if-statement

How to resolve “Syntax error on token ”else“” in Java?

人走茶凉 提交于 2020-01-10 02:15:06
问题 I am new around here but let's get straight to the question: When I was writing the following code for a class project calculator I came across a "token error". This is the full error message: Exception in thread "main" java.lang.Error: Unresolved compilation problem: Syntax error on token "else", delete this token at calculatorApplet.main(calculatorApplet.java:42) I wrote this code: import java.util.Scanner; import javax.swing.JOptionPane; public class calculatorApplet { Scanner sc = new

What is wrong with my excel formula syntax?

萝らか妹 提交于 2020-01-09 08:05:40
问题 I have made several formula to dynamically create a SQL query but to make it perfect, I want to add a condition : "if cell is not blank". I've tried a lot of different syntaxe but even with a very simple one (see exemple below), I get an error. My formula is : ="""" & SUBSTITUE(E2;"'";"''") & """" It works well and give me "Accordéoniste d''hier" I've tried =IF(E2<>"";"""" & SUBSTITUE(E2;"'";"''") & """";"") Excel display #NOM? in the cell but doesn't give me an explicit error. I've tried a

SASS ignores variables, defined in if-statement

限于喜欢 提交于 2020-01-09 07:17:01
问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

SASS ignores variables, defined in if-statement

别等时光非礼了梦想. 提交于 2020-01-09 07:15:47
问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

SASS ignores variables, defined in if-statement

谁说我不能喝 提交于 2020-01-09 07:15:09
问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

IF() statement alternative in SQLite

人走茶凉 提交于 2020-01-09 07:12:53
问题 I have the code for MySQL (perl): UPDATE pages SET rkey = rkey + 2, lkey = IF(lkey >= $key, lkey + 2, lkey) WHERE rkey >= $key I need to use this code with SQLite, but there is no support of IF() function. What I can do? 回答1: For generic SQL you can use CASE : CASE is used to provide if-then-else type of logic to SQL. Its syntax is: SELECT CASE ("column_name") WHEN "condition1" THEN "result1" WHEN "condition2" THEN "result2" ... [ELSE "resultN"] END FROM "table_name" From http://www.sqlite

Which SQL statement is faster? (HAVING vs. WHERE…)

断了今生、忘了曾经 提交于 2020-01-09 04:38:11
问题 SELECT NR_DZIALU, COUNT (NR_DZIALU) AS LICZ_PRAC_DZIALU FROM PRACOWNICY GROUP BY NR_DZIALU HAVING NR_DZIALU = 30 or SELECT NR_DZIALU, COUNT (NR_DZIALU) AS LICZ_PRAC_DZIALU FROM PRACOWNICY WHERE NR_DZIALU = 30 GROUP BY NR_DZIALU 回答1: The theory (by theory I mean SQL Standard) says that WHERE restricts the result set before returning rows and HAVING restricts the result set after bringing all the rows. So WHERE is faster. On SQL Standard compliant DBMSs in this regard, only use HAVING where you

Validate date format in a shell script

為{幸葍}努か 提交于 2020-01-08 17:42:27
问题 I have to create a Shell Script wherein one of the parameters will be the date in the format dd/mm/yyyy. My question is, how can I check if the Date passed as parameter really follows this Date Format? I tried to use the grep command as below: if echo "$1" | grep -q '^[0-3][0-9]/[0-1][0-9]/[0-9]\{4\}$' but it didn't give the correct format because the day for example can be 33, 34, (...), that is not really the correct format. Anyone know something that can really check if the date passed

Math-like chaining of the comparison operator - as in, “if ( (5<j<=1) )” [duplicate]

雨燕双飞 提交于 2020-01-08 13:25:29
问题 This question already has answers here : Is (4 > y > 1) a valid statement in C++? How do you evaluate it if so? (5 answers) Closed 2 years ago . int j=42; if( (5<j<=1) ) { printf("yes"); } else { printf("no"); } Output: yes Why does it output yes? Isn't the condition only half true? 回答1: C does not understand math-like syntax, so if(1<j<=5) is not interpreted as you expect and want; it should be if (1 < j && j <= 5) or similar. As explained in other answers, the expression is evaluated as ((1

My try catch function won't work as expected

家住魔仙堡 提交于 2020-01-08 07:40:30
问题 My program is supposed to take the given value and find the other value with it. However, my try catch statement doesn't seem to be working. It never does the first 2 if statements which are the most essential parts of the program. The 3rd one does work when you input both values. Thanks in Advance. public void calculate(View view) { EditText length_find = findViewById(R.id.feet); EditText pounds_find = findViewById(R.id.pounds); try { //int length_int = Integer.parseInt(length_find.getText()