PHP AJAX 响应的 `response.status` 返回未定义。

huangapple
go评论116阅读模式
英文:

PHP AJAX response.status return undefined

问题

Here is the translation of the non-code part of your text:

我确实有两个问题，第一个是当我尝试评估 response.status 时，它显示为 undefined；第二个是我的 "$_SESSION['promo-code'] = $arr" 也没有被创建。当输入促销码时，我得到以下内容：

AJAX

  1. function applyPromo(){
  2.     var promo = $("#promo-code-value").val().trim();
  3.     console.log(promo);
  4.     $.ajax({
  5.         url: "get-promo.php",
  6.         type: "POST",
  7.         data: { getpromo: promo },
  8.         success: function (response) {
  9.             console.log("%j", response);
  10.             console.log(response.status);
  11.             if (response.status === "success") {
  12.                 swal("Good job!!", 
  13.                     "Promo code applied Successfully!",
  14.                     "success"
  15.                     ).then(() => {
  16.                         reloadPage();
  17.                     });
  18.             } else {
  19.                 swal("Humm...", "Promo code no longer valid", "error");
  20.             }
  21.         }
  22.     })
  23. }

PHP

  1. if(isset($arr)){
  2.     //将数组添加到会话变量中
  3.     //**每次只能应用一个代码。会覆盖之前的
  4.     $_SESSION['promo-code'] = $arr; 
  5.     //成功响应
  6.     echo json_encode(array(
  7.         'status' => 'success',
  8.         'message' => $arr, //仅用于调试，将被替换为成功消息
  9.     ));
  10. } else {
  11.     echo json_encode(array(
  12.         'status' => 'error',
  13.         'message' => 'error message'
  14.     ));
  15. }

这只是为了学校项目。谢谢！

我尝试将 "success" 切换为 "error"，但它似乎根本不进行评估。我对 Ajax 或 PHP 不熟悉，但我相信我曾经看到其他人是这样做的。

英文:

I have two problems really here, the first one being that when I try to evaluate response.status I get undefined and the second one being that my "$_SESSION['promo-code'] = $arr" is also not being created. When entering a promo code, I get the following :
PHP AJAX 响应的 `response.status` 返回未定义。

AJAX

  1. 	function applyPromo(){
  2. 		var promo = $("#promo-code-value").val().trim();
  3. 		console.log(promo);
  4. 		$.ajax({
  5. 			url:"get-promo.php",
  6. 			type: "POST",
  7. 			data:{ getpromo : promo},
  8. 			success: function(response){
  9. 				console.log("%j", response);
  10. 				console.log(response.status);
  11. 				if(response.status === "success"){
  12. 					swal("Good job!!", 
  13. 						"Promo code applied Successfully!",
  14. 						"success"
  15. 						).then(() => {
  16. 							reloadPage();
  17. 						});
  18. 					}else{
  19. 					swal("Humm...", "Promo code no longervalid","error");
  20. 					}	
  21. 				}
  22. 			})
  23. 		}

PHP

  1. if(isset($arr)){
  2.         //Add array to session variable 
  3.         //** Only one code can be apply at the time . will override
  4.         $_SESSION['promo-code'] = $arr; 
  5.         //Success response 
  6.         echo json_encode(array(
  7.             'status' => 'success',
  8.             'message'=> $arr, //For debugging only , will be replaced to success message 
  9.         ));
  10.     }else {
  11.         echo json_encode(array(
  12.             'status' => 'error',
  13.             'message'=> 'error message'
  14.         ));
  15.     }

It's only for a school project. Thanks!!

Ive tried switching "success" to "error" but it is simply not evaluating .I'm new to ajax or php but I believe to have seen others makes it work that way.

答案1

得分: 0

问题之一是响应是JSON格式的。因此，您必须将其转换为JS对象，然后您就可以获取状态。以下是更新后的代码：

  1. function applyPromo(){
  2.    var promo = $("#promo-code-value").val().trim();
  3.    console.log(promo);
  4.     $.ajax({
  5.       url:"get-promo.php",
  6.       type: "POST",
  7.       data:{ getpromo : promo},
  8.       success: function(response){
  9.          let response = JSON.parse(response);
  10.          console.log("%j", response);
  11.          console.log(response.status);
  12.          if(response.status === "success"){
  13.             swal("Good job!!", 
  14.               "Promo code applied Successfully!",
  15.               "success"
  16.             ).then(() => {
  17.                reloadPage();
  18.             });
  19.          }else{
  20.             swal("Humm...", "Promo code no longer valid","error");
  21.          }   
  22.       }
  23.     })
  24.   }

问题之二是，您没有使用session_start()。以下是更新后的代码：

  1. if(isset($arr)){
  2.   //将数组添加到会话变量
  3.   //** 一次只能应用一个代码，将覆盖
  4.   session_start();
  5.   $_SESSION['promo-code'] = $arr; 
  6.   //成功响应 
  7.   echo json_encode(array(
  8.      'status' => 'success',
  9.      'message'=> $arr, //仅用于调试，将替换为成功消息 
  10.   ));
  11. }else {
  12.   echo json_encode(array(
  13.     'status' => 'error',
  14.     'message'=> 'error message'
  15.   ));
  16. }
英文:

The problem one is that response is in JSON. So you have to convert it into JS object and then you can get status. Here is the updated Code:-

  1. function applyPromo(){
  2.    var promo = $("#promo-code-value").val().trim();
  3.    console.log(promo);
  4.     $.ajax({
  5.       url:"get-promo.php",
  6.       type: "POST",
  7.       data:{ getpromo : promo},
  8.       success: function(response){
  9.          let response = JSON.parse(response);
  10.          console.log("%j", response);
  11.          console.log(response.status);
  12.          if(response.status === "success"){
  13.             swal("Good job!!", 
  14.               "Promo code applied Successfully!",
  15.               "success"
  16.             ).then(() => {
  17.                reloadPage();
  18.             });
  19.          }else{
  20.             swal("Humm...", "Promo code no longervalid","error");
  21.          }   
  22.       }
  23.     })
  24.   }

The second problem is that, you are not using session_start(). Here is the updated code:-

  1. if(isset($arr)){
  2.   //Add array to session variable 
  3.   //** Only one code can be apply at the time . will override
  4.   session_start();
  5.   $_SESSION['promo-code'] = $arr; 
  6.   //Success response 
  7.   echo json_encode(array(
  8.      'status' => 'success',
  9.      'message'=> $arr, //For debugging only , will be replaced to success message 
  10.   ));
  11. }else {
  12.   echo json_encode(array(
  13.     'status' => 'error',
  14.     'message'=> 'error message'
  15.   ));
  16. }

huangapple
  • 本文由 发表于 2023年5月14日 05:31:12
  • 转载请务必保留本文链接：https://go.coder-hub.com/76244967.html
  • ajax
  • javascript
  • json
  • jsonresponse
  • php
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定