将JSON数据发送到PHP并使用它

问题描述：

I have an issue with sending json data with ajax to a php file and then use it to upload to the server.

If I'm making echo of the variable $email_info->Name the returned data value is empty. I've tried json_decode but it doesn't do it.

This is the code I have,

Jquery:

$( document ).on('click', '#send_touch', function(){

    new_email = [];

    new_email.push({
    Name: $('#name').val(),
    Phone: $('#phone').val(),
    Email: $('#email').val(),
    Interested_in: $('#interested_in').val(),
    User_id: $('#email_user_id').val()
    });

    new_email = JSON.stringify({Email: new_email}, null, "\t");

        $.ajax({
            url: "core.php",
            type: "post",
            data: { NewMail: new_email
                  },
            success: function(data){  
                alert(data)          
            },
            error: function(){
            }   
     });    

});

PHP:

if ($_POST['NewMail']) {

 $timeStamp = time();

 $new_email = json_decode($_POST['NewMail'], true);

 $email_info = $new_email->Email[0];

 // New Email
 mysql_query("INSERT INTO " . $dbPrefix . "touches (date, user_id, name, phone, email, interested_in, seen, status) VALUES ('".safeSQL($timeStamp)."', '".safeSQL($email_info->User_id)."', '".safeSQL($email_info->Name)."','".safeSQL($email_info->Phone)."', '".safeSQL($email_info->Email)."', '".safeSQL($email_info->Interested_in)."', '0', '1')") or die("Query failed: " . mysql_error());

 echo $email_info->Name;

}

If I make an echo on the PHP side for $_POST['NewMail'] I get returned this:

{
\"Email\": [
    {
        \"Name\": \"John Doe\",
        \"Phone\": \"1234567\",
        \"Email\": \"john@doe.com\",
        \"Interested_in\": \"Text here..\",
        \"User_id\": \"1\"
    }
]
}

How can I fix this?

我有一个问题，即用ajax将json数据发送到php文件，然后用它上传到服务器。 p>

如果我正在回显变量$ email_info-＆gt; Name code>，则返回的数据值为空。我尝试过json_decode但它没有' 这是我的代码。 p>

Jquery： p>


  $  （文件）.on（'click'，'＃send_touch'，function（）{
 
 new_email = []; 
 
 new_email.push（{
 Name：$（'＃name'）。val  （），
电话：$（'＃phone'）。val（），
电子邮件：$（'＃email'）。val（），
 Interested_in：$（'＃interested_in'）。val（）  ，
 User_id：$（'＃email_user_id'）。val（）
}）; 
 
 new_email = JSON.stringify（{Email：new_email}，null，“\ t”）; 
 
 $  .ajax（{
 url：“core.php”，
 type：“post”，
 data：{NewMail：new_email 
}，
 success：function（data）{
 alert（data）\  n}，
错误：function（）{
} 
}）; 
 
}）; 
  code>  pre> 
 
  PHP： p> \  n 
 
  if（$ _POST ['NewMail']）{
 
 $ timeStamp = time（）; 
 
 $ new_email  = json_decode（$ _ POST ['NewMail']，true）; 
 
 $ email_info = $ new_email-＆gt;电子邮件[0]; 
 
 //新电子邮件
 mysql_query（“INSERT INTO”。  $ dbPrefix。  “touches（date，user_id，name，phone，email，interested_in，seen，status）VALUES（'”。。gofeSQL（$ timeStamp）。“'，'”。isfeSQL（$ email_info-＆gt; User_id）。“'，'  “.safeSQL（$ email_info-＆gt; Name）。”'，'“。safeSQL（$ email_info-＆gt; Phone）。”'，'“。。gofeSQL（$ email_info-＆gt; Email）。”'，'“。  safeSQL（$ email_info-＆gt; Interested_in）。“'，'0'，'1'）”）或死（“查询失败：”。mysql_error（））; 
 
 echo $ email_info-＆gt;名称; \  n 
} 
  code>  pre> 
 
 如果我在PHP端为 $ _ POST ['NewMail']  code>做一个回音，我会得到以下信息：   p> 
 
 
  {
 \“Email \”：[
 {
 \“Name \”：\“John Doe \”，
 \“Phone \”：  \“1234567 \”，
 \“电子邮件\”：\“john@doe.com \”，
 \“Interested_in \”：\“文字在这里.. \”，
 \“User_id \”：\  “1 \”
} 
] 
} 
  code>  pre> 
 
 我该如何解决这个问题？ p> 
  div>

答

Replace this part in PHP :

$new_email = json_decode($_POST['NewMail'], true);

by this :

if (get_magic_quotes_gpc()) {
    $_POST['NewMail'] = stripslashes($_POST['NewMail']);
}
$new_email = json_decode($_POST['NewMail'], true);

This should fix the issue.

答

you need to specify to jquery that your sending data is a JSON type. Please configure the type of data in the ajax options. For example:

$.ajax({
       url: "core.php",
       type: "json", //here comes the data's type
       data: { NewMail: new_email
       },
       success: function(data){  
        alert(data)},
        error: function(){
        }`

答

I had magic_quotes turned 'on' on my server.

I realized it when testing Hossams code:

if (get_magic_quotes_gpc()) {
    $_POST['NewMail'] = stripslashes($_POST['NewMail']);
}

So finally this got to work with this code:

if ($_POST['NewMail']) {

$timeStamp = time();

if (get_magic_quotes_gpc()) {
    $_POST['NewMail'] = stripslashes($_POST['NewMail']);
}

$new_email = json_decode($_POST['NewMail'], true);

$email_info = $new_email['Email'][0];

// New Email
mysql_query("INSERT INTO " . $dbPrefix . "touches (date, user_id, name, phone, email, interested_in, seen, status) VALUES ('".safeSQL($timeStamp)."', '".safeSQL($email_info['User_id'])."', '".safeSQL($email_info['Name'])."','".safeSQL($email_info['Phone'])."', '".safeSQL($email_info['Email'])."', '".safeSQL($email_info['Interested_in'])."', '0', '1')") or die("Query failed: " . mysql_error());

var_dump($email_info['Name']);

}

And I got the array with $email_info['Name'] instead of $email_info->Name

将JSON数据发送到PHP并使用它

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