Java 如果语句未通过,则启用Else

Java 如果语句未通过,则启用Else,java,if-statement,calculator,Java,If Statement,Calculator,因此,我是一名noob编码员(网站新手),试图自学,目前我正在制作一个简单的计算器。在Atm中,我用一个检查输入的if语句向它添加了pi。然而,当我输入一个double时,它没有任何作用,我也不知道为什么 package testing; import java.util.Scanner; public class Calculator { @SuppressWarnings("resource") public static void main(String[] args) {

因此,我是一名noob编码员(网站新手),试图自学,目前我正在制作一个简单的计算器。在Atm中,我用一个检查输入的if语句向它添加了pi。然而,当我输入一个double时,它没有任何作用,我也不知道为什么

package testing;
import java.util.Scanner;

public class Calculator {


@SuppressWarnings("resource")
public static void main(String[] args) {
    String piCheck = "pi";
    Double n1 = null;

    Scanner reader = new Scanner(System.in);

    System.out.println("Welcome to Calculator!");
    System.out.println(" ");

    System.out.println("Enter a number, ");
    System.out.println("or enter 'pi' to insert PI,");
    if (reader.nextLine().toLowerCase().contains(piCheck)) {
        n1 = Math.PI;
    } else {
        n1 = reader.nextDouble();
    }

    Scanner reader2 = new Scanner(System.in);
    System.out.println("Would you like to multiply, divide, add, subtract?");
    String uOperation = reader2.nextLine();


    int operation = 1;
    switch (uOperation.toLowerCase()) {
        case "multiply":
            operation = 1;
            break;

        case "multiplication":
            operation = 1;
            break;

        case "times":
            operation = 1;
            break;

        case "divide":
            operation = 2;
            break;

        case "division":
            operation = 2;
            break;

        case "add":
            operation = 3;
            break;

        case "addition":
            operation = 3;
            break;

        case "subtract":
            operation = 4;
            break;

        case "subtraction":
            operation = 4;
            break;

        default:
            System.out.println("Invalid operation! Rerun Calculator!");
            break;

    }


    if (operation == 1) {

        System.out.println("What do you want to multiply " + n1 + " by?");

    } else if (operation == 2) {

        System.out.println("What do you want to divide " + n1 + " by?");

    } else if (operation == 3) {

        System.out.println("What do you want to add to " + n1 + "?");

    } else if (operation == 4) {

        System.out.println("What do you want to subtract from " + n1 + "?");

    } else {
        System.out.println("Something went wrong!");
    }
    double n2 = reader.nextDouble();


    System.out.println("Here's your result!");
    double resultM = n1*n2;
    double resultD = n1/n2;
    double resultA = n1+n2;
    double resultS = n1-n2;

    if (operation == 1) {

        System.out.println(n1 + " multiplied by " + n2 + " equals:");
        System.out.println(resultM);

    } else if (operation == 2) {

        System.out.println(n1 + " divided by " + n2 + " equals:");
        System.out.println(resultD);

    } else if (operation == 3) {

        System.out.println(n1 + " plus " + n2 + " equals:");
        System.out.println(resultA);

    } else if (operation == 4) {

        System.out.println(n1 + " take away " + n2 + " equals:");
        System.out.println(resultS);

    } else {
        System.out.println("Something went wrong!");
    }
}

}您的问题是您正在读取条件中的输入,因此如果您输入一个double,else子句希望输入更多的输入,因为第一个输入已经被
reader.nextLine()使用。

您必须保存输入并重新使用它:

String input = reader.nextLine();
if (input.toLowerCase().contains(piCheck)) {
    n1 = Math.PI;
} else {
    n1 = Double.parseDouble(input);
}

您的问题是,您正在读取条件中的输入,因此如果您输入一个double,else子句希望输入更多的输入,因为第一个输入已经被
reader.nextLine()
使用

您必须保存输入并重新使用它:

String input = reader.nextLine();
if (input.toLowerCase().contains(piCheck)) {
    n1 = Math.PI;
} else {
    n1 = Double.parseDouble(input);
}

您已经在通过执行
reader.nextLine()读取输入。您只需将其解析为double,而不是执行
n1=reader.nextDouble()。欢迎使用stackoverflow!看起来你可以从你的问题中删减很多,这个问题仍然可以重现。这有助于更快地理解您的问题。请阅读。另外:你的缩进看起来有点不对劲,请修复这些东西,这样你就不会冒问题被否决的风险了!您已经在通过执行
reader.nextLine()读取输入。您只需将其解析为double,而不是执行
n1=reader.nextDouble()。欢迎使用stackoverflow!看起来你可以从你的问题中删减很多,这个问题仍然可以重现。这有助于更快地理解您的问题。请阅读。另外:你的缩进看起来有点不对劲,请修复这些东西,这样你就不会冒问题被否决的风险了!