if-statement

问题 I'm trying to do the below tutorial question. // Create a method called greatestCommonFactor // It should return the greatest common factor // between two numbers. // // Examples of greatestCommonFactor: // greatestCommonFactor(6, 4) // returns 2 // greatestCommonFactor(7, 9) // returns 1 // greatestCommonFactor(20, 30) // returns 10 // // Hint: start a counter from 1 and try to divide both // numbers by the counter. If the remainder of both divisions // is 0, then the counter is a common

问题 In bash, if I run (foo=14) And then try to reference that variable later on in my bash script: echo "${foo}" I don't get anything. How can I make bash store this variable the way I need it to? Specifically, I am using this in an if statement and checking the exit code, something kind of like: if (bar="$(foo=14;echo "${foo}"|tr '1' 'a' 2>&1)") then echo "Setting "'$bar'" was a success. It is ${bar}" else echo "Setting "'$bar'" failed with a nonzero exit code." fi 回答1: Commands enclosed in

问题 i facing issue my If statement returning always true, for each value. var radioValue = parseInt($("input[name=packageRadio]:checked").val()) alert(radioValue); // 1000, 500, -1 if ((radioValue) == -1); { alert("no"); } for every value alert("no") is always calling. 回答1: Remove semicolon after if if ((radioValue) == -1){ alert("no"); } 回答2: Remove semicolon ; after if .And also remove braces () rounded around radioValue . Like following if (radioValue == -1){ alert("no"); } 回答3: When

问题 I know that you can easily sort numbers with an array, but my assignment for class is that I need to sort four numbers in descending order using if statements and not arrays . Here is my code so far: package integersort; import java.util.Scanner; public class IntegerSort { public static void main(String[] args) { Scanner userInput = new Scanner(System.in); int firstNum, secondNum, thirdNum, fourthNum; //inputted numbers System.out.println("Enter first number:"); firstNum = userInput.nextInt()

问题 I understand from reading a couple of threads that IFERROR in Excel 2003 is preformed by an IF(ISERROR) . The code I am trying to replicate from Excel 2007+ is: =IFERROR(A1, "") I know there must be a simple solution to translating this to IF(ISERROR) , but I have had no avail. Advice would be greatly received. 回答1: The Excel 2003 equivalent of =IFERROR(A1,"") is =IF(ISERROR(A1),"") . See =IS. But neither version of that formula is very useful. A little more useful would be something like =IF

问题 My code is: data$E<-ifelse(data$D==data$B, "b", ifelse(data$D==data$C, "c", "unknowwn")) result: A B C D E 1 16 16 NA 16 b 2 20 NA 20 20 NA 3 24 NA NA 24 NA But what I want is: A B C D E 1 16 16 NA 16 b 2 20 NA 20 20 c 3 24 NA NA 24 unknowwn Dose anyone know how to solve this problem? Thanks! 回答1: data$E <- ifelse(!is.na(data$B) & data$D == data$B, "b", ifelse(!is.na(data$C) & data$D == data$C, "c", "unknowwn")) 回答2: Your condition is a bit complicated as you need to consider if either of

问题 I try to create unit test cases to check my table values are correct or wrong. This is my codes echo $a = 2/9; echo "<br>".$b=0.22222222222222; echo "<br>".gettype($a); echo "<br>".gettype($b); if($a==$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; if((string)$a==(string)$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; Why my first if condition not working? I can't find the reason. Please help me. 回答1: The test violates a cardinal rule of floating point programming: never

问题 I try to create unit test cases to check my table values are correct or wrong. This is my codes echo $a = 2/9; echo "<br>".$b=0.22222222222222; echo "<br>".gettype($a); echo "<br>".gettype($b); if($a==$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; if((string)$a==(string)$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; Why my first if condition not working? I can't find the reason. Please help me. 回答1: The test violates a cardinal rule of floating point programming: never

问题 I try to create unit test cases to check my table values are correct or wrong. This is my codes echo $a = 2/9; echo "<br>".$b=0.22222222222222; echo "<br>".gettype($a); echo "<br>".gettype($b); if($a==$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; if((string)$a==(string)$b){ echo "<br>". "equal"; }else echo "<br>". "Not equal"; Why my first if condition not working? I can't find the reason. Please help me. 回答1: The test violates a cardinal rule of floating point programming: never

问题 I hava a nested object like: let menu = { vegetarian: { vegStarter: { plainPizza: 100, redChilliPizza: 150, cheesePizza: 200, capsicumPizza: 160, onionPizza: 200, }, vegMainCourse: { pepperoniPizza: 150, mushroomsPizza: 160, extraCheesePizza: 250, blackOlivesPizza: 220, greenPeppersPizza: 180, } }, nonVegetarian: { nonVegStarter: { supremeMeatPizza: 100, meatPizza: 130, meatLoversPizza: 160, chickenPizza: 200, chilliMeatPizza: 200 }, nonVegMainCourse: { butterMeatPizza: 220, spicyChickenPizza