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Yogita Verma
5,549 PointsWhat is problem with my code? Only one condition should be printed but here each condition is being printed.
// store each exercise in a string variable
$exercise1 = 'Display "Hello World!"';
$exercise2 = 'Convert Pounds to Kilograms';
$exercise3 = 'Convert Kilograms to Pounds';
$exercise4 = 'Convert Miles to Kilometers';
$exercise5 = 'Convert Kilometers to Miles';
$exercise6 = 'Month long string of the day';
$exercise7 = 'String of the day with levels';
// create a variable containing the day of the week
$day= date('N');
var_dump($day);
if($day==1)
echo $exercise1;
if($day==2)
echo $exercise2;
if($day==3)
echo $exercise3;
if($day==4)
echo $exercise4;
if($day==5)
echo $exercise5;
if($day==6)
echo $exercise6;
if($day==7)
echo $exercise7;
2 Answers
Michael Linnane
549 PointsYou are missing your open and closing curly braces around each echo function after the if statements. Should be like so:
if($day1==1)
{
echo $exercise1;
}
If you notice on your code you have the following:
if($day==1)
echo $exercise1;
Mustafa Başaran
28,046 PointsIn conditional statements, the basic structure is as follows:
if (check condition 1) {
# some code here
} elseif (check condition2) {
# some code here
} else {
#some code here
}
you can check as many conditions as you wish in consecutive elseif statements.
else at the end checks what remains after if and elseif. I hope this is clear.