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Start your free trialYIHENG CHIU
2,761 PointsChallenge task 6 of 7
"Inside the foreach loop, we should only display one <li> tag for each flavor. The one list item should display the value for the array element that the foreach loop is considering. Remove one of the <li> elements, and change the value that gets displayed in the other to display the flavour."
I don't really understand what does it mean.
1 Answer
Oisin Kilkenny
14,213 PointsThe question is asking you to remove one of the list items and replace what will be the output.
<?php
foreach($flavors as $flavor){
?>
<li><?php echo $flavor; ?></li>
<?php
}
?>
YIHENG CHIU
2,761 PointsThank you for replying my question, however I used your code <?php foreach($flavors as $flavor){ ?> <li><?php echo $flavor; ?></li> <?php } ?> but It still does't work.
YIHENG CHIU
2,761 PointsThis is how my code looks like right now:
<?php $flavors = array(); $flavors[] = "Chocolate"; $flavors[] = "Vanilla"; $flavors[] = "Mint"; ?>
<p>Welcome to Ye Olde Ice Cream Shoppe. We sell <?php echo count($flavors) ?> flavors of ice cream.</p>
<ul>
<?php foreach($flavors as $flavor){?>
<li><?php echo $flavor; ?></li>
<?php } ?>
</ul>
Oisin Kilkenny
14,213 PointsUp to question six, this is exactly what I have.
<?php
$flavors = array("Chocolate", "Vanilla", "Damn, you stole mine.");
?>
<p>Welcome to Ye Olde Ice Cream Shoppe. We sell <?php echo count($flavors); ?> flavors of ice cream.</p>
<ul>
<?php
foreach($flavors as $flavor){
?>
<li><?php echo $flavor; ?></li>
<?php
}
?>
</ul>
YIHENG CHIU
2,761 Points<?php
$flavors = array();
$flavors[] = "Chocolate";
$flavors[] = "Vanilla";
$flavors[] = "Mint";
?>
I think here is the difference
Oisin Kilkenny
14,213 PointsPossibly, try using my code and see if it works.
YIHENG CHIU
2,761 PointsThank you very much! Now it works!
Oisin Kilkenny
14,213 PointsNo problem. :)
YIHENG CHIU
2,761 PointsYIHENG CHIU
2,761 PointsThank you very much! Now it work!!!