問題描述

這段 Java 代碼給我帶來了麻煩:

This Java code is giving me trouble:

    String word = <Uses an input>
    int y = 3;
    char z;
    do {
        z = word.charAt(y);
         if (z!='a' || z!='e' || z!='i' || z!='o' || z!='u')) {
            for (int i = 0; i==y; i++) {
                wordT  = wordT + word.charAt(i);
                } break;
         }
    } while(true);

我想檢查單詞的第三個字母是否是非元音，如果是，我希望它返回非元音及其前面的任何字符.如果是元音，則檢查字符串中的下一個字母，如果也是元音，則檢查下一個字母，直到找到非元音為止.

I want to check if the third letter of word is a non-vowel, and if it is I want it to return the non-vowel and any characters preceding it. If it is a vowel, it checks the next letter in the string, if it's also a vowel then it checks the next one until it finds a non-vowel.

例子:

word = Jaemeas 然后 wordT 必須 = Jaem

word = Jaemeas then wordT must = Jaem

示例 2:

word=Jaeoimus 然后 wordT 必須 =Jaeoim

word=Jaeoimus then wordT must =Jaeoim

問題在于我的 if 語句，我不知道如何讓它檢查那一行中的所有元音.

The problem is with my if statement, I can't figure out how to make it check all the vowels in that one line.

推薦答案

你的情況有問題.想想更簡單的版本

Your condition is flawed. Think about the simpler version

z != 'a' || z != 'e'

如果 z 是 'a' 則后半部分為真，因為 z 不是 'e' (即整個條件為真)，如果 z 為 'e' 則前半部分為真，因為 z 不是 'a' (同樣，整個條件為真).當然，如果 z 既不是 'a' 也不是 'e' 那么這兩個部分都為真.也就是說，你的條件永遠不會是假的！

If z is 'a' then the second half will be true since z is not 'e' (i.e. the whole condition is true), and if z is 'e' then the first half will be true since z is not 'a' (again, whole condition true). Of course, if z is neither 'a' nor 'e' then both parts will be true. In other words, your condition will never be false!

你可能想要 && 代替:

z != 'a' && z != 'e' && ...

或許:

"aeiou".indexOf(z) < 0

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