我在应用程序中加载了一个字符串，它可以从数字变成字母等等。我有一个简单的if语句，看看它是否包含字母或数字，但是，有些东西不太正确。下面是一个片段。
String text = "abc";
String number;
if (text.contains("[a-zA-Z]+") == false && text.length() > 2) {
number = text;
}
虽然文本变量包含字母，但条件返回为true。和&&应该eval作为两个条件都必须为真，以便处理number =文本;
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解决方案:
我能够通过使用以下代码来解决这个问题，该代码由对这个问题的评论提供。所有其他帖子也是有效的!
我使用的有效方法来自第一条评论。尽管提供的所有示例代码似乎也是有效的!
String text = "abc";
String number;
if (Pattern.matches("[a-zA-Z]+", text) == false && text.length() > 2) {
number = text;
}
亚当·博德罗吉的略有修改版本:
public class NumericStr {
public static void main(String[] args) {
System.out.println("Matches: "+NumericStr.isNumeric("20")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("20,00")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("30.01")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("30,000.01")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("-2980")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("$20")); // Should be true
System.out.println("Matches: "+NumericStr.isNumeric("jdl")); // Should be false
System.out.println("Matches: "+NumericStr.isNumeric("2lk0")); // Should be false
}
public static boolean isNumeric(String stringVal) {
if (stringVal.matches("^[\\$]?[-+]?[\\d\\.,]*[\\.,]?\\d+$")) {
return true;
}
return false;
}
}
今天不得不使用这个，所以刚刚发布了我的修改。包括货币、千位逗号或句号符号，以及一些验证。不包括其他货币符号(欧元、分)，验证逗号为每三位数。
This code is already written. If you don't mind the (extremely) minor performance hit--which is probably no worse than doing a regex match--use Integer.parseInt() or Double.parseDouble(). That'll tell you right away if a String is only numbers (or is a number, as appropriate). If you need to handle longer strings of numbers, both BigInteger and BigDecimal sport constructors that accept Strings. Any of these will throw a NumberFormatException if you try to pass it a non-number (integral or decimal, based on the one you choose, of course). Alternately, depending on your requirements, just iterate the characters in the String and check Character.isDigit() and/or Character.isLetter().
