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我在將從動態下拉列表中選擇的變量傳遞給PHP文件時遇到問題。我希望PHP選擇與變量匹配的數據庫表中的所有行。這裏是到目前爲止的代碼：將AJAX變量傳遞給PHP並在從動態下拉列表中選擇後顯示MySQL結果
select.php
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("select#type").attr("disabled","disabled");
$("select#category").change(function(){
$("select#type").attr("disabled","disabled");
$("select#type").html("<option>wait...</option>"); var id = $("select#category option:selected").attr('value');
$.post("select_type.php", {id:id}, function(data){
$("select#type").removeAttr("disabled");
$("select#type").html(data);
});
});
$("form#select_form").submit(function(){
var cat = $("select#category option:selected").attr('value');
var type = $("select#type option:selected").attr('value');
if(cat>0 && type>0)
{
var result = $("select#type option:selected").html();
$("#result").html('your choice: '+result);
$.ajax({
type: 'POST',
url: 'display.php',
data: {'result': myval},
});
}
else
{
$("#result").html("you must choose two options!");
}
return false;
});
});
</script>
</head>
<body>
<?php include "select.class.php"; ?>
<form id="select_form">
Choose a category:<br />
<select id="category">
<?php echo $opt->ShowCategory(); ?>
</select>
<br /><br />
Choose a type:<br />
<select id="type">
<option value="0">choose...</option>
</select>
<br /><br />
<input type="submit" value="confirm" />
</form>
<div id="result"></div>
<?php include "display.php"; ?>
<div id="result2"></div>
</body>
</html>
select.class.php
<?php
class SelectList
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowCategory()
{
$sql = "SELECT * FROM profession";
$res = mysql_query($sql,$this->conn);
$category = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$category .= '<option value="' . $row['id_cat'] . '">' . $row['prof_name'] . '</option>';
}
return $category;
}
public function ShowType()
{
$sql = "SELECT * FROM specialties WHERE id_cat=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$type = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$type .= '<option value="' . $row['id_type'] . '">' . $row['sp_name'] . '</option>';
}
return $type;
}
}
$opt = new SelectList();
?>
這裏就是我想通過變量Display.php的至。這個文件將從db中選擇條件，然後在select.php中打印結果。
<?php
class DisplayResults
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowResults()
{
$myval = $_POST['result'];
$sql = "SELECT * FROM specialities WHERE 'myval'=sp_name";
$res = mysql_query($sql,$this->conn);
echo "<table border='1'>";
echo "<tr><th>id</th><th>Code</th></tr>";
while($row = mysql_fetch_array($res))
{
while($row = mysql_fetch_array($result)){
echo "<tr><td>";
echo $row['sp_name'];
echo "</td><td>";
echo $row['sp_code'];
echo "</td></tr>";
}
echo "</table>";
//}
}
return $category;
}
}
$res = new DisplayResults();
?>
我真的很感謝任何幫助。請讓我知道，如果我可以提供更多的細節。
鏈接到分貝圖：http://imgur.com/YZ0SuVw
第一下拉從行業表繪製，第二從特色表。我想要做的是顯示作業表中與下拉框中選擇的專業相匹配的所有行。這將要求將下拉列表中的變量（結果）的結果轉換爲作業表中的spec_code。不確定如何做到這一點。謝謝！