取回從PHP頁面的響應，並用它在JavaScript調用Ajax後

我做一個Ajax調用，一切工作正常，我現在用的腳本是：取回從PHP頁面的響應，並用它在JavaScript調用Ajax後

$(document).ready(function() { 
    $('.sortable').sortable({ 
     stop: function(event, ui) { 
     $(ui.item).effect("highlight"); 

     //getting the id of the list element being moved and then 
     //we'll send it to the db for its id validation and 
     //updating the db upon drag and drop 
     var id = $(ui.item).attr("id"); 
     //alert(id); 

     var pos = ui.item.prevAll().length;//prevAll().length 
     var position = ++pos; 
     //alert("Moved to position: " + position);//+ "from: " + id); 
     //var x = ui.offset.left; 
     //var y = ui.offset.top; 
     //alert("left: " + x + "top: " + y); 

     $.ajax({ 
      url: "save.php", 
      data: { 
       position: position, 
       id: id 
      } 
     }); /*, success:function(result){alert(result);}*/ 
     /* $.ajax({ 
      url: "save.php", 
      data: { 
       x: x, 
       y: y 
      } 
     });*/ // data: {x: 'x', y: 'y'} ------->> for coordinates if needed*/ 

     $(ui.item).effect("highlight"); 
     } 
    }); 
})

我有一個PHP該頁面使用發送的值並在數據庫中進行必要的更改。我想知道如何從PHP腳本中獲取返回值，並通過javascript在調用頁面上使用它。 PHP代碼是：

$id1 = $_REQUEST['id']; 
    $pos1 = $_REQUEST['position'];//position where the element has been dragged to 

    $query = "select * from tab where id='$id1'"; 
    $t = mysql_query($query) or die("nothing found in the database for the provided id. ".mysql_error()); 

    if($t) 
    { 
     $r = mysql_fetch_array($t) or die("data could not be fetched from the DB ".mysql_error()); 
     $r[0];//id of the dragged <li>element = $id1 
     $r[1];//its original position 

     $e = "select * from tab where original='$pos1'"; 

     $y = mysql_query($e) or die("ERROR. ".mysql_error()); 
     $u = mysql_fetch_array($y) or die("data could not be fetched from the DB for the replaced element. ".mysql_error());     
     $id2 = $u[0];//id of the place where the dragged <li> element was dropped 
     $u[1];//original position of the place where the dragged <li> element was dropped 

     $temp1 = $r[1]; 
     $temp2 = $u[1]; 

     $temp = $temp1; 
     $temp1 = $temp2; 
     $temp2 = $temp; 

     $up = "update tab set original='$temp1' where id='$id1'"; 
     $q = mysql_query($up) or die("I query not done. ".mysql_error()); 

     $up = "update tab set original='$temp2' where id='$id2'"; 
     $q = mysql_query($up) or die("II query not done. ".mysql_error()); 
     //echo"id1: " .$id1 ." id2: " .$id2 ." r[1]: " .$temp1 ." u[1]: " .$temp2;

請幫我一把。我是AJAX的新手，所以我不太瞭解它。任何幫助表示讚賞。謝謝。

來源

2014-03-14 Utkarsh Singh

你沒有捕獲ajax響應？你有一個成功的功能，但評論它。你的代碼片段做的比你所問的要多得多 - 保持沉默，直到它只有你所關心的部分 - 這將幫助你，它會幫助別人回答。 – DrLivingston

在ajax文件中嘗試'echoing something'，然後在'ajax success'中檢測'alert（result）'..檢查它返回的結果 –

我需要將返回的值存儲在一個javascript變量中..... how我是否這樣做？ @DrLivingston –

使用json_encode onyour PHP

//$array is value you want to return. 
echo json_encode($array);

，或者如果你想從數組沒有迴音，你必須呼應JSON格式。

來源

2014-03-14 05:32:36 dhidy

取回從PHP頁面的響應，並用它在JavaScript調用Ajax後

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