如果出現錯誤消息，不希望將值插入到文本輸入

Question

下面我有一個帶有ajax代碼的jquery，它將顯示更改課程詳細信息的值，方法是插入一組輸入中的新數據（輸入集合用戶已經做出改變）到另一組輸入（輸入組，其中指出課程的當前細節）。另外，課程的下拉菜單進行更改以適應新的變化，如果它需要改變：

function submitform() {  

    $.ajax({ 
     type: "POST", 
     url: "updatecourse.php", 
     data: $('#updateCourseForm').serialize(), 
     success: function(html){ 
      $("#targetdiv").html(html); 
      //Get and store the new course number and name. 
       var newCourseNo = jQuery("#newCourseNo").val(); 
       var newCourseName = jQuery("#newCourseName").val(); 
       var newDuration = jQuery("#newDuration").val(); 

       //Set your current course number and name to your number and name. 
       jQuery("#currentCourseNo").val(newCourseNo); 
       jQuery("#currentCourseName").val(newCourseName); 
       jQuery("#currentDuration").val(newDuration); 

       //Find the currently selected course and update it. 
       var selectedOption = jQuery("#coursesDrop option:selected"); 
       var label = selectedOption.text().split(" - "); 
       selectedOption.text(newCourseNo + " - " + newCourseName); 

       $('#targetdiv').show(); 
     } 
    });   
} 

現在$("#targetdiv")是它顯示它是通過AJAX訪問從PHP頁面中檢索到的成功或錯誤消息ID：

updatecourse.php：

...//mysqli code 

echo "<span style='color: green'>Course details have been updated:<br/>" . $newcourseno . " - " . $newcoursename . "</span>"; 

}else{ 

echo "<span style='color: red'>An error has occured, Course Details have not been updated</span>"; 

} 

但我遇到的問題是，如果從PHP代碼錯誤消息在jQuery是檢索，然後我不希望課程內容，使編輯的jquery來容納當前課程文本輸入以插入新的細節。如果成功消息出現，我只希望發生這種情況。

如果出現錯誤消息，我不希望發生變化，當前課程詳細信息輸入和新課程詳細信息輸入在提交之前保持不變。

但是，這怎麼能實現呢？

Answer 1

獲得響應，JSON（這樣你可以在得到結果後操縱響應...）

試試這個，

JQUERY

$.ajax({ 
    type: "POST", 
    url: "updatecourse.php", 
    data: $('#updateCourseForm').serialize(), 
    dataType:'json'; //get response as json 
    success: function(result){ 
     if(result.errorflag){ 

      //do your stuff on getting error message 
      var newHtml="<span style='color: red'>"+result.msg+"</span>" 
      $("#targetdiv").html(newHtml); //i am displaying the error msg here 

     }else{ 
      //you got success message 

      var newHtml="<span style='color: green'>"+result.msg+"</span>" 
      $("#targetdiv").html(newHtml); 
      //Get and store the new course number and name. 
      var newCourseNo = jQuery("#newCourseNo").val(); 
      var newCourseName = jQuery("#newCourseName").val(); 
      var newDuration = jQuery("#newDuration").val(); 

      //Set your current course number and name to your number and name. 
      jQuery("#currentCourseNo").val(newCourseNo); 
      jQuery("#currentCourseName").val(newCourseName); 
      jQuery("#currentDuration").val(newDuration); 

      //Find the currently selected course and update it. 
      var selectedOption = jQuery("#coursesDrop option:selected"); 
      var label = selectedOption.text().split(" - "); 
      selectedOption.text(newCourseNo + " - " + newCourseName); 

      $('#targetdiv').show(); 
     } 
     } 
}); 

PHP

json_encode()發送效應初探爲JSON ....發送響應，與錯誤標誌陣列，以檢查它是否是成功的情況下或錯誤並且味精打印...

...//mysqli code 

    echo json_encode(array('errorflag'=>false,'msg'=>"Course details have been updated:<br/>" . $newcourseno . " - " . $newcoursename)); 

    }else{ 

    echo json_encode(array('errorflag'=>true,'msg'=>"An error has occured, Course Details have not been updated")); 

} 

Answer 2

用戶在更新的DOM EX的基礎上，JSON響應的任何標誌：

$.ajax({ 
     type: "POST", 
     url: "updatecourse.php", 
     dataType: 'json', 
     data: $('#updateCourseForm').serialize(), 
     success: function(json){ 
      $("#targetdiv").html(json.htmlContent); 
      //Get and store the new course number and name. 
     if(json.status=="success"){ 
var newCourseNo = jQuery("#newCourseNo").val(); 
       var newCourseName = jQuery("#newCourseName").val(); 
       var newDuration = jQuery("#newDuration").val(); 

       //Set your current course number and name to your number and name. 
       jQuery("#currentCourseNo").val(newCourseNo); 
       jQuery("#currentCourseName").val(newCourseName); 
       jQuery("#currentDuration").val(newDuration); 

       //Find the currently selected course and update it. 
       var selectedOption = jQuery("#coursesDrop option:selected"); 
       var label = selectedOption.text().split(" - "); 
       selectedOption.text(newCourseNo + " - " + newCourseName); 

       $('#targetdiv').show(); 
}else{ 
//show whatever mesage u want 
} 

     } 
    }); 

Answer 3

做像這樣成功阻止

success: function(html){ 
    var indexValue = html.indexOf("An error has occured"); 
    if (parseint(indexValue) < 0) { 
     return false; 
    } 

    $("#targetdiv").html(html);