使用位置从数组中获取字符串值

Question

我有类似[1,2,3,4]的职位

    String checkedids = String.valueOf(group.getCheckedIds()); //group.getCheckedIds() returns Set<Integer> positions basically

for (int i = 0; i < checkedids.length(); i++) {
    if (checkedids.contains("1")) {
        if (!TextUtils.isEmpty(valuez)) {
            valuez = valuez + "Fashion";
        } else {
            valuez = "Fashion";
        }
    } else if (checkedids.contains("2")) {
        if (!TextUtils.isEmpty(valuez)) {
            valuez = valuez + "Lifestyle";
        } else {
            valuez = "Lifestyle";
        }
    }l

}

<?xml version="1.0" encoding="utf-8"?>
<resources>

    <string-array name="dummy_text">
        <item>Fashion</item>
        <item>Lifestyle</item>
        <item>Beauty and Makeup</item>
        <item>Parenting and Kids</item>
        <item>Pets</item>
    </string-array>

</resources>

Answer 1

你需要清楚这里的各种概念。在编写代码之前要清楚自己编写的代码。我会给你正确的答案，但你必须推断出你做错了什么。

您可以直接对ID的整数集进行迭代。String.ValueOf的用法在这里似乎不对。

Set<Integer> checkedIds = group.getCheckedIds();
String[] dummyArrayValues = getResources().getStringArray(R.array.dummy_text);

for(String id: checkedIds){
    if(id > 0 && id <= dummyArrayValues.length){
         if (!TextUtils.isEmpty(valuez))
            valuez = valuez +" "+ dummyArrayValues[id-1]; //Array indexing is zero based so using id-1 to get the correct position according to your example
         else
            valuez = dummyArrayValues[id-1];
    }
}