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 } ?>
Related
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 have created navigation using mysql php. now i am trying to pass value to another page using ajax.please help me
<ul id="all_navs">
<?php $sql = mysql_query("select * from sub_category_master where main_category =1");
while($row = mysql_fetch_array($sql))
{?>
<li id="s1">
<a href="#tab_deals" onclick="showUsers('<?php echo $row['sub_category_id']; ?>')" id="insert">
<?php echo $row['sub_category_name']; ?></a>
</li>
<?php }?>
</ul>
if i understood your problem why not try something like this:
<ul id="all_navs">
<?php $sql = mysql_query("select * from sub_category_master where main_category =1");
while($row = mysql_fetch_array($sql)){
echo '<li id="s1">';
echo '<a href="#tab_deals" onclick="showUsers('.$row['sub_category_id'].')" id="insert">';
echo $row['sub_category_name'].'</a>';
echo '</li>';
}
?>
</ul>
it would help more if you told us what the problem was.
I have a simple list in html:
<ul id="sortable">
<li id="1">item1</li>
<li id="2">item2</li>
<li id="3">item3</li>
</ul>
I make it sortable with jQuery UI. This lets the user move the order of each item:
$('#sortable').sortable();
But when I get the same items from a MySql database the "sortable" doesn't work any more I cannot change the order of the items. So my question is: how to make a list sortable when the list comes from a database?
<?php
$result = mysqli_query($con, 'SELECT * FROM principal ORDER BY ordre');
while ($row = mysqli_fetch_array($result)) {
?>
<ul id="sortable">
<li id="<?php echo $row["id"] ?>"><?php echo $row["concepte"]?>;</li>
</ul>
<?php
You are generating ul elements for each record row, where as it should be one single UL having multiple LI elements, also fix the li HTML code
<?php
$result = mysqli_query($con, 'SELECT * FROM principal ORDER BY ordre');
echo '<ul id="sortable">';
while ($row = mysqli_fetch_array($result)) {
?>
<li id="<?php echo $row["id"]; ?>"><?php echo $row["concepte"]; ?></li>
<?php
}
echo '</ul>';
?>
I'm not a PHP guy but this looks like it's going to generate a ul with each iteration ? So your markup would be all screwed up
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
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>