确定字符串是否是Java中的整数[重复]

比眉伴天荒 2022-11-26 02:59 144阅读 0赞

本文翻译自：[Determine if a String is an Integer in Java \[duplicate\]][Determine if a String is an Integer in Java _duplicate]

*This question already has an answer here:* **这个问题已经在这里有了答案：**

*  *What's the best way to check if a String represents an integer in Java?* **在Java中检查字符串是否表示整数的最佳方法是什么？** *36 answers* **36个答案**

*I'm trying to determine if a particular item in an Array of strings is an integer or not.* **我正在尝试确定字符串数组中的特定项目是否为整数。**

*I am `.split(" ")'ing` an infix expression in `String` form, and then trying to split the resultant array into two arrays;* **我`.split(" ")'ing`以`String`形式插入一个中缀表达式，然后尝试将结果数组拆分为两个数组；** *one for integers, one for operators, whilst discarding parentheses, and other miscellaneous items.* **一个用于整数，一个用于运算符，同时丢弃括号和其他杂项。** *What would be the best way to accomplish this?* **做到这一点的最佳方法是什么？**

*I thought I might be able to find a `Integer.isInteger(String arg)` method or something, but no such luck.* **我以为我可以找到`Integer.isInteger(String arg)`方法或其他方法，但是没有这种运气。**

--------------------

### \#1楼 ###

参考：[https://stackoom.com/question/mP4l/确定字符串是否是Java中的整数-重复][https_stackoom.com_question_mP4l_Java_-]

--------------------

### \#2楼 ###

*Or simply* **或者简单地**

`mystring.matches("\\\\d+")`

*though it would return true for numbers larger than an int* **尽管对于大于int的数字将返回true**

--------------------

### \#3楼 ###

*Using regular expression is better.* **使用正则表达式更好。**

str.matches("-?\\d+");
    
    
    -?     --> negative sign, could have none or one
    \\d+   --> one or more digits

*It is not good to use `NumberFormatException` here if you can use `if-statement` instead.* **如果可以改为使用`if-statement`在此处使用`NumberFormatException`不好。**

--------------------

*If you don't want leading zero's, you can just use the regular expression as follow:* **如果您不想以零开头，则可以使用正则表达式，如下所示：**

str.matches("-?(0|[1-9]\\d*)");

--------------------

### \#4楼 ###

public boolean isInt(String str){
        return (str.lastIndexOf("-") == 0 && !str.equals("-0")) ? str.substring(1).matches(
                "\\d+") : str.matches("\\d+");
    }

--------------------

### \#5楼 ###

*You want to use the [Integer.parseInt(String)][Integer.parseInt_String] method.* **您要使用[Integer.parseInt（String）][Integer.parseInt_String]方法。**

try{
      int num = Integer.parseInt(str);
      // is an integer!
    } catch (NumberFormatException e) {
      // not an integer!
    }

--------------------

### \#6楼 ###

*You can use `Integer.parseInt()` or `Integer.valueOf()` to get the integer from the string, and catch the exception if it is not a parsable int.* **您可以使用`Integer.parseInt()`或`Integer.valueOf()`从字符串中获取整数，如果它不是可解析的int，则捕获该异常。** *You want to be sure to catch the `NumberFormatException` it can throw.* **您要确保捕获到它可能引发的`NumberFormatException` 。**

*It may be helpful to note that valueOf() will return an Integer object, not the primitive int.* **注意valueOf（）将返回一个Integer对象，而不是原始int可能会有所帮助。**

[Determine if a String is an Integer in Java _duplicate]: https://oldbug.net/q/mP4l/Determine-if-a-String-is-an-Integer-in-Java-duplicate
[https_stackoom.com_question_mP4l_Java_-]: https://stackoom.com/question/mP4l/%E7%A1%AE%E5%AE%9A%E5%AD%97%E7%AC%A6%E4%B8%B2%E6%98%AF%E5%90%A6%E6%98%AFJava%E4%B8%AD%E7%9A%84%E6%95%B4%E6%95%B0-%E9%87%8D%E5%A4%8D
[Integer.parseInt_String]: http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html#parseInt%28java.lang.String%29