Need a little help at JS.

Question

In my this code...
switch(true)
{case (I=0): var Oyp=7;
case(I>0)&&(I<70000): var Oyp =6;
case(I>=70000)&&(I<200000): var Oyp=5;
case(I>=200000)&&(I<500000): var Oyp =4;
case(I>=500000)&&(I<900000): var Oyp=3;
case(I>=900000)&&(I<1356000): var Oyp=2;
case(I>=1356000)&&(I<1761000): var Oyp=1; } document.write (Oyp);
Suppose the value of I is 345180, still the value of Oyp will be printed as undefined. Why? Thanks for answering.

Answer 1

You cant use condition in switch

Answer 2

Exactly: you can use condition expression in 'case' statement, but the 'switch' expression value will be compared to the value of each 'case' expression... Typically, switch-case statements are use to compare a single variable value to only once value at each case:

switch (myvar) {
case value:
// do something if myvar == value
// and run each next case block
// until a 'break' keyword occurs
// most of the time, each case block will end with it
// in all cases, no more 'case' statement will be tested
case result_of_expression:
// do something if myvar == result of expression
// as a condition is an expression, you can use one, but your 'myvar' will be compared necessarly to boolean value true or false
}

So, in your code, at first case statement, condition of nexted instruction execution is equivalent to:

if (true == (l=0))

... wich is false (and where you probably do a first mistake: you doesn't compared l to 0, but assign 0 to l... so the 'result' of the assignement-expression is the value of l, and 0 is not equivalent to true -- and almost, l is now equal to zero for next 'case' statements ;P). Anyway, if l is already equal to 0, the 'case' statement will always be false ^^ And if by magical it would be true, all different case block will be running, as you don't use 'break' keyword...

The second 'case' statement could be writen:

if (true == ((l>=0)&&(l<70000))

... which is actually always true, as l is now equal to zero ^^ So Oyp is set to 6, but next to 5, then 4... and finally will carrying 1 at last 'case' block... if Oyp is declared outside the 'switch' scope (closure), else Oyp is unknown (undefined) after exiting the 'switch' closure ;)

Answer 3

you have to define oyp outside the switch scope

Answer 4

Okay thanks chrizzhigh

Answer 5

can you mark it as answer ^^
and no probs keep on coding