Java，确保字符串只包含字母数字、空格和破折号

Java, Make sure a String contains only alphanumeric, spaces and dashes

提问by Lancelot

In Java, I need to make sure a String only contains alphanumeric, spaceand dashcharacters.

在 Java 中，我需要确保 String 只包含alphanumeric,space和dash字符。

I found the class org.apache.commons.lang.StringUtilsand the almost adequate method isAlphanumericSpace(String)... but I also need to include dashes.

我找到了类org.apache.commons.lang.StringUtils和几乎足够的方法isAlphanumericSpace(String)……但我还需要包含破折号。

What is the best way to do this? I don't want to use Regular Expressions.

做这个的最好方式是什么？我不想使用正则表达式。

回答by Skip Head

You could use:

你可以使用：

StringUtils.isAlphanumericSpace(string.replace('-', ' '));

回答by Varkhan

Hum... just program it yourself using String.chatAt(int), it's pretty easy...

嗯...只需使用String.chatAt( int)自己编程，这很容易...

Iterate through all charin the string using a position index, then compare it using the fact that ASCII characters 0 to 9, a to z and A to Z use consecutive codes, so you only need to check that character x numerically verifies one of the conditions:

使用位置索引遍历字符串中的所有字符，然后使用 ASCII 字符 0 到 9、a 到 z 和 A 到 Z 使用连续代码的事实进行比较，因此您只需要检查字符 x 以数字方式验证其中一个条件：

• between '0' and '9'
• between 'a' and 'z'
• between 'A and 'Z'
• a space ' '
• a hyphen '-'

• 在“0”和“9”之间
• 'a' 和 'z' 之间
• 在“A”和“Z”之间
• 空间 ' '
• 连字符“-”

Here is a basic code sample (using CharSequence, which lets you pass a String but also a StringBuilder as arg):

这是一个基本的代码示例（使用 CharSequence，它允许您传递一个 String 和一个 StringBuilder 作为 arg）：

public boolean isValidChar(CharSequence seq) {
    int len = seq.length();
    for(int i=0;i<len;i++) {
        char c = seq.charAt(i);
        // Test for all positive cases
        if('0'<=c && c<='9') continue;
        if('a'<=c && c<='z') continue;
        if('A'<=c && c<='Z') continue;
        if(c==' ') continue;
        if(c=='-') continue;
        // ... insert more positive character tests here
        // If we get here, we had an invalid char, fail right away
        return false;
    }
    // All seen chars were valid, succeed
    return true;
}

回答by araqnid

Just iterate through the string, using the character-class methods in java.lang.Character to test whether each character is acceptable or not. Which is presumably all that the StringUtils methods do, and regular expressions are just a way of driving a generalised engine to do much the same.

只需遍历字符串，使用 java.lang.Character 中的字符类方法来测试每个字符是否可以接受。这大概是 StringUtils 方法所做的全部，而正则表达式只是驱动通用引擎执行相同操作的一种方式。

回答by MStodd

You have 1 of 2 options: 1. Compose a list of chars that CAN be in the string, then loop over the string checking to make sure each character IS in the list. 2. Compose a list of chars that CANNOT be in the string, then loop over the string checking to make sure each character IS NOT in the list.

您有 2 个选项中的 1 个： 1. 编写一个可以在字符串中的字符列表，然后循环检查字符串以确保每个字符都在列表中。2. 编写一个不能在字符串中的字符列表，然后循环检查字符串以确保每个字符都不在列表中。

Choose whatever option is quicker to compose the list.

选择任何可以更快地编写列表的选项。

回答by r3flss ExlUtr

Definitely use a regex expression. There's no point in writing your own system when a very comprehensive system in place for this exact task. If you need to learn about or brush up on regex then check out this website, it's great: http://regexr.com

绝对使用正则表达式。当有一个非常全面的系统来完成这个确切的任务时，编写自己的系统是没有意义的。如果您需要了解或复习正则表达式，请查看此网站，它很棒：http: //regexr.com

I would challenge yourself on this one.

我会在这个问题上挑战自己。