[英]MySQL drop downs
我有一張名為餐廳的表,其中包含餐廳的詳細信息,例如名稱,地址等。
我希望創建一個評論部分,以允許用戶通過表格為每個餐廳評分和評論。 目前,我在窗體中有2個下拉菜單,一個是“縣”,另一個是“餐館名”,我希望用戶選擇一個縣,然后在下一個下拉菜單中僅填充該縣的餐館。
做這個的最好方式是什么?
如果您需要更多信息,請告訴我。 目前,我的兩個下拉列表都填充了mySql表中的數據,但它顯示了當前的所有餐廳。
drop.php
<?php
mysql_connect("localhost", "B00606958", "uHmB4jRw") or die("Error connecting to database: ".mysql_error());
/*
localhost - location of mysql server,
if connection fails it will stop loading the page and display an error
*/
mysql_select_db("b00606958") or die(mysql_error());
/* tutorial_search is the name of database we've created */
?>
<?php
$query_parent = mysql_query("SELECT DISTINCT County FROM restaurants") or die("Query failed: ".mysql_error());
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Dependent DropDown List</title>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#parent_cat").change(function() {
$(this).after('<div id="loader"><img src="images/loading.gif" alt="loading subcategory" /></div>');
$.get('loadsubcat.php?parent_cat=' + $(this).val(), function(data) {
$("#sub_cat").html(data);
$('#loader').slideUp(200, function() {
$(this).remove();
});
});
});
});
</script>
</head>
<body>
<form method="get">
<label for="category">County</label>
<select name="parent_cat" id="parent_cat">
<?php while($row = mysql_fetch_array($query_parent)): ?>
<option value="<?php echo $row['County']; ?>"><?php echo $row['County']; ?></option>
<?php endwhile; ?>
</select>
<br/><br/>
<label>Restaurant</label>
<select name="sub_cat" id="sub_cat"></select>
</form>
</body>
</html>
loadsubcat.php
mysql_connect("localhost", "B00606958", "uHmB4jRw") or die("Error connecting to database: ".mysql_error());
/*
localhost - location of mysql server,
if connection fails it will stop loading the page and display an error
*/
mysql_select_db("b00606958") or die(mysql_error());
/* tutorial_search is the name of database we've created */
$parent_cat = $_GET['parent_cat'];
$query = mysql_query("SELECT RestaurantID FROM restaurants WHERE County = {$parent_cat}");
while($row = mysql_fetch_array($query)) {
echo "<option value='$row[RestaurantID]'>$row[RestaurantName]</option>";
}
?>
表結構:
Tbl名稱=餐廳
1個餐廳名稱varchar(255)
2 AddressLine1 varchar(100)
3 AddressLine2 varchar(100)
4鎮varchar(50)
5縣瓦爾查爾(50)
6郵政編碼varchar(7)
7電話varchar(11)
8電子郵件varchar(50)
9網站varchar(100)
10 NoOfDishes int(255)
11 RestaurantID bigint(20)
12級整數(5)
您有兩種方法可以做到這一點:
1)不要在每個頁面中編寫connection code
。 寫一頁並在需要的每一頁中包含。 我創建了一個頁面,即dbConnection.php頁面。
使用此代碼。 而且,如果出現任何錯誤。 隨便問。
dbConnection.php
<?php
$con = mysql_connect("localhost", "B00606958", "uHmB4jRw") or die("Error connecting to database: ".mysql_error());
mysql_select_db("b00606958",$con) or die(mysql_error());
?>
drop.php
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Dependent DropDown List</title>
</head>
<body>
<?php include('dbConnection.php');?> //connection here.
<form method="get">
<label for="category">County</label>
<select name="parent_cat" id="parent_cat">
<?php
$query_parent = mysql_query("SELECT DISTINCT County FROM restaurants") or die("Query failed: ".mysql_error());
while($row = mysql_fetch_array($query_parent)): ?>
<option value="<?php echo $row['County']; ?>"><?php echo $row['County']; ?></option>
<?php endwhile; ?>
</select>
<br/><br/>
<label>Restaurant</label>
<div class="RestaurantDiv"></div>
</form>
</body>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#parent_cat").change(function() {
$(this).after('<div id="loader"><img src="images/loading.gif" alt="loading subcategory" /></div>');
var county = $(this).val();
$.ajax({url:"loadsubcat.php?parent_cat="+county,cache:false,success:function(result){
$(".RestaurantDiv").html(data);
$('#loader').slideUp(200, function() {
$(this).remove();
});
}});
});
});
</script>
</html>
loadsubcat.php
<?php
include('dbConnection.php');
$parent_cat = $_GET['parent_cat'];
$query = mysql_query("SELECT * FROM restaurants WHERE County = {$parent_cat}");
?>
<select name="sub_cat" id="sub_cat"></select>
<?
while($row = mysql_fetch_array($query)) {?>
<option value="<?echo $row['RestaurantID'];?>"><?php echo $row['RestaurantName']?></option>
<?}?>
</select>
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