jQuery ajax請求：如何在成功函數中訪問發送的數據？

Question

所以我試圖實現以下，但我無法弄清楚如何使這項工作。

$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: { myVar: "hello" }, 
    success: function(response) { 
    console.log('received this response: '+response); 
    console.log('the value of myVar was: '+data.myVar); // <<<< data.myVar is not accessible from here 
    console.log('the value of myVar was: '+myVar); // <<<< myVar is not accessible from here 
    } 
}); 

是否有訪問的.success()功能的myVar值的方法嗎？ 我可以以某種方式獲取原始的data對象，該對象是在這個ajax請求中發送的，在.success()函數中？

希望得到您的解決方案。 謝謝！

Answer 1

一般來說，如果你想能夠多次引用數據，您需要確保它在正確的範圍內。您在.ajax()內傳遞的json數據對象的範圍是ajax函數。如果您希望能夠引用data:以外的值，例如您在中調用.ajax()的範圍，最簡單的方法就是將其分配給變量。例如

myData = { myVar: "hello" }; 
 
$.ajax({ 
 
    url: "whatever.php", 
 
    method: "POST", 
 
    data: myData, 
 
    success: function(response) { 
 
    console.log('received this response: '+response); 
 
    $("#response").html(response); 
 
    console.log('the value of myVar was: '+myData.myVar); // <<<< data.myVar is not accessible from here 
 
    $("#myVar").html(myData.myVar); 
 
    } 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<p id="response"></p> 
 
<p id="myVar"></p>

Answer 2

只需將您要發佈的數據首先存儲在變量中。

var data = {myVar: "hello"} 
$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: data, 
    success: function(response) { 
    console.log('received this response: '+response); 
    console.log('the value of myVar was: '+data.myVar); 
    } 
}); 

Answer 3

創建一個數據對象，這樣就可以在請求

var postData ={ myVar: "hello" }; 

$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: postData , 
    success: function(response) { 
    console.log('received this response: '+response); 
    console.log('the value of myVar was: '+postData.myVar); 

Answer 4

使我的評論到答案的不同部分，然後訪問...

號有沒有辦法從成功處理程序中訪問$.ajax()的參數。您當然可以將該參數分配給更高範圍的變量，然後傳遞該變量。然後，您將可以從success處理程序中訪問該變量。

相反，你可以這樣做：

var sentData = { myVar: "hello" }; 
$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: sentData, 
    success: function(response) { 
    console.log('received this response: ' + response); 
    console.log('the value of myVar was: ' + sentData.myVar); 
    } 
}); 

Answer 5

你總是可以做類似如下：

var myData = '{ myVar: "hello" }'; 

$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: myData, 
    success: function(response) { 
     console.log('the value of myVar was: ' + myData.myVar); 
    } 
}); 

甚至更​​好

var myData = { 
    myVar: "hello" 
}; 

$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: JSON.stringify(myData), 
    success: function(response) { 
     console.log('the value of myVar was: ' + myData.myVar); 
    } 
}); 

Answer 6

您可以使用this訪問整個對象。所以你可以這樣做：

$.ajax({ 
    url: "whatever.php", 
    method: "POST", 
    data: { myVar: "hello" }, 
    success: function(response) { 
    console.log('received this response: '+response); 
    console.log('the value of myVar was: '+this.data.myVar); 
    } 
});