我有两个文件 SearchHotelUsingXMLhttp.html ,其他文件是 getHotelDetails.php ,它位于 webservices 这是一个目录。 现在对于给定的酒店搜索我创建了一个按钮搜索,从搜索按钮我调用java脚本函数searchHotel(str)。从这个javacript函数我使用xmlHttp请求调用getHotelDetails.php,但是我收到以下消息。
注意:第2行的C:\ xampp \ htdocs \ Experiements \ webservices \ getHotelDetails.php中的未定义索引:q
城市名称
SearchHotelUsingXMLhttp.html
<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
<head>
<script type="text/javascript">
function searchHotel(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("POST","webservices/getHotelDetails.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<h1>Hotels</h1>
<form>
City: <select name="hotel_city" >
<option value="Bangalore"> Bangalore</option>
<option value="Chennai"> Chennai</option>
</select>
<br/><br/>
<input type="button" id="myBtn" value="Search Hotel" onclick="searchHotel(this.value)">
<!--
<input type="button" id="myBtn" value="Search Hotel" onclick="searchHotel(this.value)">
-->
</form>
<br>
<div id="txtHint"><b>Hotel info will be listed here.</b></div>
</body>
</html>
getHotelDetails.php
<?php
$q = $_POST['q'];
$con = mysqli_connect('localhost','root','3456','my_db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM hotel WHERE city = '$q'";
$result = mysqli_query($con,$sql);
echo "<table border='1'>
<tr>
<th>city</th>
<th>name</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['city'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
答案 0 :(得分:2)
您将参数作为GET参数发送。你应该从$ _GET
中获得PhP例如:
JavaScript的:
xmlhttp.open("GET","webservices/getHotelDetails.php?q="+str,true);
xmlhttp.send();
PHP的:
<?php
$q = $_GET['q'];
如果你想使用POST,它会有点复杂。你可以在这里找到这个例子: Send POST data using XMLHttpRequest