How to generate lists using PHP & MySQL

Question

I was wondering how would go about displaying five images in a row using lists. For example, how would I generate the following XHTML code below using PHP & MySQL?

Here is the XHTML code.

<ul>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
</ul>

<ul>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
    <li><a href="#" title=""><img src="" /></a></li>
</ul>

Here is my PHP & MySQL code so far.

$url = array();
$title = array();

$dbc = mysqli_query($mysqli,"SELECT *
                             FROM images
                             GROUP BY images.id");

if (!$dbc) {
    print mysqli_error($mysqli);
} else {
    while($row = mysqli_fetch_array($dbc)){ 
        $url[] = $row["url"];
        $title[] = $row["title"];
    }
}

GSto · Accepted Answer · 2010-06-29 19:51:08Z

2

just echo it out it your while loop: EDIT: noticed your example has a new ul every 5 images, code has been adjusted to handle that.

if (!$dbc) {
    print mysqli_error($mysqli);
} else {
    $rowCount = 0;
    while($row = mysqli_fetch_array($dbc)){ 
        if($rowCount % 5 == 0){
         echo "<ul>";
        }
       echo "<li><a href='" .$row["url"]. "' title='".$row['title']."'>";
        echo "<img src='".$row['src']."'></a></li>";
       if($rowCount % 5 == 4) {
         echo "</ul>";
       }
       $rowCount++;
    }

}

edited Jun 29, 2010 at 19:51

answered Jun 29, 2010 at 19:44

GSto

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8 Comments

pOp Over a year ago

would this work if I only have six images? would it add a new <ul> tag after 5 images?

Eric Petroelje Over a year ago

You would still need to split them into groups of 5 like his example. A counter and some modulo division should do the trick.

pOp Over a year ago

Is there a better way to do this or is this the best way?

Eric Petroelje Over a year ago

@GSto - problem now is that it won't close the <ul> tag if the number of items isn't an exact multiple of 5.

BojanG Over a year ago

Your list element might not close properly.

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Your Common Sense · Accepted Answer · 2010-06-29 20:02:02Z

you've got it perfect! now you have just to add an HTML part. Though, separate arrays is not good, it would be better to use a single one

$DATA = array();
while($row = mysqli_fetch_array($dbc)){ 
  $DATA[] = $row;
}

and then you're going to print out HTML

<ul>
<? foreach ($DATA as $row): ?>
    <li><a href="<?=$row['url']?>" title="<?=$row['title']?>"><img src="" /></a></li>
<? endforeach ?>
</ul>

though if you need to split your lists into smaller parts, you will need another code

there are many ways to do it. the easiest one is to add some markers to array.
and than use it at the template:

<? foreach ($DATA as $row): ?>
  <? if($row['ul']): ?><ul><? endif ?>
      <li><a href="<?=$row['url']?>" title="<?=$row['title']?>"><img src="" /></a></li>
  <? if($row['/ul']): ?></ul><? endif ?>
<? endforeach ?>

Eric Petroelje · Accepted Answer · 2010-06-29 20:20:17Z

0

The easiest way to do this without having lots of crazy statements to determine when to split the list is to take advantage of array_chunk:

<?php
$url = array();
$title = array();

$dbc = mysqli_query($mysqli,"SELECT *
                             FROM images
                             GROUP BY images.id");

if (!$dbc) {
    print mysqli_error($mysqli);
} else {
    while($row = mysqli_fetch_array($dbc)){ 
        $rows = $row;
    }
}


$row_groups = array_chunk($rows, 5);

foreach($row_groups as $row_group ) {
   echo "<ul>";
   foreach( $row_group as $row ) {
?>
       <li><a href="<?=$row['url'] ?>" title="<?=$row['title'] ?>"><img src="<?=$row['url'] ?>"/></a></li>';
<?php
   }
   echo "</ul>";
}
?>

edited Jun 29, 2010 at 20:20

answered Jun 29, 2010 at 20:00

Eric Petroelje

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6 Comments

Your Common Sense Over a year ago

The OP has the code already template ready. But you all pull him down to the usual templateless approach, used by the millions of ignorant PHP lemmings.

Eric Petroelje Over a year ago

@Col - He has the code in a template? I didn't see anything about a template engine in his question? Could you enlighten me as to what engine he is using? smarty? xtemplate?

Your Common Sense Over a year ago

it doesn't really matter. All templates are very common. Even with xtpl he have to parse template in the view part, not in the controller.

sarsnake Over a year ago

@Col - I applaud the use of templates correction on your side, BUT, the question was NOT about template use. Please realize that many people here are learning new skills - which is better done gradually. First, php developer needs to know how to do it without templates, the add the templating engine of choice ( i prefer Smarty).

Eric Petroelje Over a year ago

@gnomixa - Agreed. I could just as well respond to every PHP question with a quip about "another lemming using PHP who should be using ASP.NET"

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