函數socialbookmarksTableData(data)被另一個函數調用來生成一個表的內容 - data是一個JSON對象。在函數內部,我調用2個使用getJSON和POST(以json作爲返回對象)的其他函數來獲取一些數據。問題是:雖然函數正確執行,但我得到了2個變量(bookmarkingSites,bookmarkCategories)的未定義值。請幫助解決方案。jquery控件執行回調函數
function socialbookmarksGetBookmarkCategories(bookmarkID)
{
var toReturn = '';
$.post("php/socialbookmark-get-bookmark-categories.php",{
bookmarkID: bookmarkID
},function(data){
$.each(data,function(i,categID){
toReturn += '<option value ="' + data[i].categID + '">' + data[i].categName + '</option>';
})
return toReturn;
},"JSON");
}
function socialbookmarksGetBookmarkSites()
{
var bookmarkingSites = '';
$.getJSON("php/socialbookmark-get-bookmarking-sites.php",function(bookmarks){
for(var i = 0; i < bookmarks.length; i++){
//alert(bookmarks[i].id);
bookmarkingSites += '<option value = "' + bookmarks[i].id + '">' + bookmarks[i].title + '</option>';
}
return bookmarkingSites;
});
}
function socialbookmarksTableData(data)
{
var toAppend = '';
var bookmarkingSites = socialbookmarksGetBookmarkSites();
$.each(data.results, function(i, id){
var bookmarkCategories = socialbookmarksGetBookmarkCategories(data.results[i].bookmarkID);
//rest of the code is not important
});
$("#searchTable tbody").append(toAppend);
}
現在我得到一個空字符串。 – user253530 2010-04-08 04:51:50