Question

我正在尝试为一个项目设置模态，在这个项目中，客户可以更新自己喜欢的球队，出生城市和他们的体型。在没有输入一些信息时，我希望将错误消息返回给他们，但是回显的唯一错误信息是“由于没有提交新信息，城市未更新。”

模态代码如下：

<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-  labelledby="myModalLabel" aria-hidden="true">
 <div class="modal-dialog" role="document">
 <div class="modal-content">
  <div class="modal-header">
    <button type="button" class="close" data-dismiss="modal" aria-label="Close">
      <span aria-hidden="true">&times;</span>
      </button>
     <h4 class="modal-title" id="myModalLabel">Modal title</h4>
    </div>
     <div class="modal-body">
    <form>
  <div class="form-group">
    <label for="city">You were born in:</label>
   <input type="text" class="form-control" id="city"  placeholder="city">

   </div>
   <div class="form-group">
   <label for="favorite_team">Favorite team:</label>
    <input type="text" class="form-control" id="favoriteTeam"  placeholder="favorite team">

   </div>
   <div class="form-group">
   <label for="size">size:</label>
    <input type="text" class="form-control" id="size"  placeholder="size">

   </div>

    <div class="alert alert-success" id="successAlert" role="alert"  style="display: none"></div>
   <div class="alert alert-danger" id="updateFail" style="display:none" >          </div>

    </form>
    </div>
     <div class="modal-footer">
     <button type="button" class="btn btn-secondary" data-dismiss="modal"  >Close</button>
     <button type="button" class="btn btn-primary" id="saveBtn"  onClick="Changes()">Save changes</button>
   </div>
   </div>
   </div>
   </div>

我的ajax代码是：

  function Changes(){
   $.ajax({
           type: "POST",
          url: "../php_parsers/update_parse.php",
          data: "city=" + $("#city").val() + "&favoriteTeam=" +    $("#favoriteTeam").val() + "&size=" + $("#size").val(),
            success: function(result) {
              if (result == "success") {

                $("#successAlert").html(result).show();

                } else {

                $("#updateFail").html(result).show();

                }
             }

           })

           }

我的php代码是update_parse.php：

<?php
// AJAX CALLS THIS UPDATE CODE TO EXECUTE
$error = "";
$success = "";
if(!$_POST['city']){
$error .= "City was not updated as there was not new information submitted.";
} else if(!$_POST['team']){
$error .= "Team was not updated as there was not new information submitted.";
} else if(!$_POST['size']){
    $error.= "size was not updated as there was not new information submitted.";
    } 

    if($error == ""){
        echo $error;
        }else {
            $success = "update successful so far";
        echo $success;


            }


 ?>

欢迎所有反馈。

Answer 1

您正在使用if else if构造进行检查，最多只会报告一个错误。将其更改为独立的if语句

if(!$_POST['city']){
    $error .= "City was not updated as there was not new information submitted.";
} 
if(!$_POST['team']){
    $error .= "Team was not updated as there was not new information submitted.";
} 
if(!$_POST['size']){
    $error.= "size was not updated as there was not new information submitted.";
}

Answer 2

尝试像这样建立你的AJAX

$.ajax({
    url: "../php_parsers/update_parse.php",
    type: "POST",
    data: { 
       city:         $("#city").val(), 
       favoriteTeam: $("#favoriteTeam").val(), 
       size:         $("#size").val() 
    }
  }).done(function(result) {
     if (result == "success") {
        $("#successAlert").html(result).show();
     } else {
        $("#updateFail").html(result).show();
     }
})

也不是你的回归有点时髦

if($error != ""){  <------------- Change
  echo $error;
}else {
  $success = "update successful so far";
  echo $success;
}

Answer 3

尝试将您的参数设置为此

var param= {
    "city" : $("#city").val(),
    "favoriteTeam" : $("#favoriteTeam").val(),
    "size" : $("#size").val()
 };

然后在你的ajax属性data中使它像这样：

$.ajax({
    type: "POST",
    url: "../php_parsers/update_parse.php",
    data: param,
    success: function(result) {
      if (result == "success") {
        $("#successAlert").html(result).show();
        } else {
        $("#updateFail").html(result).show();
        }
     }
})

Ajax从PHP接收错误的错误

3 个答案: