PHP用于解析JSON并添加到数据库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

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。

The Question is:

4 个解决方案

#1

更多相关文章

随机推荐