PHP用于解析JSON并添加到数据库mysql
[英]PHP for parsing JSON and adding to database mysql
I am making a web app (android) with phonegap and jquery mobile.
我正在制作一个带有phonegap和jquery mobile的网络应用程序（android）。
I am trying to send three fields from an html form as json, to a php page which will decode the json string/object (im new to json, ajax, jquery) and add the three fields as a mysql query to a database on my localhost.
我试图将一个html表单中的三个字段作为json发送到一个php页面，该页面将解码json字符串/对象（即时为json，ajax，jquery）并将三个字段作为mysql查询添加到我的数据库中本地主机。
My html page looks like this:
我的html页面如下所示：
<script type="text/javascript">
$(document).ready(function(){
$('#btn').bind('click', addvalues);
});
function addvalues(){
$.ajax({
url: "connection.php",
type: "POST",
data: "id=" + $("#id").val()+"&name=" + $("#name").val() + "&Address=" + $("#Address").val(),
datatype:"json",
success: function(status)
{
if(status.success == false)
{
//alert a failure message
}
else {
//alert a success message
}
}
});
}
</script>
</head>
<body>
<div data-role="header">
<h1>My header text</h1>
</div><!-- /header -->
<div data-role="content">
<form id="target" method="post">
<label for="id">
<input type="text" id="id" placeholder="ID">
</label>
<label for="name">
<input type="text" id="name" placeholder="Name">
</label>
<label for="Address">
<input type="text" id="Address" placeholder="Address">
</label>
<input type="submit" id "btn" value="Add record" data-icon="star" data-theme="e">
</form>
</div>
</body>
The Question is:
How exactly do i extract the three fields (ID, name, Address) from the string that i have sent to my php file (connection.php)? connection.php is hosted by my local server.
我究竟如何从我发送到我的php文件（connection.php）的字符串中提取三个字段（ID，名称，地址）？ connection.php由我的本地服务器托管。
I am familiar with making connections to database, as also with adding queries to mysql. I only need help with extracting the three fields, as i am new to ajax and jquery and json.
我熟悉与数据库的连接，以及向mysql添加查询。我只需要提取三个字段的帮助，因为我是ajax和jquery和json的新手。
As of now, this is ALL i have done in connection.php:
截至目前，这是我在connection.php中做的所有事情：
<?php
$server = "localhost";
$username = "root";
$password = "";
$database = "jqueryex";
$con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
mysql_select_db($database, $con);
//I do not know how to use the json_decode function here
//And this is how, i suppose, we will add the values to my table 'sample'
$sql = "INSERT INTO sample (id, name, Address) ";
$sql .= "VALUES ($id, '$name', '$Address')";
if (!mysql_query($sql, $con)) {
die('Error: ' . mysql_error());
} else {
echo "Comment added";
}
mysql_close($con);
?>
Please add the relevant code in this file and help me out. I will be highly obliged. :)
请在此文件中添加相关代码并帮助我。我将非常感激。 :)
4 个解决方案
#1
3
what you want to do this this
你想做什么呢
$(document).ready(function () {
$('#btn').on('click', function () {
$.ajax({
url: "connection.php",
type: "POST",
data: {
id: $('#id').val(),
name: $('#name').val(),
Address: $('#Address').val()
},
datatype: "json",
success: function (status) {
if (status.success == false) {
//alert a failure message
} else {
//alert a success message
}
}
});
});
});
then in your php do this
然后在你的PHP做到这一点
//set variables from json data
$data = json_decode($_POST); // Or extract(json_decode($_POST) then use $id without having to set it below.
$id = $data['id'];
$name = $data['name'];
$Address = $data['Address'];
//And this is how, i suppose, we will add the values to my table 'sample'
$sql = "INSERT INTO sample (id, name, Address) ";
$sql .= "VALUES ($id, '$name', '$Address')";
if (!mysql_query($sql, $con)) {
die('Error: ' . mysql_error());
} else {
echo "Comment added";
}
mysql_close($con);
be sure you sanitize those inputs before you insert them though.
确保在插入之前清理这些输入。
#2
1
use:
使用：
$id = json_decode($_POST['id']);
$name = json_decode($_POST['name']);
$Address = json_decode($_POST['Address']);
$sql .= "VALUES ($id, '$name', '$Address')";
in place of :
取代：
$sql .= "VALUES ($id, '$name', '$Address')";
#3
0
use $id_json_encoded = json_encode($_POST['id']); to encode the posts of the form then use jquery script like this
使用$ id_json_encoded = json_encode（$ _ POST ['id']）;要对表单的帖子进行编码，然后使用这样的jquery脚本
<script type="text/javascript">
$(document).ready(function() {
$（document）.ready（function（）{
$("#audit").click(function(){
$.post("/audit_report/"+$('#branch').val()+"/"+$('#ttype').val()+"/"+$('#from').val()+"/"+$('#to').val(),
{
targetDiv:"divswitcher"
}
,
function(data){
$("#imagediv").html('<img src="/public/images/spinner_white.gif"> loading...');
//alert(data);
$("#bodydata").html(data);
$("#imagediv").html('');
// $("#divswitcher").html(data);
});
});
});
#4
0
this is the complete code that encodes form data and send it to the server using ajax
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<script type="text/javascript" charset="utf-8" src="jquery.js"></script>
<script type="text/javascript" language="javascript" src="jquery-ui-1.8.16.custom.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$("#btn").click(function(){
var encoded = json_encode("/audit_report/"+$('#id').val()+"/"+$('#name').val()+"/"+$('#Address').val());
$.post(encoded,
{
targetDiv:"divswitcher"
}
,
function(data){
$("#imagediv").html('<img src="spinner_white.gif"> loading...');
//alert(data);
$("#bodydata").html(data);
$("#imagediv").html('');
// $("#divswitcher").html(data);
});
});
});
</script>
</head>
<body>
<div data-role="header">
<h1>My header text</h1>
</div><!-- /header -->
<div data-role="content">
<form id="target" method="post">
<label for="id">
<input type="text" id="id" name="id" >
</label>
<label for="name">
<input type="text" id="name" name="name" >
</label>
<label for="Address">
<input type="text" id="Address" name="Address" >
</label>
<input type="button" id="btn" name="btn" value="Add record" />
</form>
</div>
</body>
</html>
<div id="bodydata" class="class">
</div>
注意！
本站翻译的文章，版权归属于本站，未经许可禁止转摘，转摘请注明本文地址：https://www.itdaan.com/blog/2012/07/06/6f2d3d82102a58102bf0e3d604d5b0ac.html。