if() statement

[Anon]

In my script I have a part where the code looks something like this:

if($type == "1" or $type == "3"):
echo "one or three";
elseif($type == "2" or $type == "3"):
echo "two or three";
endif;

This works fine when I use one or two but when I try 3, only the fist one shows up. What am I doing wrong?

[[deleted]]

You should use () around the statements in your IF:

if (($type == "1") or ($type=="3"))

Remember, you are checking wether $type="1" is true, or $type="3" is true.
Without parenthesis you are checking wether
"$type=="1"or$type=="3" is true, which is different.

[e-wizard]

Rather use the switch - statement for this purpose:

Instead of:
if ($i == 0) {
print "i equals 0";
}
if ($i == 1) {
print "i equals 1";
}
if ($i == 2) {
print "i equals 2";
}

try using

switch ($i) {
case 0:
print "i equals 0";
break;
case 1:
print "i equals 1";
break;
case 2:
print "i equals 2";
break;
}

or i.e

switch ($i) {
case 0:
case 1:
case 2:
print "i is less than 3 but not negative";
break;
case 3:
print "i is 3";
}

or

A special case is the default case. This case matches anything that wasn't matched by the other cases. For example:

switch ($i) {
case 0:
print "i equals 0";
break;
case 1:
print "i equals 1";
break;
case 2:
print "i equals 2";
break;
default:
print "i is not equal to 0, 1 or 2";
}

Returning to your specific question:

Both if-clauses contain a 'true' statement

--A-clause---------------------------
if (a==1 or a==3)
print (" a=1 ,could also be 3");
--B-clause----------------------------
elseif (a==2 or a == 3)

print("a=2 , could also be 3");

Input->2
A-clause = FALSE
B-clause = TRUE, a is 2 or 3 !

Input->3
A-clause = TRUE, a is 1 or 3
B-clause = TRUE, a is 2 or 3

because:

when using 'or', only one of the compared
has to be true in order fpr the if clause
to become true

cu

e-wizard

[Anon]

If you want both echo statements to be executed when the input is 3, you have to replace the elseif-statement with another if-statement like so

if($type == \"1\" or $type == \"3\"):
echo \"one or three\";
if($type == \"2\" or $type == \"3\"):
echo \"two or three\";

This is because an elseif-statement is only executed if the conditions of the previous if-statement are not met.

When you have 3 as the input in your code though, the conditions of the first if-statement are met and the elseif won\'t get executed.