The drop down menu only show the first loop with its first item from database and then every items of the first item shown in the first loop. There is no second item listed of the first loop. Any idea to help me with this loop???
<ul class="pureCSSMenum">
<li class="pureCssMenui">E-Policy
<ul class="pureCssMenum">
<?
$sql = "SELECT * FROM e_category";
$result = mysql_query($sql);
while ($epolicy = mysql_fetch_array($result)) {
?>
<li class="pureCssMenui"><a class="pureCssMenui" href="#"><?=$epolicy['e_cat_name']?></a>
<ul class="pureCssMenum">
<?php
$sql = "SELECT * FROM e_subcategory";
$result = mysql_query($sql);
while ($epolicysub = mysql_fetch_array($result)) {
?>
<li class="pureCssMenui"><a class="pureCssMenui" href="#"><?=$epolicysub['e_subcat_name']?></a></li>
<? } ?>
</ul>
<? } ?>
</li>
</ul>
</li>
</ul>
You must use two result variables $result1, $result2 for outer and inner variable, rest all is fine i think.
Hope this will help
Related
I want to print an attribute (i.e. active) on the first element of an array.
The ugly way is to use a variable:
<ul class="nav nav-pills">
<?php $firstItem = true; ?>
<?php while ($row = $res_sections->fetchArray()) { ?>
<li class="nav-item">
<a class="nav-link<?php if ($firstItem) echo ' active' ?>" data-toggle="pill" href="#nav<?= $row['section']?>"><?php echo "{$row["section"]}"?></a>
</li>
<?php $firstItem = false; ?>
<?php } ?>
</ul>
I wonder if there's something more elegant.
Is there a way to know if $row is the first element (or any) of the fetchArray() result?
EDIT
Here where fetchArray() comes from:
$db = new SQLite3("/db.sqlite");
$sql_sections = "SELECT DISTINCT section FROM settings";
$res_sections = $db->query(SQLite3::escapeString($sql_sections));
I've been trying to build a dropdown menu but I'm not getting my desired results. Here's my code:
<?php require_once 'core/init.php'?>
<?php
$sql = 'SELECT * FROM categories WHERE parent = 0';
$pquery = mysqli_query($db,$sql);
?>
<?php while($parent = mysqli_fetch_assoc($pquery)):?>
<?php
$parent_id = $parent['id'];
$sql2 = 'SELECT * FROM categories WHERE parent = "parent_id"';
$cquery = mysqli_query($db,$sql2);
?>
<li class='dropdown'>
<a href='#' class='dropdown-toggle' data-toggle='dropdown'>
<?php echo $parent['id'];?><span class='caret'</span</a>
<ul class='dropdown-menu' role='menu'>
<?php while($child = mysqli_fetch_assoc($cquery)):>
<li><a href='#'><?php echo $child['parent'];?></a>
</li>
<?php endwhile; ?>
</ul>
</li>
<?php endwhile;?>
My DB is like this and the result is this.
Try with console phpmyadmin before write codes in page.
I think your codes is wrong.
Your code must like this
$parent_id = $parent['id'];
$sql2 = 'SELECT * FROM categories WHERE parent = '.$parent_id;
I suggest to you for use ajax to be beautiful again.
I am stuck with my current knowledge of PHP. I want to make a menu where the info is stored into a database. What I want to achieve is the following:
<?php
$names= array();
$aliases= array();
$items = array();
?>
<?php while($row = mysqli_fetch_array($menu, MYSQLI_ASSOC)){
$names[] = $row['page'];
$aliases[] = $row['alias'];
$items[] = $row['item'];
}?>
<div id="menu">
<ul>
<?php foreach($names as $page_name){
<?php echo $page_name;?>
?>
*** IF THERE ARE SUB MENU ITEMS, SHOW THEM ***
<ul>
<li>
*** NAME OF SUB ITEM***
</li>
</ul>
</ul>
</div>
I have absolutely no idea if i am on the right pad. The foreach works fine for the names or aliasses only, but i want to sort of combine them. Then i want to check if there is a submenu for the hover, but also set the parent item on class "active".
Since it is hard for me to explain it in English, please ask me any question needed so i can provide more info.
Thanks in advance!
First of all, you don't need three different arrays, only $names array would be enough. Second, change your while() loop in the following way,
while($row = mysqli_fetch_array($menu, MYSQLI_ASSOC)){
$names[] = array('page_name' => $row['page'], 'alias' => $row['alias'], 'item' => $row['item']);
}
This will create a multidimensional array comprising of name, alias and sub-items.
And finally, display the entire menu in the following way,
<div id="menu">
<ul>
<?php foreach($names as $page_array){ ?>
<?php echo $page_array['page_name']; ?>
<?php
if(!empty($page_array['item'])){
?>
<ul>
<li>
<?php echo $page_array['item']; ?>
</li>
</ul>
<?php
}
} ?>
</ul>
</div>
You are overcomplicating things:
<?php
echo '<div id="menu">
<ul>';
while($row = mysqli_fetch_array($menu, MYSQLI_ASSOC)){
$names[] = $row['page'];
$aliases[] = $row['alias'];
$items[] = $row['item'];
echo '<li class="pagename">'.$row['page'].'';
if ('checkforsubitem') {
echo '<ul>
<li>
'.$row['name_of_subitem'].'
</li>
</ul>';
}
}
echo '</li>
</ul>
</div>';
?>
However, this assumes some things about the submenu-bits. Normally, you would either fetch those items within the same query, and have them available to you via $row, but you can also do separate queries for those items based on a connection from the $row['id'] or something, and pull those subitems that corresponds to that id, and run another while-loop around the secondary <li>-items.
I'm new to coding with php and using MySQL. I am having trouble to display a list of categories by their ID so that each category is displayed individually as a heading. Instead I got it to display a category name but its only echoing out a category name twice that's the same. Here is my code...
$sql= "SELECT * FROM categories ";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query))
{
$id=$row['id'];
$cat_name=$row['cat_name'];
}
?>
<ul class="nav nav-list">
<li class="nav-header"><?php echo $cat_name;?></li>
<li class="nav-header"><?php echo $cat_name;?></li>
</ul>
Your li tag is outside the while loop. So the $id and $cat_name is only the last record in the DB, then you echo them twice. That's way you got the same name twice.
Try echo the li tag in the loop (but not the ul):
<ul class="nav nav-list">
<?php
while($row = mysql_fetch_array($query))
{
$id=$row['id'];
$cat_name=$row['cat_name'];
echo '<li class="nav-header">' .$cat_name. '</li>';
}
?>
</ul>
You can use this code as-is :
$sql= "SELECT * FROM categories ";
$query = mysql_query($sql); ?>
<ul class="nav nav-list">
<?php while($row = mysql_fetch_array($query)) : ?>
<li class="nav-header"><?php echo $row['cat_name'];?></li>
<?php endwhile; ?>
</ul>
This will loop through your records and for each record, it will print an entire li with the required data.
Note that separating your PHP code from your HTML code like this has several benefits. It will be better colored in your editor and it is also easier to integrate.
The reason you are printing the same value out twice is because $cat_name is a string variable and will only hold one value at a time you may want to save the items an array and loop at a seperate time, like such
<?php
$sql= "SELECT * FROM categories ";
$query = mysql_query($sql);
$category = array();
while($row = mysql_fetch_array($query))
{
$array = array(
id => $row['id'],
name => $row['cat_name']
);
array_push($category,$array);
}
?>
<ul class="nav nav-list">
<?php
foreach($category as $c)
{
echo '<li class="nav-header">'.$c['name'].'</li>';
};
?>
</ul>
$sql= "SELECT * FROM categories ";
$query = mysql_query($sql);
while($row = mysql_fetch_assoc($query))
{
?>
<ul class="nav nav-list">
<li class="nav-header"><?php echo $row['id']; ?></li>
<li class="nav-header"><?php echo $row['cat_name']; ?></li>
</ul>
<?php } ?>
EDIT FIXED by user #jeroen
You have to move the li and a tags inside the while loop.
Thank you for this, F4LLCON
So the final code (not so clean, but works now):
<?php
$query = mysql_query("SELECT * FROM `apps` ");
while ($query_row = mysql_fetch_assoc($query))
{
?>
<li>
<?php
$meer = $query_row['TITLE'];
$desc_inject = '';
$sub_string = substr($desc_inject, 0, 200);
echo $sub_string." " . '' . '' . $meer . '';
?>
</li>
<br />
<?php
}
?>
I want a drop-down menu that will show the TITLES out of my MySQL database.
I know how to do everything concerning the ID etc.
The only problem is that the PHP version of the drop-down menu will SELECT all the TITLES in my database and parse it as ONE link.
So instead of
Home
>About
>Contact
>Another
it will look like
Home
>About
Contact
Another
With other words
Home
>About Contact Another
It will not make multiple drop-down links but only one drop-down link
If you for example do:
<div>
<?php
echo $query_row['TITLE'];
?>
</div>
It will make a individual <div></div> for every TITLE in my database, so I thought this method would also work for the drop-down links..
Does anybody know how to fix it so it will make individual <li></li> for every TITLE?
Here is a normal drop-down menu:
<li id="media">
<ul>
<li id="1a">About</li>
<li id="1b">Contact</li>
<li id="1b">Another</li>
</ul>
</li>
And here is the PHP drop-down menu:
<li id="media">
<ul>
<li>
<a href="">
<?php
$query = mysql_query("SELECT * FROM `apps` ");
while ($query_row = mysql_fetch_assoc($query))
{
echo $query_row['TITLE'];
?>
<br />
<?php
}
?>
</a>
</li>
</ul>
</li>
You have to move the li and a tags inside the while loop.
echo '<li>' . $query_row['TITLE'] . '</li>';
<li id="media">
<ul>
<?php
$query = mysql_query("SELECT * FROM `apps` ");
while ($query_row = mysql_fetch_assoc($query))
{
echo "<li><a href=''>" . $query_row['TITLE'] ."</a></li>";
}
</ul>
</li>