我似乎无法获得以下代码来制作包含来自 mysql 数据库的数据的下拉菜单。 “包括(‘connect.php’)；”连接到 mysql 数据库，我知道它在不同的页面上工作。有什么建议么？
下面是完整的代码。 列出客户
<BODY>
<H1>Find Customer's Albums Page</H1>
From a dropdown list of customers, a user should be able to pick a customer and see a list of albums (all fields in the CD table) purchased by that customer.
<HR>
<FORM ACTION="listCustomer.php" METHOD="POST"/>
Customer:
<select name="mydropdownCust">
<option value="101">101</option>
<option value="102">102</option>
<option value="103">103</option>
<option value="104">104</option>
<option value="105">105</option>
<option value="106">106</option>
<option value="107">107</option>
<option value="108">108</option>
<option value="109">109</option>
<option value="110">110</option>
</select>
<BR>
<?php
include('connect.php');
$query = "SELECT Cnum, CName FROM Customer";
$result = mysql_query ($query);
echo "<select name=dropdown value=''>Dropdown</option>";
while($r = mysql_fetch_array($result))
{
echo "<option value=$r["Cnum"]>$r["CName"]</option>";
}
echo "</select>";
?>
<BR>
<INPUT TYPE="SUBMIT" Value="Submit"/>
</FORM>
<FORM ACTION="listMenu.html" METHOD="POST"/>
<INPUT TYPE="SUBMIT" Value="Main Menu"/>
</FORM>
</BODY>
</HTML>
最佳答案
<?php
include('connect.php');
$query = "SELECT Cnum, CName FROM Customer";
$result = mysql_query ($query);
echo "<select name='dropdown' value=''><option>Dropdown</option>";
while($r = mysql_fetch_array($result)) {
echo "<option value=".$r['Cnum'].">".$r['CName']."</option>";
}
echo "</select>";
?>
从外观上看，您缺少一个开始选项标签，因此它只是将“Dropdown”输出为一行文本。
编辑
为了完全透明，因为我没有connect.php，所以我不得不添加自己的数据库连接。我的整个页面看起来是这样的:
<?
//Adding to display errors.
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<HTML>
<HEAD>
</HEAD>
<BODY>
<H1>Find Customer's Albums Page</H1>
From a dropdown list of customers, a user should be able to pick a customer and see a list of albums (all fields in the CD table) purchased by that customer.
<HR>
<FORM ACTION="listCustomer.php" METHOD="POST"/>
Customer:
<select name="mydropdownCust">
<option value="101">101</option>
<option value="102">102</option>
<option value="103">103</option>
<option value="104">104</option>
<option value="105">105</option>
<option value="106">106</option>
<option value="107">107</option>
<option value="108">108</option>
<option value="109">109</option>
<option value="110">110</option>
</select>
<BR />
<?php
// BEGIN ADDED CONNECTION HACKY GARBAGE
$con=mysql_connect("localhost","root","root");
// Check connection
if (mysqli_connect_errno($con)) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$selected = mysql_select_db("sample",$con)
or die("Could not select examples");
// END ADDED CONNECTION HACKY GARBAGE
$query = "SELECT Cnum, CName FROM Customer";
$result = mysql_query ($query);
echo "<select name='dropdown' value=''><option>Dropdown</option>";
while($r = mysql_fetch_array($result)) {
echo "<option value=".$r['Cnum'].">".$r['CName']."</option>";
}
echo "</select>";
?>
<BR />
<INPUT TYPE="SUBMIT" Value="Submit"/>
</FORM>
<FORM ACTION="listMenu.html" METHOD="POST"/>
<INPUT TYPE="SUBMIT" Value="Main Menu"/>
</FORM>
</BODY>
</HTML>
关于PHP-从数据库中获取并存储在下拉菜单html中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16701692/