错误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！