2014-02-10 69 views
1

腳本:如何正確發佈數據與Ajax到div?

function buttonBuild(id, building, nick) 
    { 
     $("#BuildedBox").ajax({ 
     type: "POST", 
     url: "BlockEditor/build.php", 
     data: 'block_id=' + id + '&building=' + building + '&nick=' + nick, 
     cache: false, 
     success: function(response) 
     { 
     alert("Record successfully updated"); 
     $.load("#BuildedBox") 
     } 
    }); 
} 

build.php:

include_once("$_SERVER[DOCUMENT_ROOT]/db.php"); 

$block_id = $_GET['block']; 
$building = $_GET['building']; 
$nick = $_GET['nick']; 

echo"$block_id - $building - $nick"; 

的index.php:

<a href=\"#\" onClick=\"buttonBuild(k152, digger, Name);\" >[BUILD]</a> 
<div id="BuildedBox"></div> 

似乎我的腳本不會工作。我做錯了什麼?

+1

什麼是$(「#BuildedBox」)。ajax({'和'$ .load(「#BuildedBox」)'?你想要達到什麼目標? – Satpal

+0

我試圖從鏈接發佈數據到DIV ..實時.. – user3292639

回答

2

檢查了這一點

function buttonBuild(id, building, nick) 
    { 
     $.ajax({ 
     type: "POST", 
     url: "BlockEditor/build.php", 
     data: 'block_id=' + id + '&building=' + building + '&nick=' + nick, 
     cache: false, 
     success: function(response) 
     { 
     alert("Record successfully updated"); 
     /***************/ 
     $("#BuildedBox").html(response); 
     /***************/ 
     } 
    }); 
} 
+0

'$(「#BuildedBox」)。ajax()'??? –

+0

ohh沒有看到...現在糾正...只是糾正了什麼OP張貼:D – Abhidev

+0

helpfull太男人,謝謝;) – user3292639

1
var weightd = $("#weight").val(); 
       var user_id = 43; 
       $.ajax({ 
       type: "POST", 
       url:"<?php bloginfo('template_directory')?>/ajax/insert.php", 
       data: { weight:weightd,user_ids:user_id}, 
       success:function(result){ 
        $("#result1").html(result); 

       }); 


<div id="result1">Result div</div> 
+0

如果你找到解決方案,請親切的答案 –

0

變化$.load("#BuildedBox")$("#BulderBox").html(response)

當您通過ajax向腳本提供數據時,提供的數據會進入「響應」變量。當你想把這些數據寫入div時,你必須使用「.html」方法。

更容易以這種方式使用的 「負載」:從服務器

function buttonBuild(id, building, nick) 
    { 
     $("#BuildedBox").load("BlockEditor/build.php?block_id=" + id + "&building=" + building + "&nick=" + nick); 
} 

的 「負載」 方法加載的數據和寫入結果的HTML入元件:https://api.jquery.com/load/

編輯:

As @ a-wolff在評論中說,要在加載中使用POST,您應該這樣構造:

function buttonBuild(id, building, nick) 
    { 
     $("#BuildedBox").load("BlockEditor/build.php",{ 
     block_id:id, 
     building:building, 
     nick:nick 
     }); 
} 
+1

你可以傳遞數據作爲對象加載( )方法來使用POST,並避免緩存問題 –

+0

謝謝你..我認爲這是有用的..要測試它 – user3292639