Question

def countVowels(str1):
    str1_array = str1.split(" ")
    vA = ["A", "a"]
    vE = ["E", "e"]
    vI = ["I", "i"]
    vO = ["O", "o"]
    vU = ["U", "u"]

    vA_count = 0
    for vA in str1_array:
        vA_count = vA_count + 1
    vE_count = 0
    for vE in str1_array:
        vE_count = vE_count + 1
    vI_count = 0
    for vI in str1_array:
        vI_count = vI_count + 1    
    vO_count = 0
    for vO in str1_array:
        vO_count = vO_count + 1    
    vU_count = 0
    for vU in str1_array:
        vU_count = vU_count + 1

    print("A, E, I, O, and U  appear, respectively, ", vA_count, vE_count, vI_count, vO_count, "and", vU_count, "times.")

onItemSelected的情况0内的条件“ if（height_spinner.getSelectedItemPosition（）== 1）”未得到执行，仅条件“ if（height_spinner.getSelectedItemPosition（）== 0）”已执行。
onItemSelected案例0内的条件“ if（height_spinner.getSelectedItemPosition（）== 1）”未得到执行，仅条件“ if（height_spinner.getSelectedItemPosition（）== 0）”已被执行。

请提出修改建议并解决我的问题

Answer 1

这是因为当您处于位置0（又称为情况0）时：

 @Override
        public void onItemSelected(AdapterView parent, View view, int position, long id) {
            switch (position){
                case 0:
                    weight_et.setHint("Weight(kg)");
                    weight_et1.setVisibility(View.GONE);
                    weight_et1.setEnabled(false);
                    weight_et.setEnabled(true);

                    if (height_spinner.getSelectedItemPosition()==0) {...}

代码：height_spinner.getSelectedItemPosition()等于0（在位置1，它将等于1，依此类推）

因此条件：height_spinner.getSelectedItemPosition() == 1不会在情况0时被调用

条件“如果（if（height_spinner.getSelectedItemPosition（）== 1）”未得到执行

1 个答案: