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Sample code sharing for operating JSON with PHP+JQUERY

黄舟Original: 2017-03-24 09:20:241825browse

This article mainly introduces the PHP+JQUERY method of operating JSON, and analyzes php+jQuery combined with ajax to implement json format data based on specific examples. For related operation skills, friends in need can refer to

This article describes the method of operating JSON with PHP+JQUERY. Share it with everyone for your reference, the details are as follows:

json.html Code:

<html>
<head>
<meta http-equiv="Content-Type" content="text/html;charset=utf-8">
<title>PHP Json传输数据</title>
</head>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript">
$(function(){
$("#submit").click(function(){
var text = $("input").serialize();
$.ajax({
&#39;type&#39;:"POST",
&#39;url&#39;:&#39;json_encode.php&#39;,
&#39;dataType&#39;:&#39;json&#39;,
&#39;data&#39;:text,
success:insertData
});
});
});
function insertData(data){
var str = "姓名="+data.name+"<br/>性别="+data.sex+"<br/>年龄="+data.age;
$("#view").html(str);
}
</script>
<body>
姓名：<input name="name" id="name" type="text" value=""><br/>
性别：<input name="sex" id="sex" type="text" value=""><br/>
年龄：<input name="age" id="age" type="text" value="">
<input type="submit" name="submit" id="submit" value="提交">
<p style="font-size:14px;" id="view">
</p>
</body>
</html>

json_encode.php Code

<?php
header("Content-type:text/html;charset=utf8");
include("Json.php");
$name = $_POST[&#39;name&#39;];
$sex = $_POST[&#39;sex&#39;];
$age = $_POST[&#39;age&#39;];
$json_arg = array(&#39;name&#39;=>$name,&#39;sex&#39;=>$sex,&#39;age&#39;=>$age);
$json = new JSON;
$json_result = $json->encode($json_arg);
echo $json_result;
?>

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