如何傳遞字符串數組中的字符串資源引用android

我想傳入字符串數組中的字符串資源值，但它沒有在數組中取值。將採取空值如何傳遞字符串數組中的字符串資源引用android

下面是我的代碼：

prepaid = getResources().getString(R.string.lbl_prepaid); 
dth = getResources().getString(R.string.lbl_dth); 

public String[] addedRTHomeList(){ 
     String[] homeList = {prepaid,dth,"PostPaid","Utility Services","Complaint Entry","Complaint Status","Recharge Status",chat,"Topup Request","Redeem Discount","Private Bus Booking","ST Bus Booking","Hotel Booking","Flight Booking","Money Transfer","Settings","Reports","DTH Activation"}; 
     return homeList; 
    } 


for(int i = 0; i < ba.addedRTHomeList().length ; i++) 
     { 
      md = new MenuDetail(); 
      md.setMenuName(ba.addedRTHomeList()[i]); 
      md.setImageId(ba.RTDrwableListThemeRed()[i]); 

      RTmenuListRed.add(md); 

      RTMenuThemeRed.put(ba.RTMenuCode()[i],RTmenuListRed); 
     }

在for循環，當我訪問addedRTHomeList數組的第一個值，它取空值

來源

2016-07-15 Varshil shah

爲什麼你不想使用<字符串數組>資源？ –

你不能這個'homeList = {prepaid，dth}'你只能用Array.combine –

Thanx合併..如果可能請分享例子.. –

取而代之的是，你可以在創建<string-array>你的string.xml。像，

<string-array name="your_array_name"> 
    <item>Element 1</item> 
    <item>Element 2</item> 
    <item>Element 3</item> 
    <item>Element 4</item> 
    . 
    . 
    . 

</string-array>

在這之後，你可以在Java中獲得的，

String[] mTestArray = = getResources().getStringArray(R.array.your_array_name);

這或許有助於你的問題。

來源

2016-07-15 05:37:25

使用AarryList而不是簡單的數組因爲arraylist具有簡單的動態實現。

當你創建數組時你已經告訴了它的長度。

public String[] addedRTHomeList(){ 
     String[] homeList = {prepaid,dth,"PostPaid","Utility Services","Complaint Entry","Complaint Status","Recharge Status",chat,"Topup Request","Redeem Discount","Private Bus Booking","ST Bus Booking","Hotel Booking","Flight Booking","Money Transfer","Settings","Reports","DTH Activation"}; 
     return homeList; 
    }

像上面homeList是17大小，你不能在18指數增加值，因爲18是不存在的，之所以你收到此錯誤

解決方案：

使用字符串ArrayList的例子使用

ArrayList<String> homeList = new ArrayList<String>(); 

homeList.add("PostPaid"); 
homeList.add("Complaint Status"); 
homeList.add("Recharge Status"); 

. 
. 
}

從ArrayList中獲取值homeList.get(index)簡單地說。

現在，如果您添加了值，它將創建新的索引並將其添加到最後。運氣最好。

來源

2016-07-15 05:39:47

由於您試圖將變量初始化爲類變量，您將收到異常。我在addedRTHomeList()方法裏面移動了初始化，發現沒有問題。

試試這個代碼，

public String[] addedRTHomeList(){ 
    String prepaid = getResources().getString(R.string.lbl_prepaid); 
    String dth = getResources().getString(R.string.lbl_dth); 
    String chat = getResources().getString(R.string.lbl_chat); 

    return new String[]{prepaid,dth,"PostPaid","Utility Services","Complaint Entry","Complaint Status","Recharge Status",chat,"Topup Request","Redeem Discount","Private Bus Booking","ST Bus Booking","Hotel Booking","Flight Booking","Money Transfer","Settings","Reports","DTH Activation"}; 
} 

String[] items = ba.addedRTHomeList(); 
for(int i = 0; i < items.length ; i++){ 
    md = new MenuDetail(); 
    md.setMenuName(items[i]); 
    md.setImageId(ba.RTDrwableListThemeRed()[i]); 

    RTmenuListRed.add(md); 

    RTMenuThemeRed.put(ba.RTMenuCode()[i],RTmenuListRed); 
}

來源

2016-07-15 05:41:34

它不是沒有工作 –

沒有錯誤..它採取空 –

不，在string.xml中值不爲空.. –

如何傳遞字符串數組中的字符串資源引用android

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