Question

我正在尝试在Java中创建一个方法，该方法将识别一行文本中的非字母数字符号并将其删除。但是，我可以使用的工具有限。我必须使用循环和扫描程序来读取文本，并且只使用charAt和substring方法来删除有问题的字符。这是我提出的代码

public String sanitize2(){
        Scanner sanitizer = new Scanner(input.getText());
        String toBeSanitized = sanitizer.nextLine();
        int length = toBeSanitized.length();
        for (int i=0;i < toBeSanitized.length(); i++){
            if ((toBeSanitized.charAt(i) >= 'a' && toBeSanitized.charAt(i) <= 'z') ||
                (toBeSanitized.charAt(i) >= 'A' && toBeSanitized.charAt(i) <= 'Z') ||
                (toBeSanitized.charAt(i) >= '0' && toBeSanitized.charAt(i) <= '9')) 
            {
                   toBeSanitized = toBeSanitized.substring(0,i) + toBeSanitized.substring(i+1);
            }
        }
        input.setText(toBeSanitized);
        return toBeSanitized;
    }

但是当我在输入字段中插入文本时，它完全给出了错误的答案。任何人都能明白我在这里做错了吗？

Answer 1

如果您不需要使用所有给定的工具，那么我认为这可能有助于解决您的问题。

    String toBeSanitized = "(*this*)* text is ^0nly for t3st!ng!";
    String sanitizedStr = "";
    int length = toBeSanitized.length();
    for (int i=0;i < toBeSanitized.length(); i++){
        if ((toBeSanitized.charAt(i) >= 'a' && toBeSanitized.charAt(i) <= 'z') ||
        (toBeSanitized.charAt(i) >= 'A' && toBeSanitized.charAt(i) <= 'Z') ||
        (toBeSanitized.charAt(i) >= '0' && toBeSanitized.charAt(i) <= '9') || 
        toBeSanitized.charAt(i) == ' ') // optional
        {
            sanitizedStr += toBeSanitized.charAt(i);
        }
    }

    System.out.println(sanitizedStr);

我还包括空格，但如果你愿意，可以删除它。希望它有所帮助。

在Java中对单个字符进行子串

1 个答案: