Is it possible to sort MySql results with select dropdown list?

Question

I'm a project about an online publication system and in this website there are users . I have already get MySql results on my website, for some reason, the results are random. I want to sort users database result with a drop down list using the select, option tag...what should I use to get this function works ? Here is my code for what I want to sort by :

    <form>
    <label>Sort By:</label>
        <select name="sort" id="sort" style="float: left;">
            <option value="ID">Id Number</option>
            <option value="name">Name A-Z</option>
            <option value="surname">Name Z-A</option>
            <option value="title">Academic Title</option>
            <option value="department">Departments</option>

        </select>
    </form>

Sure. The script can add ORDER BY to the query, and base the parameters to this option on $_GET['sort']. — Barmar
– Barmar, Commented Apr 19, 2014 at 15:27
You can use sql order by, or you get result and store to array and use php to sort. — lighter
– lighter, Commented Apr 19, 2014 at 15:27
Dear Mr @Barmar if I post my code could you please help me and provide me with more detail? Thanks in advance. — Ahmed Abdulrahman
– Ahmed Abdulrahman, Commented Apr 19, 2014 at 15:38
Be a real programmer and try it yourself. If you can't get it working, show what you tried and we'll help you fix it. — Barmar
– Barmar, Commented Apr 19, 2014 at 15:46

Vishnuraj V · Accepted Answer · 2014-04-19 22:08:59Z

1

Try this

<html>
<head>
<script>
function sortResult(str)
{
if (str=="")
  {
  document.getElementById("result").innerHTML="";
  return;
  } 
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("result").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","results.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<form>
<select name="sortby" onchange="sortResult(this.value)">
            <option value="ID">Id Number</option>
            <option value="name">Name A-Z</option>
            <option value="surname">Name Z-A</option>
            <option value="title">Academic Title</option>
            <option value="department">Departments</option>
</select>
</form>
<br>
<div id="result"><b>Results will be listed here.</b></div>

</body>
</html>

And your results.php

<?php
$q = $_GET['q'];

$con = mysqli_connect('localhost','username','password','my_db');
if (!$con)
  {
  die('Could not connect: ' . mysqli_error($con));
  }

mysqli_select_db($con,"database");
$sql="SELECT * FROM tablename ORDER BY ".$q;

$result = mysqli_query($con,$sql);

echo "<table border='1'>
<tr>
<th>ID</th>
<th>Name</th>
<th>Surname</th>
<th>Title</th>
<th>Department</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" .$row['ID']. "</td>";
  echo "<td>" .$row['name']. "</td>";
  echo "<td>" .$row['surname'] ."</td>";
  echo "<td>" .$row['title']. "</td>";
  echo "<td>" .$row['department']. "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
?>

edited Apr 19, 2014 at 22:08

answered Apr 19, 2014 at 21:23

Vishnuraj V

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2 Comments

Ahmed Abdulrahman Over a year ago

Ohh Thank you bro for you help I really appreciate it , Can i send you MySql code that i have used ?? @vishnurajv

Strawberry Over a year ago

You already have the result set, so there's no point hitting the database again.

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