抓住两个价值而不是一个 [英] Grabbing Two Values Instead Of One
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问题描述
然而,到目前为止,我的脚本运行良好;我发现需要传递两个值。不只是一个。现在我从tblLocations传递RestID。我还需要来自同一个表的CityID。如何传递这两个值?
<?php require(' config.php'); ?>
// 这是基于地点,城市,区域
$( document )。ready( function ()
{
$( 。Doggie)。change(函数()
{
var LocationString = ' Lid =' + $( this )。val();
$ .ajax ({
类型: POST,
url: ajax_city.php,
data:LocationString,
cache: false ,
成功: function (html){
$(跨度> .Kitty)。html(html);
}
});
});
$(' .Kitty')。live( 更改, function (){
var LocationString = ' Lid =' + $( this )。val();
$ .ajax({
type: POST,
url: ajax_area.php,
data:LocationString,
cache: false ,
成功: function (html){
$( .Pig)。html(html);
}
});
});
});
< / script>
< / head >
< body>
< div id = frame1>
< label>地点:< / label >
< select name = Doggie class = Doggie id = Doggie>
<选择选项= 选中> - 选择地点 - < / 选项 >
<?php
$ sql = mysql_query( SELECT tblLocations.RestID as Lid,tblRestaurants .RestName as name
FROM tblRestaurants INNER JOIN tblLocations ON tblRestaurants.RestID = tblLocations.RestID
GROUP BY tblLocations.RestID,tblRestaurants.RestName
ORDER BY tblRestaurants.RestName ASC) ;
while ($ row = mysql_fetch_array($ sql))
{
echo ' < option value ='。$ row [' Lid']。' >'。$ row [< span class =code-string>' name']。' < /选项>'跨度>;
}?>
< / 选择 >
< label>城市:< / label >
< select name = Kitty class = Kitty id = < span class =code-string> Kitty>
<选择选项= 选择> - 选择城市 - < / 选项 >
< / 选择 >
< label>面积:: < / label >
< select name = Pig class = 猪 id = Pig>
<选择选项= 选中> - 选择区域 - < / 选项 >
< / 选择 >
< / div >
< / body >
< / html >
其中一个PHP文件
<?php
require(' config。 PHP跨度>);
if($ _ POST [' Lid'])
{
$ Lid = $ _ POST [' Lid'];
$ sql = mysql_query( SELECT tblLocations.RestID为LID,tblAreas.AreaName为名称
FROM tblLocations INNER JOIN tblAreas ON tblLocations.AreaID = tblAreas.AreaID
WHERE tblLocations.RestID = $ Lid
GROUP BY tblLocations.RestID,tblAreas.AreaName
ORDER BY tblAreas.AreaName ASC跨度>);
echo ' <选择的选项=已选择> - 选择区域 - < / option>';
while($ row = mysql_fetch_array($ sql))
{
echo ' < option value ='。$ row [' Lid']。' >'。$ row [< span class =code-string>' name']。' < /选项>'跨度>;
}
}
?>
解决方案
( document )。ready( function ()
{
( 。Doggie)。change(< span class =code-keyword> function ()
{
var LocationString = ' Lid =' +
( this 跨度>)VAL();
I have this script that works very well so far, however; its come to light that I need two values passed along. Not just one. Right Now I am passing the RestID from the tblLocations. I also need the CityID from the same table. How can I pass the two values along?
<?php require('config.php'); ?>
//This is based on Place, City, Area
$(document).ready(function()
{
$(".Doggie").change(function()
{
var LocationString = 'Lid='+ $(this).val();
$.ajax({
type: "POST",
url: "ajax_city.php",
data: LocationString,
cache: false,
success: function (html) {
$(".Kitty").html(html);
}
});
});
$('.Kitty').live("change",function(){
var LocationString = 'Lid='+ $(this).val();
$.ajax({
type: "POST",
url: "ajax_area.php",
data: LocationString,
cache: false,
success: function (html) {
$(".Pig").html(html);
}
});
});
});
</script>
</head>
<body>
<div id="frame1">
<label>Place :</label>
<select name="Doggie" class="Doggie" id="Doggie">
<option selected="selected">--Select Place--</option>
<?php
$sql = mysql_query("SELECT tblLocations.RestID as Lid, tblRestaurants.RestName as name
FROM tblRestaurants INNER JOIN tblLocations ON tblRestaurants.RestID = tblLocations.RestID
GROUP BY tblLocations.RestID, tblRestaurants.RestName
ORDER BY tblRestaurants.RestName ASC");
while($row=mysql_fetch_array($sql))
{
echo '<option value="'.$row['Lid'].'">'.$row['name'].'</option>';
} ?>
</select>
<label>City :</label>
<select name="Kitty" class="Kitty" id="Kitty">
<option selected="selected">--Select City--</option>
</select>
<label>Area: :</label>
<select name="Pig" class="Pig" id="Pig">
<option selected="selected">--Select Area--</option>
</select>
</div>
</body>
</html>
And one of the PHP files
<?php
require('config.php');
if($_POST['Lid'])
{
$Lid=$_POST['Lid'];
$sql=mysql_query("SELECT tblLocations.RestID as LID, tblAreas.AreaName as name
FROM tblLocations INNER JOIN tblAreas ON tblLocations.AreaID = tblAreas.AreaID
WHERE tblLocations.RestID = $Lid
GROUP BY tblLocations.RestID, tblAreas.AreaName
ORDER BY tblAreas.AreaName ASC");
echo '<option selected="selected">--Select Area--</option>';
while($row=mysql_fetch_array($sql))
{
echo '<option value="'.$row['Lid'].'">'.$row['name'].'</option>';
}
}
?>
解决方案
(document).ready(function() {
(".Doggie").change(function() { var LocationString = 'Lid='+
(this).val();
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