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Running a for-loop does not display all the items

By : armandops
Date : July 31 2020, 08:00 AM
I wish this help you You can create a result string and append the flavours and finally return the resulting string.
code :
def list_of_flavors(self):
    """Display all the ice cream flavors."""
    result = f'We have many exotic flavors, for instance'
    for flavor in self.flavors:
        result += flavor 
    return result

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PHP - display 'X' items from foreach loop with link to display next 'X' items

By : user2140875
Date : March 29 2020, 07:55 AM

how to count listbox items while running a for each loop in vb.net?

By : Dan
Date : March 29 2020, 07:55 AM
I hope this helps . I have a list of links and need to check some data in it. , Insert a counter variable inside the For Each and use it.
code :
Dim i as integer = 0
For Each item In myListBox.Items
    // your code goes here
    // do you need to check if your code went ok?
    i = i+1
    label.Text = string.Format("Processing item {0} of {1}", i, myListBox.Items.Count)
Next item 

Foreach Loop - Display items differently, first 3 item, items 4-6 etc

By : Sergey Saprykin
Date : March 29 2020, 07:55 AM
it fixes the issue Im hoping this will be an easy one.. , Use a for loop instead of foreach.
code :
@for ( int i = 0; i < Model.ContentList.Count; i++ )
   var Article = Model.ContentList[i];
   if ( i < 3 ){
      // for Items 1 - 3
   } else if ( i < 7 ){
      // for Items 4 - 7
   } else {
      // for Items 7 +

trying to get foreach loop to display items with subitems correctly in list items

By : user3732917
Date : March 29 2020, 07:55 AM
this one helps. Under the first "foreach" loop you should reset your $subs array to make sure it's empty. Otherwise it gets down to the second loop, sees the old "$subs" from the previous loop, and loops through it again:
code :
foreach ( $menuitems as $item ) {    
    $subs = array();
    if ($item ['submenu']) {
        $subs [$submenuI] = $item ['submenu'];
$html = new cacheHTML ( 'topmenu' );
if (! $html->isCached) {
    $menuitems = getMenuLevelsArray ( 25, 1, 1 );
    $submenuI = "0";

    foreach ( $menuitems as $item ) {
        if ($item ['submenu']) {
            $subs [$submenuI] = $item ['submenu'];

        echo '<li';
            echo 'class="dropdown"';
        echo ">
        <a href='{$item['url']}'";

            echo 'class="dropdown-toggle" data-toggle="dropdown"    role="button" aria-expanded="false"';
        echo ">{$item['text']}";

            echo '<span class="caret"></span>';
        echo '</a> <!-- submenu begins here -->

        foreach($subs as $submenuI => $menu){
            echo '<ul class="dropdown-menu" role="menu">

            for($a = 0; $a < count ( $menu ); $a ++) {
                echo "<a href='{$menu[$a]['url']}'";
                    echo "class='last'"; 
                echo ">{$menu[$a]['text']}</a>
            echo '</li>

        echo '</li>';



$html->show(); ?>

Loop through PHP function, display userdata from Mysql - limit to display to 25 items and add next button

By : havill
Date : March 29 2020, 07:55 AM
I hope this helps you . Jquery pagination by default will only break-up the given table into several tabs on the page, that is all the data is actually loaded. There are some that let you do an AJAX request to fetch the next page, to work in conjunction with your PHP code.
Ideally, you want to limit the query using LIMIT (number of rows in the result object) and OFFSET (Start from row X ) which will give you number of records starting from your offset row up until the limit, and then use logic on the PHP side to determine the maximum number of pages.
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