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Php 为什么使用AJAX提交的表单会重定向到下一页,并且错误/成功消息不会显示在同一页上的警报中?_Php_Jquery_Ajax_Form Submit - Fatal编程技术网

Php 为什么使用AJAX提交的表单会重定向到下一页,并且错误/成功消息不会显示在同一页上的警报中?

Php 为什么使用AJAX提交的表单会重定向到下一页,并且错误/成功消息不会显示在同一页上的警报中?,php,jquery,ajax,form-submit,Php,Jquery,Ajax,Form Submit,我正在为我的网站使用PHP、Smarty、jQuery、AJAX等。以下是我使用AJAX提交的表单的HTML代码: <form name="question_issue_form" id="question_issue_form" action="http://localhost/xyz/pqr/web/control/modules/questions/question_issue.php"> <input type="hidden" name="form_sub

我正在为我的网站使用PHP、Smarty、jQuery、AJAX等。以下是我使用AJAX提交的表单的HTML代码:

<form name="question_issue_form" id="question_issue_form" action="http://localhost/xyz/pqr/web/control/modules/questions/question_issue.php">
      <input type="hidden" name="form_submitted" id="form_submitted" value="yes"/>
      <input type="hidden" name="op" id="op" value="question_issue"/>
      <input type="hidden" name="question_id" id="question_id" value="35718"/>

      <table class="trnsction_details" width="100%" cellpadding="5">
        <tbody>    
          <tr>
            <td></td>
            <td>
              <input type="checkbox" name = "que_issue[]" value = "Question is wrong" id ="chkQueWrong">Question is wrong</input>
            </td>
          </tr>
          <tr>
            <td></td>
            <td><input type="checkbox" name = "que_issue[]" value = "Answers are wrong" id ="chkAnsWrong">Answers are wrong</input></td> 
          </tr>
          <tr>
            <td></td>
            <td><input type="checkbox" name = "que_issue[]" value = "Question direction is incorrect" id ="chkDirIncorrect">Question direction is incorrecct</input></td>                
          </tr>
          <tr>
            <td></td>
            <td><input type="checkbox" name = "que_issue[]" value = "Other" id ="chkOther">Other</input></td>          
          </tr>
          <tr>
            <td></td>
            <td class="set_message" style="display:none;"><textarea name="que_issue_comment" id = "que_issue_comment" rows="4" cols="25" maxlength="100"></textarea></td>      
          </tr>
          <tr>
            <td></td>
            <td><input type="submit" name="submit" value="Submit" id="report_question_issue" class="c-btn submit_form"/></td>
          </tr>
        </tbody>
      </table>
    </form>
我提交此表单的文件(question_issue.PHP)的PHP代码如下:

$(document).ready(function() {
$('#question_issue_form').submit(function() {
var ans = confirm("Are you sure to report the question issue?");
    if (!ans) { 
      return false;
    }
var post_url = $(this).attr('action');
$.ajax({
        type: "POST",
        url: post_url,
        data: $('#question_issue_form').serialize(),
        dataType: 'json',
        success: function(data) { alert(data);
          var error = data.error_message;
          if(error)
            alert(error);
          else {
            alert("Question issue has been reported successfully.");
            $.colorbox.close();
          }
        }
      });
    });
  });
<?php 
  require_once("../../includes/application-header.php");

  $objQuestionIssue = new QuestionIssue(); 

  prepare_request();
  $request = $_POST ;
$user_type = $_SESSION[SESSION_NAME_CONTROL][STAFF_TYPE];

  if($user_type == 'super_admin' || $user_type == 'admin' || $user_type == 'data_entry_operator' || $user_type == 'owner' || $user_type == 'faculty' || $user_type == 'content_development_head' || $user_type == 'test_admin' || $user_type == 'student_admin')
    $requested_user_type = 'staff';
  else       
    $requested_user_type = 'student';

    $form_data = array();
    $form_data['question_id']        = $request['question_id'];
    $form_data['reported_site_id']   = SITE_ID;
    $form_data['reported_user_type'] = $requested_user_type;
    $form_data['reported_user_id']   = $_SESSION[SESSION_NAME_CONTROL][STAFF_ID];
    $form_data['que_issue']          = implode(",", $request['que_issue']);
    $form_data['que_issue_comment']  = $request['que_issue_comment'];
    $form_data['que_issue_date']     = time();
switch( $op ) { 
        case "question_issue":
          if($request['form_submitted']=='yes') {
                $ret = $objQuestionIssue->InsertQuetionIssue($form_data, $question_issue_error_messages);
                /*If condition : If there are any errors in submission of a report question form*/
                if(!$ret) {
                    $error_msg  = $objQuestionIssue->GetAllErrors();
                    $data = array();
                    $data['error_message'] = $error_msg['error_msgs'];
                    $data = json_encode($data);
                    echo $data;
                    die;
                /*Else condition : If there is no error in submission of a report question form*/   
                } else {
                    $data = array();
                    $data['success_message'] = "success";
                    $data = json_encode($data);
                    echo $data;
                    die;
                }
            } else {
                  $smarty->assign('question_id', $request['question_id']);
                  $file_to_show = 'question-issue.tpl';
                }           
                $smarty->display($file_to_show);
            break;  
            die;
  }  
?>

我面临的问题是,当我单击确认警报的Ok按钮时,表单会被提交,但json格式的错误消息或成功消息会出现在一个空白屏幕上


实际上,它们应该出现在弹出式警报中,页面不应该被重定向到其他URL。但是错误/成功消息打印在一个空白的白色页面上,页面也被重定向到question_issue.php。有人能帮我避免这些事情,并将错误/成功消息显示在同一页面上的警告框中吗?

我认为您应该防止表单的默认行为,因此它不会真正提交,而只要求json:

$('#question_issue_form').submit(function(e) {
    e.preventDefault();
    // your code

我试过你的把戏。它正在工作,但在显示错误/成功警报消息之前,警报消息会显示“[object object]”和“Ok”按钮。这是什么?如何避免?@PHPLover您有一个
警报(数据)在您的代码中。这可能就是PHP返回的对象,因为
data
是JSON对象。这就像你试图提醒一个数组而不是字符串。你可能想进入某些财产。是的,那是我的错。谢谢你纠正了。你的答案已经被接受,并且也被投了赞成票。干杯