Ajax Post Value未傳遞到php文件

Question

我在將變量傳遞到php頁面時遇到問題。

這是下面的代碼：

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);


}

我想念什么嗎？ 實際數據正在保存到數據庫中，但$ first和$ second值未傳遞到php文件。

Answer 1

您正在使用POST類型，請在POST中進行檢索：

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

或更改您的JQuery調用：

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

Answer 2

因為您正在傳遞數據引發POST方法並嘗試使用GET所以請更改這兩行

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

或者只是將您的方法更改為jQuery中的GET

type: "GET"

Answer 3

您正在ajax中使用類型：“ POST”，並嘗試使用$ _GET獲取，請嘗試

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

Answer 4

並且有一種方法可以像這樣傳遞數據

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

   }
});

在那里你可以使用

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

希望能幫助到你。