Java-查看字符串中是否包含任何字符 [英] Java - See if a string contains any characters in it

问题描述

我遇到的问题是，当我检查字符串是否包含任何字符时，它只会查看第一个字符而不是整个字符串。例如，我希望能够输入 123abc，并且可以识别字符，因此失败。我还需要将字符串的长度设置为11个字符，并且由于我的程序只能使用1个字符，因此不能再进行任何操作了。

The problem i'm having is when i check to see if the string contains any characters it only looks at the first character not the whole string. For instance I would like to be able to input "123abc" and the characters are recognized so it fails. I also need the string to be 11 characters long and since my program only works with 1 character it cannot go any further.

到目前为止，这是我的代码：

Here is my code so far:

public static int phoneNumber(int a)
{
   while (invalidinput)
            {
        phoneNumber[a] = myScanner.nextLine();

        if (phoneNumber[a].matches("[0-9+]") && phoneNumber[a].length() == 11 )
        {   
            System.out.println("Continue");
            invalidinput = false;
        }
        else
        {
            System.out.print("Please enter a valid phone number: ");
        }

    }

    return 0;
}

例如为什么我不带支票去查看phoneNumber.length（ ）它仍然只注册1个字符，因此如果我输入 12345，它仍然会失败。我只能输入 1，程序才能继续。

For instance why if i take away the checking to see the phoneNumber.length() it still only registers 1 character so if i enter "12345" it still fails. I can only enter "1" for the program to continue.

如果有人可以向我解释这是如何工作的，那就太好了

If someone could explain how this works to me that would be great

推荐答案

您的正则表达式和如果条件错误。像这样使用它：

Your regex and if condition is wrong. Use it like this:

 if ( phoneNumber[a].matches("^[0-9]{11}$") ) {
    System.out.println("Continue");
    invalidinput = false;
 }

这将只允许 phoneNumber [a] 为11个字符，仅包含数字 0-9

This will only allow phoneNumber[a] to be a 11 character long comprising only digits 0-9

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