Question

我最近一直在研究jQuery，今天我正在使用AJAX。我现在要做的只是将日期选择器中的日期发送到PHP页面，该页面将发回该日期。我稍后会改进PHP页面以完成我需要做的事情。这是HTML页面的代码：

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="generator" content="AlterVista - Editor HTML"/>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js">
    $(document).ready(function() {
        $("#foo").submit(function(){
            var serializedData = $(this).serialize();
            $.post('demo.php', serializedData, function(response) {
                // Log the response to the console
                console.log("Response: "+response);
                $("#result").text() = response;
            });
        });
    });
  </script>
  <title></title>
</head>
<body>
    <form id='foo'>
        <input type='date' id='dataI' name='dataI'>
        <input type='submit' value='send'>
        <p id='result'></p>
    </form>
</body>
</html>

这是我制作的简单php页面：

<?php
    echo $_POST["dataS"];
?>

但是当我尝试发送数据时没有任何反应。我究竟做错了什么？我尝试使用我在这里看到的堆栈溢出的解决方案，但它不起作用。

Answer 1

我附上代码示例;

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="generator" content="AlterVista - Editor HTML"/>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
  
  <script type="text/javascript">
    function validate(){
		
		var serializedData = $('#foo').serialize();
            $.post('ajax.php', serializedData, function(response) {
                // Log the response to the console
                console.log("Response: "+response);
                $("#result").html(response);
            });
return false;

	}
  </script>
  <title></title>
</head>
<body>
    <form id='foo' action="#" onsubmit="return validate()">
        <input type='date' id='dataI' name='dataI'>
        <input type='submit' value='send'>
        <p id='result'></p>
    </form>
</body>
</html>

Ajax代码文件;

 echo $_POST["dataI"];

Answer 2

尝试添加return false;之后

$.post('demo.php', serializedData, function(response) {
            // Log the response to the console
            console.log("Response: "+response);
            $("#result").text() = response;
        });

语句。

 $(document).ready(function() {
        $("#foo").submit(function(){
            var serializedData = $(this).serialize();
            $.post('demo.php', serializedData, function(response) {
                // Log the response to the console
                console.log("Response: "+response);
                $("#result").text() = response;
            });
             return false;
            });
        });

jQuery / AJAX交互的问题

2 个答案: