how to display output from mysql in random mannerusing php

Question

I want to display output from mysql using php in random manner. by default whole array will be displayed sequentially.

echo "<table border='1'>

</tr>
<tr>
<th>User ID</th>
<th>User </th>
<th>Site </th>
<th>IP Address</th>
<th>Date</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
 echo "<td>" . $row['userID'] . "</td>";
  echo "<td>" . $row['user_name'] . "</td>";
  echo "<td>" . $row['site_name'] . "</td>";
  echo "<td>" . $row['ip_address'] . "</td>";
  echo "<td>" . $row['date'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

I want to output to be displayed randomly on each access.

i.e. row1 is displayed on first access on top of output. row10 is displayed on 2nd access on top of output. row13 is displayed on 3rd access on top of output.

thanks for suggestions and help.

fortune · Accepted Answer · 2014-06-03 09:50:17Z

2

you can use mysql rand() for the purpose:

For eg:

SELECT * FROM myTable ORDER BY RAND();

answered Jun 3, 2014 at 9:50

fortune

3,3721 gold badge23 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

shabeer · Accepted Answer · 2014-06-03 09:50:01Z

1

You can use ORDER BY RAND() in your sql query

SELECT * FROM `table_name` ORDER BY RAND()

You can also shuffle your php array using shuffle($my_array);

You can also use

function shuffle_assoc($list) { 
  if (!is_array($list)) return $list; 

  $keys = array_keys($list); 
  shuffle($keys); 
  $random = array(); 
  foreach ($keys as $key) { 
    $random[$key] = $list[$key]; 
  }
  return $random; 
}

answered Jun 3, 2014 at 9:50

shabeer

1,0649 silver badges18 bronze badges

Collectives™ on Stack Overflow

how to display output from mysql in random mannerusing php

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related