[英]What am I doing wrong in the code?I am making a jquery/json call to call the data from the database
I have pasted the code for index.php and jsonPhp.php.I am new to JSON and learning json with ajax.Here,I am trying to get the data from the server using json.When I click on the link SERVER DATA, The data from the server must appear without re-loading the page using jQuery/json.I have written the code but I dont get it working.Please help. 我已经粘贴了index.php和jsonPhp.php的代码。我是JSON的新手,并且使用ajax学习json。在这里,我正尝试使用json从服务器获取数据。当我单击链接SERVER DATA,服务器中的数据必须在不使用jQuery / json重新加载页面的情况下出现。我已经编写了代码,但是我没有使它工作。请帮助。 Thanks.
谢谢。
<head>
<title>JSON WITH PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.js"
type="text/javascript"></script>
<script type="text/javascript">
< ![CDATA[
$(function () {
$('#click').click(function () {
$.post('jsonPhp.php', function (data) {
//$("#content").html(data)+'<br>';
var pushedData = jQuery.parseJSON(data);
var htmlData = "";
//loop through using jQuery
$.each(pushedData, function (i, field) {
htmlData = htmlData + '-' + field.id + ',' + field.place + ',' + field.description + '<br>';
});
$('#content').html(htmlData);
});
});
});]] >
</script>
</head>
<body>Click on the link below to get the data from the Server Dynamicallly!
<br
/>
<a href="#" id="click">Server Data</a>
<div id="content"></div>
</body>
<?php
$db = mysql_connect("localhost","root","")or die(mysql_error());
mysql_select_db("places",$db) or die(mysql_error());
if(isset($_POST['place']))
$place=$_POST['place'];
if(isset($_POST['description']))
$description=$_POST['description'];
$myrows = array();
$result = mysql_query("SELECT * FROM search");
while( $row = mysql_fetch_assoc($result) ) {
$myrows[] = $row;
}
echo json_encode($myrows);
?>
Your jQuery post doesn't post the necessary items. 您的jQuery帖子未发布必要的项目。
Be careful using this code. 使用此代码时要小心。
尝试指定随POST请求发送的参数,结果如下:
$.post('jsonPhp.php', { place:'myplace', /* other params */ }, function (data) { ...
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