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检查字符串是否只包含指定的字符

[英]check if a string only contains specified characters

I want to check that a string only contains letters, a-zA-Z , numbers 0-9 , and operators (+=-*/^) 我想检查一个字符串只包含字母, a-zA-Z ,数字0-9和运算符(+=-*/^)

For example, the following expression would be permissible: 3 + 5(x^2) 例如,以下表达式是允许的: 3 + 5(x^2)

For example, the following expression would not be permissible: 3 $+ 5(x^2) 例如,下面的表达式将容许: 3 $+ 5(x^2)

Would I use the matches function to do this? 我会使用匹配功能来执行此操作吗?

I tried: 我试过了:

// contains only operators, numbers, or letters
if(!exp.matches("[(+=-*/^)a-zA-Z0-9]")) { 
    return false;
}

I also tried escaping the asterisk, but it didn't work. 我也试过逃避星号,但它没有用。

You should have escaped - or put it at the start/end of the character class, else it creates a range . 您应该已经转义-或者将其放在角色类的开头/结尾,否则它会创建一个范围 Note that [=-*] range is invalid : 请注意, [=-*]范围无效

在此输入图像描述

Also, you need a quantifier, + to match 1 or more chars, * to match 0 or more. 此外,您需要一个量词, +匹配1个或多个字符, *匹配0或更多。

Use 采用

if(!exp.matches("[a-zA-Z0-9+=*/^()-]+")) { 
        return false;
}

If you do not need to match ( and ) , remove them from the character class. 如果您不需要匹配() ,请从字符类中删除它们。

Also, since the String#matches() requires a full string match, no anchors at the start and end of the regular expression are necessary. 此外,由于String#matches()需要完整的字符串匹配,因此不需要在正则表达式的开头和结尾处使用锚点。

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