Question

我在将变量传递给php页面时遇到问题。

以下是代码：

var varFirst = 'something'; //string
var varSecond = 'somethingelse'; //string

$.ajax({  
   type: "POST",  
   url: "test.php",  
   data: "first="+ varFirst +"&second="+ varSecond,  
       success: function(){  
          alert('seccesss');

   }
});

PHP：

$first = $_GET['first']; //This is not being passed here
$second = $_GET['second']; //This is not being passed here

$con=mysqli_connect("localhost","root","pass","mydb"); 

if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

mysqli_query($con,"INSERT INTO mytable (id, first, second) VALUES ('', $first, $second)");

mysqli_close($con);


}

我错过了什么？实际数据保存到数据库BUT $ first，$ second值没有传递给php文件。

Answer 1

您正在使用POST类型，在POST中检索它：

$first = $_POST['first'];
$second = $_POST['second'];

或者更改你的JQuery调用：

$.ajax({  
   type: "GET",  
   url: "test.php",  
   data: "first="+ varFirst +"&second="+ varSecond,  
       success: function(){  
          alert('seccesss');
   }
});

Answer 2

这是因为你正在传递数据抛出POST方法，并尝试使用GET来改变这两行

$first = $_POST['first']; //This is not being passed here
$second = $_POST['second']; //This is not being passed here

或者只是将您的方法更改为jquery中的GET

type: "GET"

Answer 3

您在ajax中使用类型：“POST”并尝试使用$ _GET获取，请尝试

$first = $_REQUEST['first']; //This is not being passed here
$second = $_REQUEST['second'];

Answer 4

还有一种传递数据的方法

$.ajax({  
   type: "POST",  
   url: "test.php",  
   data: {first: varFirst,second: varSecond},  
       success: function(){  
          alert('seccesss');

   }
});

你可以使用

$_POST['first'];
$_POST['second'];

希望它有所帮助。

Ajax Post Value没有传递给php文件

4 个答案: