我正在尝试构建我的第一个jQuery Web应用程序,但是我碰到了一个障碍,似乎无法弄清楚这一点。将php链接到jquery
我有一个PHP页面和一个HTML页面。 HTML页面有一个带有三下拉列表的表单。 PHP页面连接到数据库,但我不确定如何从php页面传递查询结果来填充html/javascript页面上的下拉列表。
这是我的代码到目前为止。
HTML:
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#selector").submit(function() {
$.ajax({
type: "GET",
url: "DBConnect.php",
success: function(msg){
alert(msg);
}
});
var select_car_make = $('#select_car_make').attr('value');
var select_car_model = $('#select_car_model').attr('value');
var select_car_year = $('#select_car_year').attr('value');
alert("submitted");
}); //end submit
});
</script>
<h1 style="alignment-adjust:center">Car information:</h1>
<hr />
<div id="results">
<form action="get.php" id="selector" method="get" name="sizer">
<table width="451" height="70" border="0">
<th width="145" height="66" scope="row"><label for="select_car_make"></label>
<div align="center">
<select name="select_car_make" id="select_car_make" onchange="">
</select>
</div></th>
<td width="144"><label for="select_car_model"></label>
<div align="center">
<select name="select_car_model" id="select_car_model">
</select>
</div></td>
<td width="140"><label for="select_car_year"></label>
<div align="center">
<select name="select_car_year" id="select_car_year">
</select>
</div></td>
</tr>
</table>
<input name="completed" type="submit" value="submit" />
</form>
这里是PHP页面:
<?php
$DBConnect = mysqli_connect("localhost", "root", "password", "testing")
or die("<p>Unable to select the database.</p>" . "<p> Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect)) . "<p>";
echo "<p>Successfully opened the database.</p>";
$SQLString1 = " SELECT car_make FROM cars";
$QueryResult = mysqli_query($DBConnect, $SQLString1)
您可以将HTML页面保存为PHP页面,然后将其保存在HTML标签标签中,并使用PHP来填充数据库中的数据。 – NightHawk 2011-06-03 18:52:49