2011-06-03 55 views
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下面是由json_encode()在ajax頁面生成的代碼。現在,我想檢索數據(標題,ISBN,作者...)。任何人都可以幫我一把嗎?我看過jQuery.parseJSON函數,但感到困惑。從json_encode訪問數據

{ 
    "9780077225957": { 
     "Items_Data": { 
      "Title": "Developing Management Skills: What Great Managers Know and Do", 
      "Data_Source": "Amazon", 
      "Item_ID": "1329", 
      "ISBN": "9780077225957", 
      "Authors": "Timothy Baldwin", 
      "Edition": "1", 
      "Year": "2007", 
      "Publisher": "McGraw-Hill/Irwin", 
      "Amazon_Thumb_URL": "http://ecx.images-amazon.com/images/I/41pVf7GKujL._SL160_.jpg" 
     } 
    } 
} 
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你是從$ .getJSON中檢索這個嗎?你想用這些數據做什麼?創建元素?對象? – natedavisolds 2011-06-03 03:37:41

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@natedavisolds它目前是純ajax,但我想我可以在我的ajax上添加dateType:json。我將它們用於純HTML。 – 2011-06-03 04:04:54

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所以你要基於返回的JSON在DOM中創建一個元素?是對的嗎? – natedavisolds 2011-06-03 04:12:06

回答

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在JavaScript變量只是將分配工作,但我認爲你需要在第一個改變9780077225957關鍵,因爲你不會使用(。)運算符的訪問。我已將其更改爲Book9780077225957並正常工作。下面是我用

var book = { "Book9780077225957" : { "Items_Data": { "Title": "Developing Management Skills: What Great Managers Know and Do", "Data_Source": "Amazon", "Item_ID": "1329", "ISBN": "9780077225957", "Authors": "Timothy Baldwin", "Edition": "1", "Year": "2007", "Publisher": "McGraw-Hill/Irwin", "Amazon_Thumb_URL": "http://ecx.images-amazon.com/images/I/41pVf7GKujL.SL160.jpg" } } }; 

alert(book.Book9780077225957.Items_Data.Title); 
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雖然我認爲你需要你$ .getJSON通過Ajax檢索數據:) – Tapos 2011-06-03 04:21:21

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我會希望做類似的代碼:

var Books = { 
    toElement: function(data) { 
     if (data) { 

      $book = $('<div></div>') 
       .addClass('book'); 

      $('<span></span>').className("title").text(data.Title)appendTo($book); 
      $('<span></span>').className("authors").text(data.Authors)appendTo($book); 
      $('<span></span>').className("isbn").text(data.ISBN)appendTo($book); 

      $book.appendTo($('#books'));    
     } 
    } 
}; 

$(function() { 
    $("#Button1").click(function() { 
     var params = []; 
     $.getJSON("book_url.js", params, function(returnedJSON) { 
      Books.toElement(returnedJSON.Items_Data);   
     }); 

    }); 
}); 

我不知道在JSON的第一要素是什麼。它可能不會被返回JSON.Items_Data。出於某種原因,我認爲第一級被忽略。但我可以考慮其他事情。

希望這會有所幫助。