PHP- 从数据库中获取并存储在下拉菜单 html 中
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PHP- Fetch from database and store in drop down menu html
提问by Sean Cleveland
I can't seem to get the following code to make a dropdown menu that contains data from a mysql database. The "include('connect.php');" connects to the mysql database and I know it works on separate pages. Any suggestions?
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。“包含（'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？
Below is the entire code. listCustomer
下面是整个代码。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>
回答by stslavik
<?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>";
?>
From the looks of things, you're missing an opening option tag, so it's just outputting "Dropdown" as a line of text.
从外观上看，您缺少一个开始选项标签，因此它只是将“下拉列表”作为一行文本输出。
Edit
编辑
Just to be completely transparent, because I did not have connect.php, I had to add my own DB connections. My whole page looked thusly:
为了完全透明，因为我没有connect.php，我不得不添加自己的数据库连接。我的整个页面看起来如此：
<?
//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>
回答by mspir
First off, you are missing an option opening tag, as correctly mentioned by stslavik. But this is not causing the issue here as it seems (it's auto-corrected by the browser - in my tests atleast).
首先，您缺少一个选项开始标签，正如 stslavik 正确提到的那样。但这并没有像看起来那样导致问题（它是由浏览器自动更正的 - 至少在我的测试中）。
Secondly, this wont work (problem causer):
其次，这行不通（问题原因）：
echo "<option value=$r["Cnum"]>$r["CName"]</option>";
You should use
你应该使用
echo "<option value=".$r["Cnum"].">".$r["CName"]."</option>";
or, as I always prefer single quotes to enclose echo or print output strings:
或者，因为我总是喜欢用单引号将 echo 或打印输出字符串括起来：
echo '<option value='.$r['Cnum'].'>'.$r['CName'].'</option>';
Third alternative (complex syntax: What does ${ } mean in PHP syntax?)
第三种选择（复杂的语法：${} 在 PHP 语法中是什么意思？）
echo "<option value={$r["Cnum"]}>{$r["CName"]}</option>";
回答by Leandro Villagran
assuming you get data from the database try this
假设你从数据库中获取数据试试这个
echo "<option value={$r['Cnum']}>{$r['CName']}</option>";
回答by RonEskinder
try,
尝试，
echo "<option value=' . $r['Cnum'] . '>' . $r['CName'] . '</option>";
instead of
代替
echo "<option value=$r[Cnum]>$r[CName]</option>";
