Remove separator of last menu - php

Question

I want to remove last separator after Login Menu.
Here is my code and output.
<?php
if (!empty($topmenu) && !empty($menulist)) {
foreach ($topmenu as $mainparent) {
$arry = getmenuvalue($mainparent->id, $menulist, MAINURL);
if (isset($mainparent->children) && !empty($mainparent->children)) {
echo '<li class="dropdown"> ' . $arry['name'] . '<span class="caret"> </span>';
echo '</li>';
echo '<li> | </li>';
} else {
echo '<li>' . $arry['name'] . '</li>';
echo '<li> | </li>';
}
}
}
?>
Result of this code is
Home | Register | Login |
I want to Remove last Separator after Login menu.
I want result like this.
Home | Register | Login

Answer 1

Try This
<?php
if (!empty($topmenu) && !empty($menulist)) {
$count = count($topmenu);
$i = 1;
foreach ($topmenu as $mainparent) {
$arry = getmenuvalue($mainparent->id, $menulist, MAINURL);
if (isset($mainparent->children) && !empty($mainparent->children)) {
echo '<li class="dropdown"> ' . $arry['name'] . '<span class="caret"> </span>';
echo '</li>';
if($count != $i)
echo '<li> | </li>';
} else {
echo '<li>' . $arry['name'] . '</li>';
if($count != $i)
echo '<li> | </li>';
}
$i++;
}
}
?>

Answer 2

You can follow this example:
<?
$array = array('One','Two','Three'); // your array
$count = count($array); // check the array count
$i = 1; // use incremental
foreach ($array as $value) {
$separator = ($i == $count ? '' : '|'); // compare if last index use empty else separator
echo $value. $separator; // print separator with value
$i++; // +1 in every iteration.
}
?>
Result:
One|Two|Three
UPDATE 1:
Example with your code
<?
if (!empty($topmenu) && !empty($menulist)) {
$count = count($topmenu);
$i = 1;
foreach ($topmenu as $mainparent) {
$arry = getmenuvalue($mainparent->id, $menulist, MAINURL);
if (isset($mainparent->children) && !empty($mainparent->children)) {
echo '<li class="dropdown"> ' . $arry['name'] . '<span class="caret"> </span>';
echo '</li>';
} else {
echo '<li>' . $arry['name'] . '</li>';
//echo '<li> | </li>';
}
if($i == $count ? '' : '<li> | </li>');
$i++;
}
}
?>
Side Note:
I am not sure about the $topmenu check count($topmenu); if you get the count it will work. Try it.

Answer 3

You can collect your list items and concat them using implode method:
<?php
if (!empty($topmenu) && !empty($menulist)) {
$listItems = array();
foreach ($topmenu as $mainparent) {
$arry = getmenuvalue($mainparent->id, $menulist, MAINURL);
if (isset($mainparent->children) && !empty($mainparent->children)) {
$listItems[] = '<li class="dropdown"> ' . $arry['name'] . '<span class="caret"> </span></li>';
} else {
$listItems[] = '<li>' . $arry['name'] . '</li>';
}
}
echo implode('<li> | </li>', $listItems);
}
?>

Answer 4

You can do something like this
<?php
if (!empty($topmenu) && !empty($menulist)) {
foreach ($topmenu as $key => $mainparent) {
$arry = getmenuvalue($mainparent->id, $menulist, MAINURL);
if (isset($mainparent->children) && !empty($mainparent->children)) {
echo '<li class="dropdown"> ' . $arry['name'] . '<span class="caret"> </span>';
echo '</li>';
//echo '<li> | </li>';
} else {
echo '<li>' . $arry['name'] . '</li>';
//echo '<li> | </li>';
}
end($topmenu);
if ($key !== key($topmenu)) {
echo '<li> | </li>';
}
}
}
?>