How to retrieve data in LIFO manner form MySQL???

Hi all,

I have a doubt on how retrieve data form MySQL database in Last-in-First Out manner. I mean to say Recent entry should be on top of the page.


waiting for reply.................


Thank you in advance

 

SELECT * FROM table ORDER BY something [b]DESC[/b]

Code (markup):

"something" is usually "ID"...

 

It is giving this warning
"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in"

This is my code
$query = "SELECT * FROM datatable ORDER BY Sl.no DESC";
Sl.no is autoincremented for every entry.

 

Either i use ASC or DESC i am getting the same result. Why????

 

Plese reply

 

what's your table structure?

better post few lines form you code form query to fetch array

 

My table has a fields Sl.no, subject, title, url, email, date, text


$query = "SELECT * FROM datatable ORDER BY 'Sl.no' DESC ";
$result = mysql_query($query);

while ($row = mysql_fetch_array($result, MYSQL_NUM))
{
if($row != NULL)
{
echo "<b>The subject is" .$row[1]. "</b><br>";
}

echo "<div class='class2'>";
echo "<a href='$row[3]' target='_blank'>$row[2]</a></div>";
echo "<h3>$row[4]$row[5]&nbsp;&nbsp;submitted on $row[6]</h3>";
echo "<p>$row[7]</p>";

Is there any thing wrong??
Either i use DESC or ASC i am getting the same results .

 

Try this:

SELECT * FROM datatable ORDER BY no DESC

Is SI referring to a different table? If so, you need to do a join.

 

It is giving this warning
"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in"

Sl.no is a field in my table.

 

can't fix it yet. Any solution please???

 

mysql_query($query);

Code (markup):

returns not valid mysql resource, but FALSE.
Catch the error and throw error msg!
Like: if (!$result) die (mysql_error());

 

One solution could be donot use '.' in sl.no

Other can be change the field name for testing to SRNO

other can be take another field instead of sl.no and check if it works like change it to subject

Regards

Alex

 

It is returning valid resource. ASC or DESC nothing is working. Just returns Hower the data is entered into data base.

 

I tried buddy, Still not working!!!!

 

Replace:

$query = "SELECT * FROM datatable ORDER BY 'Sl.no' DESC ";


with:

$query = "SELECT * FROM datatable ORDER BY `Sl.no` DESC";

 

it gives
"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in"

I got to fix this immediately, please help me.

 

Hey Buddies,

It is working.....

Thanks, Thanks, Thanks, Thanks

A special Thanks to "SoKickIt"

 

Okay, so i want to do this exact same thing, but i only want to show the last three or so entries into the table, not all of them, any way i can do this? Thanks for your help in advance

 

Just add "LIMIT 3" at the end.

 

thanks i thought that would work, but just wanted to make sure.