Ajax submit form and receive json method-JS Tutorial-php.cn

Ajax submit form and receive json method

小云云

Release： 2017-12-19 14:09:03

Original

1851 people have browsed it

After clicking the button, the data is submitted to the server in the form of a form and the return data from the server is received. The page does not refresh during the process. This article mainly introduces the example code of Ajax submitting a form and receiving json. It is very good and has reference value. Friends in need can refer to it. I hope it can help everyone.

html code

<html xmlns="http://www.w3.org/1999/xhtml">
 <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
 <script src="https://cdn.bootcss.com/jquery/1.12.4/jquery.min.js"></script>
 <script src="./testajaxjs.js"></script>
 <head>
 </head>
 <body>
  <form id="form1">
   <p>xingming:<input type="text" name="xingming"/></p>
   <p>nianling:<input type="text" name="nianling"/></p>
  </form>
  <button type="button" id="mybt" onclick="mysubmmit()">
   ajax提交
  </button>
 </body>
</html>

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js code

function mysubmmit(){
 $.ajax({
  type: "POST",
  url: "testajaxend.php",
  data: $(&#39;#form1&#39;).serialize(),
  async: false,
  success: function(databack){
   //console.log("chenggong");
   console.log(databack);
  },
  error: function(request){
   console.log("shibaile");
  }
 });
}

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Backend php code

<?php
  $name = $_POST[&#39;xingming&#39;];
  $age = $_POST[&#39;nianling&#39;];
  $myarray = array("name"=>$name, "age"=>$age);
  $myjson = json_encode($myarray);
  echo $myjson;
?>

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