编写一种方法来删除 Java 字符串中的元音
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原文地址: http://stackoverflow.com/questions/19329307/
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Writing a method to remove vowels in a Java String
提问by user2864813
I am a beginner of programming, and am writing a Java method to remove vowel in Strings, but I do not know how to fix this error: ";" expected:
我是编程初学者，正在编写 Java 方法来删除字符串中的元音，但我不知道如何解决此错误";" expected：：
public String disemvowel(String s) {
boolean isVowel(char c);
if (c == 'a') {
return true;
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}
回答by cmd
A much simpler approach would be to do the following:
一个更简单的方法是执行以下操作：
String string = "A really COOL string";
string = string.replaceAll("[AaEeIiOoUu]", "");
System.out.println(string);
This will apply the regular expression, [AaEeIiOoUu]to string. This expression will match all vowels in the character group [AaEeIiOoUu]and replace them with ""empty string.
这会将正则表达式[AaEeIiOoUu]应用于string. 此表达式将匹配字符组中的所有元音[AaEeIiOoUu]并将它们替换为""空字符串。
回答by Vallabh Patade
String str= "Your String";
str= str.replaceAll("[AEIOUaeiou]", "");
System.out.println(str);
回答by Makoto
You've got a lotof syntax errors.
你有很多语法错误。
boolean isVowel(char c);- not sure what you're doing with this. if you want it as a separate method, separate it out (and don't place a semicolon after it, which would be invalid syntax.else if ifis invalid syntax. If you're doing anelse if, then you only need the oneif.- Even if the code would compile,
for (int z = 0; z <= l; z++)will cause you to step off of the String. Remove the<=in favor of<. isVowel == "false"is never going to work. You're comparing a String to a boolean. You want!isVowelinstead.
boolean isVowel(char c);- 不确定你在做什么。如果你想把它作为一个单独的方法，把它分开（不要在它后面放分号，这将是无效的语法。else if if是无效的语法。如果你正在做一个else if，那么你只需要一个if。- 即使代码会编译，
for (int z = 0; z <= l; z++)也会导致您离开字符串。删除<=支持<. isVowel == "false"永远不会工作。您正在将字符串与布尔值进行比较。你想要!isVowel。
Putting the syntax errors aside, think of it like this.
抛开语法错误，这样想。
- You have a string that contains vowels. You wish to have a string that doesn't contain vowels.
- The most straightforward approach is to iterate over the String, placing all non-vowel characters into a separate String, which you then return.
- 您有一个包含元音的字符串。您希望有一个不包含元音的字符串。
- 最直接的方法是遍历字符串，将所有非元音字符放入一个单独的字符串中，然后返回该字符串。
Interestingly enough, the half-method you have there can accomplish the logic of determining whether something is or isn't a vowel. Extract that to its own method. Then, call itin your other method. Do take into account capital letters though.
有趣的是，您在那里使用的半方法可以完成确定某事物是否为元音的逻辑。将其提取到它自己的方法中。然后，在您的其他方法中调用它。不过要考虑大写字母。
I leave the rest as an exercise to the reader.
我把剩下的留给读者作为练习。
回答by Josh M
You could try something like this:
你可以尝试这样的事情：
public static String removeVowels(final String string){
final String vowels = "AaEeIiOoUu";
final StringBuilder builder = new StringBuilder();
for(final char c : string.toCharArray())
if(vowels.indexOf(c) < 0)
builder.append(c);
return builder.toString();
}
回答by Hot Licks
Here is your code, without changing any logic, but unscrambling the isVowel method:
这是您的代码，没有更改任何逻辑，但对 isVowel 方法进行了解读：
public String disemvowel(String s) {
// Removed the "isVowel" method from here and moved it below
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
// Note that the "isVowel" method has not been called.
// And note that, when called, isVowel returns a boolean, not a String.
// (And note that, as a general rule, you should not compare strings with "==".)
// So this area needs a lot of work, but we'll start with this
boolean itIsAVowel = isVowel(s.charAt(z));
// (I made the variable name "itIsAVowel" to emphasize that it's name has nothing to do with the method name.
// You can make it "isVowel" -- the same as the method -- but that does not in any way change the function.)
// Now take it from there...
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}
// You had this line ending with ";"
boolean isVowel(char c) {
if (c == 'a') {
return true;
// Note that you coded "if if" on the lines below -- there should be only one "if" per line, not two
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
// You were missing this final return
return false;
}
(Yes, I know this should be a comment, but you can't put formatted code in a comment.)
（是的，我知道这应该是注释，但是您不能将格式化的代码放在注释中。）
