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[英]how can I fetch data from php to my jquery script in a desired json format?
[英]How can I send JSON data from a PHP script to be used by jQuery?
我对某些JSON数据有疑问。 我不知道如何获取PHP生成的一些数据并将其转换为可在jQuery脚本中使用的数据。 我需要的功能是:我需要能够单击页面上的图像,并且根据所选元素,我需要显示数据库的结果。
这是我得到的HTML页面:
<html>
<head>
<title>pippo</title>
<script><!-- Link to the JS snippet below --></script>
</head>
<body>
Contact List:
<ul>
<li><a href="#">
<img src="contacts/pippo.png" onclick="javascript:change('pippo')"/>pippo
</a></li>
<li><a href="#">
<img src="contacts/pluto.png" onclick="javascript:change('pluto')"/>pluto
</a></li>
<li><a href="#">
<img src="contacts/topolino.png" onclick="javascript:change('topolino')"/>topolino
</a></li>
</ul>
</body>
</html>
这是被称为的PHP代码:
<?php
include('../dll/config.php');
$surname = $_POST['surname'];
$result = mysql_query("select * from profile Where surname='$surname'") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$_POST['name'] = ucfirst($row['name']);
$_POST['tel'] = $row['telephone'];
$_POST['companymail'] = $row['companymail'];
$_POST['mail'] = $row['email'];
$_POST['fbid'] = $row['facebook'];
}
?>
这是我正在使用的Ajax JavaScript代码:
<script type="text/javascript">
function change(user) {
$.ajax({
type: "POST",
url: "chgcontact.php",
data: "surname="+user+"&name=&tel=&companymail=&mail=&fbid",
success: function(name,tel,companymail,mail,fbid){
alert(name);
}
});
return "";
}
</script>
有人告诉我,这个JS代码段可以满足我的要求:
$.getJSON('chgcontact.php', function(user) {
var items = [name,surname,tel,companymail,email,facebook];
$.each(user, function(surname) {
items.push('surname="' + user + "'name='" + name + "'telephone='" + telephone + "'companymail='" + companymail + "'mail='" + mail + "'facebook='" + facebook);
});
/*
$('<ul/>', {
'class': 'my-new-list',
html: items.join('')
}).appendTo('body');
*/
});
但是我不清楚-我不知道该如何使用它或在代码中的什么位置添加它。
您将必须在PHP脚本中创建正确的JSON字符串,然后在脚本末尾echo
该字符串。
一个简单的例子:
$person = new stdClass;
$result = mysql_query("select * from profile Where surname='$surname'")
or die(mysql_error());
while ($row = mysql_fetch_array( $result )) {
$person->name = ucfirst($row['name']);
$person->tel = $row['telephone'];
$person->companymail = $row['companymail'];
$person->mail = $row['email'];
$person->fbid = $row['facebook'];
}
echo json_encode($person);
我尝试通过此处经过更正和注释的代码来解释您的代码有几个问题:
<html>
<head><title>pippo</title>
<!-- added link to jQuery library -->
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.4/jquery.min.js"></script>
<!-- javascript can go here -->
<script type="text/javascript">
$.ajax({
type: "POST",
url: "chgcontact.php",
// use javascript object instead of `get` string to represent data
data: {surname:user, name:'', tel:'', companymail:'', mail:'', fbid:''},
success: function(data){
// removed name,tel,companymail,mail,fbid
alert(JSON.parse(data));
}
});
return "";
}
</script>
</head>
<body>
Contact List:
<ul>
<!-- removed `javascript` form onclick handler -->
<li><a href="#"><img src="contacts/pippo.png" onclick="change('pippo')"/>pippo</a></li>
<li><a href="#"><img src="contacts/pluto.png" onclick="change('pluto')"/>pluto</a></li>
<li><a href="#"><img src="contacts/topolino.png" onclick="change('topolino')"/>topolino</a></li>
</ul>
</body>
</html>
<?php
$surname = $_POST['surname'];
$result = mysql_query("select * from profile Where surname='$surname'")
or die(mysql_error());
while ($row = mysql_fetch_array( $result )){
// create data object
$data = new stdClass();
// add values to data object
$data->name = ucfirst($row['name']);
$data->tel = $row['telephone'];
$data->companymail = $row['companymail'];
$data->mail = $row['email'];
$data->fbid = $row['facebook'];
// send header to ensure correct mime type
header("content-type: text/json");
// echo the json encoded data
echo json_encode($data);
}
?>
所有代码都未经测试,但是您应该能够看到我在每个步骤中所做的事情。 祝好运。
并扩展Brian Driscoll的答案。 您将需要使用user.name
格式来从返回的$.getJSON("blah", function(user){});
访问name
字段$.getJSON("blah", function(user){});
所以...
items.push('surname="'+user+"'name='"+user.name+"'telephone='"+user.telephone+"'companymail='"+user.companymail+"'email='"+user.email+"'facebook='"+user.facebook+);
以您创建的这种格式,它只会插入看起来很长的丑陋字符串,因此您可能要花一些时间使它看起来更好。 祝好运!
POST到PHP页面的JSON通常不在$_POST
变量中,而是在$HTTP_RAW_POST_DATA
。
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