Java 使用 parse.string 返回字符串输入

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/19202622/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-12 15:00:01  来源:igfitidea点击:

Return string Input with parse.string

javastringjoptionpane

提问by Makominami

I'm having an issue with a java program trying to get a string input from a joptionpane menu with a prompt box. With returning a string input. I don't know if im going about it all wrong by trying to use

我在尝试从带有提示框的 joptionpane 菜单中获取字符串输入的 java 程序遇到问题。返回一个字符串输入。我不知道我是否通过尝试使用来解决所有问题

String.parseString(input)

Im very much a beginner with this so any help would have to be as simple as possible or a correction outright.

我是一个初学者,所以任何帮助都必须尽可能简单或彻底纠正。

   private static String getStringInput (String prompt) {
         String input = EZJ.getUserInput(prompt);
         return String.parseString(input);
   }


UseCalls.java:27: error: cannot find symbol
         return String.parseString(input);
                      ^
 symbol:   method parseString(String)
 location: class String
 1 error

Here is a sample of the menu Im trying to use it with

这是我尝试使用的菜单示例

    do {

        userInput = mainMenu();

        if (userInput.equals("1")) {
            String name = getStringInput("Name?");
            String address = getStringInput("Address?");
            call[numCalls++] = new Call();
        }
        } while (!userInput.equals("0"));


}

Here is the EZJ mini method

这是 EZJ 迷你方法

public class EZJ {

public static String getUserInput (String prompt) {
    return JOptionPane.showInputDialog(prompt);
}
public static void dialog(String inputValue) {
    JOptionPane.showMessageDialog ( null, inputValue );
}

}

采纳答案by Computoguy

You don't need to parse the string, it's defined as a string already.

您不需要解析字符串,它已经定义为字符串。

Just do:

做就是了:

    private static String getStringInput (String prompt) {
     String input = EZJ.getUserInput(prompt);
     return input;
    }

回答by Valerij Bielskij

As you see in an error UseCalls.java:27: error: cannot find symbol return String.parseString(input);there is no method parseStringin Stringclass. There is no need to parse it as long as JOptionPane.showInputDialog(prompt);already returns a string.

正如您在错误中看到的那样,类中UseCalls.java:27: error: cannot find symbol return String.parseString(input);没有方法。只要已经返回一个字符串就不需要解析它。parseStringStringJOptionPane.showInputDialog(prompt);

回答by sahil dhawan

If you're really bent upon converting Integer to String value, I suggest use String.valueOf(YourIntegerVariable). More details can be found at: http://www.tutorialspoint.com/java/java_string_valueof.htm

如果您真的很想将整数转换为字符串值,我建议使用 String.valueOf(YourIntegerVariable)。可以在以下位置找到更多详细信息:http: //www.tutorialspoint.com/java/java_string_valueof.htm