sql select current images code

Question

Here is my code

@$sql="select * from mp_images where id='5' and status='0'";
@$query=mysql_query($sql);
while(@$row=mysql_fetch_array($query))
{
    @$image=$row ['photo'];
?>
<img src="image/<?php echo $image; ?>" width="360" height="150">
<?php
}
?>

I want to display current latest images after I have added it into database

@VivekSadh: How would the OP be getting an error when he's suppressing them? — eggyal
– eggyal, Commented Jul 15, 2013 at 8:11
@eggyal Thanks for info. I never knew that @ is used to suppress error message. Thanks :) — Vivek Sadh
– Vivek Sadh, Commented Jul 15, 2013 at 8:15
Do you mean that you want to get the image of the latest inserted or updated record? — Dinup Kandel
– Dinup Kandel, Commented Jul 15, 2013 at 8:25
Please don't use the deprecated mysal_* extension anymore. deprecated means it's no longer supported, and will be removed some time in the future. Switch to mysqli_* or PDO. Also: never suppress errors. They're still being issued (which slows you down) and debugging is made a lot harder. Errors are there for a reason, use them to your advantage. Think of them as a fire alarm: you can use earplugs to not hear the warning and burn to a crisp, or you can get a fire extinguisher to stop the fire burning your house down. I know which one I'd prefer... — Elias Van Ootegem
– Elias Van Ootegem, Commented Jul 15, 2013 at 8:31

Teemu Tapanila · Accepted Answer · 2013-07-16 11:56:35Z

1

Just order them by id

@$sql="select * from mp_images where status='0' order by id desc limit 1";

edited Jul 16, 2013 at 11:56

answered Jul 15, 2013 at 8:08

Teemu Tapanila

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1 Comment

5zellsb Over a year ago

Hi..thanks for your answer, but I just want to show only 1 recent/latest image. Just now it show the list. How to do that thanks...

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