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Start your free trialJason Brown
9,626 PointsProgram always returns today's date, no matter how many times I change the $day variable ...
I can't get this program to return anything but today's date. Please help! : "" <?php // Store each exercise in a string variable. $exercise1 = "this"; $exercise2 = "isnt"; $exercise3 = "working"; $exercise4 = "i"; $exercise5 = "want"; $exercise6 = "it"; $exercise7 = "to work";
$day = date('N');
if ($day == 1) { echo $exercise1; } elseif ($day == 2) { echo $exercise2; } elseif ($day == 3) { echo $exercise3; } elseif ($day == 4) { echo $exercise4; } elseif ($day == 5) { echo $exercise5; } elseif ($day == 6) { echo $exercise6; } else { echo $exercise7; } $day = 3; $day = date(2);
// create a variable containing the day of the week. // Use if statement to test for the day of the week. // display corresponding exercise string.
$day = 1; $day = 1; ?> ""
1 Answer
Eric Butler
33,512 PointsHey Jason,
You need to move your different $day
declarations up above the if
block, because right now the only $day
value being fed into all your if
statements is the one above it, which is $day = date('N');
. If you moved your $day = 3
or $day = 1
statements up so they're just underneath your $day = date('N')
statement, you'd echo different outputs as desired.
Hope this helps!
Jason Brown
9,626 PointsJason Brown
9,626 PointsGah! Of course! I wouldn't do this if I was actually writing a program ... I just wasn't thinking about that because I was fooling around in workspaces. Ohhh boy ... Thanks so much for your response. A valuable lesson learned, either way. Thanks again.