问题描述
我在整个应用程序中加载了一个字符串，它从数字变为字母等等.我有一个简单的 if 语句来查看它是否包含字母或数字，但是有些东西不能正常工作.这是一个片段.
I have a string that I load throughout my application, and it changes from numbers to letters and such. I have a simple if statement to see if it contains letters or numbers but, something isn't quite working correctly. Here is a snippet.
String text = "abc";
String number;
if (text.contains("[a-zA-Z]+") == false && text.length() > 2) {
number = text;
}
尽管 text 变量确实包含字母，但条件返回为 true.和 && 应该评估为两个条件都必须为 true 才能处理 number = text;
Although the text variable does contain letters, the condition returns as true. The and && should eval as both conditions having to be true in order to process the number = text;
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解决方案:
我能够通过使用此问题的评论提供的以下代码来解决此问题.所有其他帖子也有效！
I was able to solve this by using this following code provided by a comment on this question. All other post are valid as well!
我使用的有用的东西来自第一条评论.尽管提供的所有示例代码似乎也有效！
What I used that worked came from the first comment. Although all the example codes provided seems to be valid as well!
String text = "abc";
String number;
if (Pattern.matches("[a-zA-Z]+", text) == false && text.length() > 2) {
number = text;
}
推荐答案
如果您要将数字作为文本处理，请更改:
If you'll be processing the number as text, then change:
if (text.contains("[a-zA-Z]+") == false && text.length() > 2){
到:
if (text.matches("[0-9]+") && text.length() > 2) {
与其检查字符串是否不包含字母字符，而是检查以确保它仅包含数字.
Instead of checking that the string doesn't contain alphabetic characters, check to be sure it contains only numerics.
如果您确实想使用数值，请使用 Integer.parseInt() 或 Double.parseDouble() ，正如其他人在下面解释的那样.
If you actually want to use the numeric value, use Integer.parseInt() or Double.parseDouble() as others have explained below.
作为旁注，将布尔值与 true 或 false 进行比较通常被认为是不好的做法.只需使用 if (condition) 或 if (!condition).
As a side note, it's generally considered bad practice to compare boolean values to true or false. Just use if (condition) or if (!condition).
这篇关于Java String - 查看一个字符串是否只包含数字而不包含字母的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，WP2