How to retrieve data from mysql php and show in table from dropdown

Question

Hey guys i am a newbie to php.What problem i am facing is i have created a dropdown which is populated from the data from database using this code.This is working fine for me and it is populating dropdown too

    include('connect.php');
    $query="select * from faculty";
    $result=mysql_query($query);
    while($row = mysql_fetch_assoc($result))
    {$dropdown.="\r\n<option value='{$row['Designation']}'>{$row['Designation']}      </option>";}
    echo "<select>".$dropdown."</select>";

Solution i want is,when a user selects a value from dropdown,result should be retrieved from database and should be displayed in table.Please help me guys

i have 6 columns in my table and i am creating a dropdown in which i am giving designation column values .So wen a user select electrical branch from it,table should display all the rows for faculty members who are from electrical branccs — dhruvan
– dhruvan, Commented Feb 11, 2013 at 9:22
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — h2ooooooo
– h2ooooooo, Commented Feb 11, 2013 at 9:55

Ankur · Accepted Answer · 2013-02-11 09:26:05Z

0

You have to basically :
1) Perform a form sending to some server side script(PHP in your case) when there is a change in the dropdown selection (use onchange event for the dropdown) and fetch the values from the db ,
2) Tell the server side script to spit out an html string which contains the table containing the desired information.3)
3) Output the string on your page.

This will do a page refresh.

If you donot want to have a page refresh, resort to using Ajax. P.S. I recommend using some framework such as jQuery in case you need to use Ajax

answered Feb 11, 2013 at 9:26

Ankur

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3 Comments

dhruvan Over a year ago

i ll b very thankfull if u can provide me a link to the tutorial which will help me

Ankur Over a year ago

You can have a look into these links: 1) stackoverflow.com/questions/5704180/… 2) daniweb.com/web-development/php/threads/422052/…

Ankur Over a year ago

Answers should be marked accepted if the problem gets solved. This shows that SO is being useful in solving problems.

GBRocks · Accepted Answer · 2013-02-11 09:46:59Z

0

What you have to do is something like this:

In your Html:

<select onchange="fetchContent()">
    <option id="1_Designation">abcd</option>
    <option id="2_Designation">1234</option>
    <option id="3_Designation">lkjh</option> 
</select>

In your javascript:

fetchContent()
{
    id = $(this).id;
    $.ajax({
      type: "POST",
      url: "/path/content.php?id="+id,
      success: function(response) {
         $("#tableRow").html(response);
      }
   });
}

In content.php you will have to get the value of id and then do the necessary data retrieval and then return the data.

$id = $_POST['id'];
//retrieve the data to $data
echo $data;

answered Feb 11, 2013 at 9:46

GBRocks

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1 Comment

Ankur Over a year ago

Nice way to give a feel of things :) I just wish to enhance this answer by changing:

$.ajax({ type: "POST", url: "/path/content.php?id="+id, success: function(response) { $("#tableRow").html(response); } })

to $.post({ "/path/content.php", {id: id}, function(response) { $("#tableRow").html(response); } })

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