将javascript变量传递给PHP使用AJAX

I have 1 javascript variable and i want to pass i to my php file. i search about it and i find that to use ajax but i don't really know how use it ! here is my code . where am i do wrong ? my .js file :

我有1个javascript变量,我想把我传递给我的php文件。我搜索它,我发现使用ajax但我真的不知道如何使用它!这是我的代码。我哪里做错了?我的.js文件:

var message1 = message.message;

        var jq = document.createElement('script');
        jq.src = "https://code.jquery.com/jquery-1.10.2.js";
        document.querySelector('head').appendChild(jq);

            $(document).ready(function() {
               $.ajax({
                        type: "POST",
                        url: 'http://localhost/a.php',
                        data: { newMessages : message1 },
                        success: function(data)
                        {
                            alert("success!");
                        }
                    });
         });

my a.php file :

我的a.php文件:

<?php
  if(isset($_POST['newMessages']))
  {
      $uid = $_POST['newMessages'];
      echo $uid;
  }
?>

2 个解决方案

#1

Listen onload event of the asynchronously loaded script and execute your function in callback function.[Ref]

监听异步加载脚本的onload事件,并在回调函数中执行函数。[Ref]

Try this:

function loadScript(src, callback) {
  var s,
    r,
    t;
  r = false;
  s = document.createElement('script');
  s.type = 'text/javascript';
  s.src = src;
  s.onload = s.onreadystatechange = function() {
    if (!r && (!this.readyState || this.readyState == 'complete')) {
      r = true;
      callback();
    }
  };
  t = document.getElementsByTagName('script')[0];
  t.parentNode.insertBefore(s, t);
}

var message1 = "My message";
loadScript('https://code.jquery.com/jquery-1.10.2.js', function() {

  $(document).ready(function() {
    $.ajax({
      type: "POST",
      url: 'http://localhost/a.php',
      data: {
        newMessages: message1
      },
      success: function(data) {
        alert("success!");
      }
    });
  });
})

#2

-1

The php file should return data in json format. You can add complete, always, error to your Ajax to better catch the result

php文件应以json格式返回数据。您可以向Ajax添加完整的始终错误,以便更好地捕获结果

#1