2012-11-05 24 views
0

我有一個這個函數在onClick發生時被觸發+使用ref。jQuery,檢查.detach變量是否爲空或已分離數據存儲?

我設法讓.detach成功工作。

但是當我的函數第二次運行時,我想檢查detachedMini是否實際上包含任何東西,以便我可以重新附加它,然後再拆下下一批。

bikeFilter = function (y) { 

    var detachedMini, 
     detachedMain; 

    bikeDetach = function() { 

     detachedMini = $('#bike-mini-slider .bike[data-group!=' + y + ']'); 
     detachedMain = $('#bike-main-slider .bike[data-group!=' + y + ']'); 

     detachedMini.detach(); 
     detachedMain.detach(); 

    }; 

    if (!detachedMini == 0) { 

     $('#bike-mini-slider .bike').append(detachedMini); 
     $('#bike-main slider .bike').append(detachedMain); 

    }      

    bikeDetach(); 

}; 

我已經試過這...

if (!detachedMini == 0) { 

但我不認爲它的正常工作。有任何想法嗎?

回答

2

使用jQuery對象的length屬性:

if (detachedMini.length) { 
    $('#bike-mini-slider .bike').append(detachedMini); 
    $('#bike-main slider .bike').append(detachedMain); 
} 
+0

嗯謝謝 - 我得到未定義的錯誤長度,所以也許我已經我的範圍混淆了。 – Joshc

0

嘗試

if (detachedMini.length != 0) 
0

如何:

bikeFilter = function (y) { 

    var detachedMini = null, 
     detachedMain = null; 

    bikeDetach = function() { 

     detachedMini = $('#bike-mini-slider .bike[data-group!=' + y + ']').detach(); 
     detachedMain = $('#bike-main-slider .bike[data-group!=' + y + ']').detach(); 

    }; 

    if (typeof(detachedMini) != "null") { 

     $('#bike-mini-slider .bike').append(detachedMini); 
     $('#bike-main slider .bike').append(detachedMain); 

    }      

    bikeDetach(); 

};