Get data from database and use loop in php

Question

I am working on college project and i am stuck at the moment.

I have a table and it have multiple records. Records are saved by user. I want to show all data stored by specific user on webpage (under a separate div). I am able to fetch all data from database by using below code.

$prj= mysql_query("select * from project where uid=$uid");
        $record= mysql_fetch_assoc($prj);
        $project_name = "$record['project_name']";
        $short_dis= "$record['short_dis']";
        $poster= "$record['poster']";

mysql_close($prj);

Here are some php code.

    <div class="media services-wrap55 wow fadeInDown">
    <img class="img-responsive" src="uploads/poster/photo-original_827156899.jpg"><br>
 <a href="#"> <h4 class="media-heading">second project for testing</h4></a>
 <p>    by user</p>
 <p> second project for testing  second project for testing  second project for testing  second project for testing.</p>

 <a class="" href="iska-link.html">More ... </a>
 <p>
 <div id="progressBar2" class="tiny-green"><div></div></div><p>
 <div class="counter-inner"><div id=example1" data-countdown="07/02/2015 00:11:00"></div></div><p>
 <div class="col-sm-11 text-right">
   <div class="entry-meta">

 <span><i class="fa fa-comment"></i><a href="comments">  2 Comments </a></span>
 <span><i class="fa fa-thumbs-up"></i><a href="#"> 56 </a></span>
 <span><i class="fa fa-thumbs-down"></i><a href="#"> 56 </a></span>
 <span><i class="fa fa-star"></i><a href="#"> 56 Fav </a></span>
  </div>
  </div>
 </div>
</div>

As you can see these are static data. I want to put dynamic data from database in this.

Is there any advise how to get the data in div by loop.

I can show one record by using echo. I am not looking for entire code just wanted database part and some sample while using it in php.

If you can, you should stop using mysql_* functions. They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and consider using PDO, it's really not hard. — Jay Blanchard
– Jay Blanchard, Commented Jun 25, 2015 at 18:28
Manual first, Community last. Go read: php.net/manual/de/mysqli-result.fetch-array.php and comments. — Jens A. Koch
– Jens A. Koch, Commented Jun 25, 2015 at 18:29
I will try to use mysqli/PDO. At the moment i don't have idea about them. As i am still learning. But i will definitely try. However, if you advice on this question then it would be nice. Thanks — Roxx
– Roxx, Commented Jun 25, 2015 at 18:30
@JensA.Koch I already tried but i am not able to show it on webpage. because i think i need to generate div at runtime as well. — Roxx
– Roxx, Commented Jun 25, 2015 at 18:32

Death-is-the-real-truth · Accepted Answer · 2015-06-25 18:37:38Z

4

I didn't know your table structure, so a sample code is given so that you can take help of it and do accordingly:-

    <?php
error_reporting(E_ALL);
ini_set('display_errors',1);
$conn = mysqli_connect('hostname','username','password','databasename');

$prj= mysqli_query($conn,"select * from project where uid=$uid") or die(mysqli_error($conn));
        $record = array();
        while($row = mysql_fetch_assoc($prj)){
            $record[] = $row;
        }
mysql_close($conn);
?>
<?php foreach($record as $rec){?>
 <div class="media services-wrap55 wow fadeInDown">
    <img class="img-responsive" src="uploads/poster/<?php echo $rec['image_path'];?>"><br> // assuming that image_path is the field in table where you putted image name
 <a href="#"> <h4 class="media-heading"><?php echo $rec['project_name'];?></h4></a>// assuming that project_name is the field in table where you putted project name
 <p><?php echo $rec['user_name'];?></p>// assuming that user_name is the field in table where you putted user name
 <p><?php echo $rec['project_description'];?></p>// assuming that project_description is the field in table where you putted project name
<!-- in the same way you can do for others also -->
 <a class="" href="iska-link.html">More ... </a>
 <p>
 <div id="progressBar2" class="tiny-green"><div></div></div><p>
 <div class="counter-inner"><div id=example1" data-countdown="07/02/2015 00:11:00"></div></div><p>
 <div class="col-sm-11 text-right">
   <div class="entry-meta">

 <span><i class="fa fa-comment"></i><a href="comments">  2 Comments </a></span>
 <span><i class="fa fa-thumbs-up"></i><a href="#"> 56 </a></span>
 <span><i class="fa fa-thumbs-down"></i><a href="#"> 56 </a></span>
 <span><i class="fa fa-star"></i><a href="#"> 56 Fav </a></span>
  </div>
  </div>
 </div>
</div>
<?php } ?>

Note:- stop using mysql_*, it is officially deprecated. use mysqli_* or PDO.

answered Jun 25, 2015 at 18:37

Death-is-the-real-truth

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4 Comments

Roxx Over a year ago

Thanks for the answer. Appreciate it. Voted up. Will test and let you know.

Death-is-the-real-truth Over a year ago

sure but change variables and values according to your table structure. i think you understand what i want to say.

Roxx Over a year ago

I understood. One thing i want to know is in your code you put all the table data in array($record[] = $row;). Do i need to follow the same or i need to define table fields?

Death-is-the-real-truth Over a year ago

if you want to show limited number of fields then define it like $record['user_name'] = $row['user_name']; and so on . if you want to show all then no need.

JasonJensenDev · Accepted Answer · 2015-06-25 18:36:55Z

4

First off, you should be using PDO to query your database instead of mysql_query(). You code as it is now allows for SQL injection.

Here's how I would do it:

$db = new PDO('mysql:host=CHANGE_THIS_TO_YOUR_HOST_NAME;
    dbname=CHANGE_THIS_TO_YOUR_DATABASE',
    'CHANGE_THIS_TO_YOUR_USERNAME',
    'CHANGE_THIS_TO_YOUR_PASSWORD');

$sql='select * from project where uid = :uid';
$query = $db->prepare($sql);
$query->bindValue(':uid', $_REQUEST['uid']);
$query->execute();
$projects = $query->fetchAll();

Then to display it in a loop, do this:

foreach ($projects as $project) {
   echo '<div class="project">';
   echo '<span class="project-name">'. $project['project_name'] .'</span>';
   echo '<span class="project-dis">'. $project['short_dis'] .'</span>';
   echo '<span class="project-poster">'. $project['poster'] .'</span>';
   echo '</div>';
}

answered Jun 25, 2015 at 18:36

JasonJensenDev

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1 Comment

Roxx Over a year ago

Thanks for the answer. Appreciate it. Voted up. Will test and let you know.

Jens A. Koch · Accepted Answer · 2015-06-25 18:42:29Z

// database query to get result
$result = mysqli_query($conn, "select * from project where uid=$uid")

// new array for all rows 
$rows = array();

// run mysql_fetch_assoc() in a loop (iterate) to get all rows from 
// the whole result set and reassign them to the prepared and 
// empty $rows array
// so that you can later iterate rows again to output your divs
while($row = mysql_fetch_assoc($result)){
    $rows[] = $row;
}

mysql_close($conn);

// now iterate the $rows array to output data for each $row
<?php foreach($rows as $row){ ?>

    // debug output to see what data is in a row
    var_dump($row);

    // print value of a key from the row array
    echo $row['some_keyname'];

   // use stringconcatenation to build your html output, e.g. div    
   echo '<div>' . $row['some_keyname'] . '</div>';

<?php } ?>

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