[英]PHP, MySql drop down list only showing one record per selection
我是新手,我正在尝试创建一个动态下拉列表以从数据库检索数据。 问题是,第一个下拉列表中的每个选择仅给我一个下拉列表项(第二个列表)。 有人请帮忙。 这是代码。
<?php
require_once("dbcontroller.php");
$query ="SELECT * FROM campus";
?>
<html>
<head>
<TITLE>Campus and Faculty Select</TITLE>
<head>
<style>
body{width:610px;}
.frmDronpDown {border: 1px solid #F0F0F0;background-color:#C8EEFD;margin: 2px 0px;padding:40px;}
.demoInputBox {padding: 10px;border: #F0F0F0 1px solid;border-radius: 4px;background-color: #FFF;width: 50%;}
.row{padding-bottom:15px;}
</style>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
function getcampus_id(val) {
$.ajax({
type: "POST",
url: "get_faculty.php",
data:'campus_id='+val,
success: function(data){
$("#faculty-list").html(data);
}
});
}
function selectcampus_id(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
</script>
</head>
<body>
<div class="frmDronpDown">
<div class="row">
<label>Country:</label><br/>
<select name="campus" id="campus-list" class="demoInputBox" onChange="getcampus_id(this.value);">
<option value="">Select Country</option>
<?php
$query ="SELECT * FROM campus";
$results = mysqli_query($con, $query);
//loop
foreach ($results as $campus){
?>
<option value="<?php echo $campus["campus_id"]; ?>"> <?php echo $campus["name"]; ?></option>
<?php
}
?>
</select>
</div>
<div class="row">
<label>State:</label><br/>
<select name="faculty" id="faculty-list" class="demoInputBox">
<option value="">Select State</option>
</select>
</div>
</div>
</body>
</html>
get_faculty.php
<?php
require_once("dbcontroller.php");
if(!empty($_POST["campus_id"])) {
$campus_id = $_POST["campus_id"];
$query ="SELECT * FROM faculty WHERE faculty_id = $campus_id";
$results = mysqli_query($con, $query);
?>
<option value="">Select Campus</option>
<?php
foreach($results as $faculty) {
?>
<option value="<?php echo $faculty["faculty_id"]; ?>"><?php echo $faculty["faculty_name"]; ?></option>
<?php
}
}
?>
和dbcontroller.php
<?php
$username = "root";
$password = "";
$host = "localhost";
$dbname = "registration";
$con = mysqli_connect($host, $username, $password) or die("Could not Connect");
mysqli_select_db($con, $dbname);
?>
使用while($row = mysqli_fetch_assoc($result)
从数据库获取所有记录。
<?php
require_once("dbcontroller.php");
if(!empty($_POST["campus_id"])) {
$campus_id = $_POST["campus_id"];
$query ="SELECT * FROM faculty WHERE faculty_id = $campus_id";
$results = mysqli_query($con, $query);
?>
<option value="">Select Campus</option>
<?php
while($row = mysql_fetch_assoc($results)) {
?>
<option value="<?php echo $row ["faculty_id"]; ?>"><?php echo $row ["faculty_name"]; ?></option>
<?php
}
}
?>
foreach
处理数组! $result
变量不是其mysqli_result
对象的数组
因此,您需要get
每个结果行,然后使用该行数据。 为此使用while
循环。
当您像这样编码PHP和HTML以使用
while () :
. . .
endwhile;
语法,因为这样可以更轻松地查看循环等从何处开始和结束。 否则,您可能会在这种代码中迷失方向。
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
require_once("dbcontroller.php");
if(!empty($_POST["campus_id"])) {
$campus_id = $_POST["campus_id"];
$query ="SELECT * FROM faculty WHERE faculty_id = $campus_id";
$result = mysqli_query($con, $query);
if ( $result === false ) {
echo mysqli_error($con);
exit;
}
echo '<option value="">Select Campus</option>';
while ( $faculty = mysqli_fetch_assoc($result) ) :
echo '<option value="' . $faculty['faculty_id'] . '">';
echo $faculty['faculty_name'];
echo '</option>';
endwhile;
}
?>
当然,我假设您未显示的代码在此代码前后都有必需的
<select....>
和</select>
标记
在dbcontroller
不到任何错误,但我会做
<?php
$username = "root";
$password = "";
$host = "localhost";
$dbname = "registration";
$link = mysqli_connect($host, $username, $password, $dbname);
if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') '
. mysqli_connect_error());
}
作为mysqli_select_db();
当第一个数据库成功连接到第二个数据库时,实际上是切换到第二个数据库
问题在这里,
if(!empty($_POST["campus_id"])) {
$campus_id = $_POST["campus_id"];
$query ="SELECT * FROM faculty WHERE faculty_id = $campus_id";
$result = mysqli_query($con, $query);
if ( $result === false ) {
echo mysqli_error($con);
exit;
}
该
$query ="SELECT * FROM faculty WHERE faculty_id = $campus_id";
应该
$query ="SELECT * FROM faculty WHERE campus_id = $campus_id";
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