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;
}
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().
你可以使用正则表达式。匹配
if(text.matches("\\d*")&& text.length() > 2){
System.out.println("number");
}
或者你可以使用像Integer.parseInt(String)或更好的Long.parseLong(String)这样的onversions来获得更大的数字
比如:
private boolean onlyContainsNumbers(String text) {
try {
Long.parseLong(text);
return true;
} catch (NumberFormatException ex) {
return false;
}
}
然后用:
if (onlyContainsNumbers(text) && text.length() > 2) {
// do Stuff
}
