Zymic Webmaster Forums > Show Data From Mysql Only If There Is Data In The Table

Full Version: Show Data From Mysql Only If There Is Data In The Table

Zymic Webmaster Forums > Web Design & Development > Server Side Scripting > PHP

MrTouz

Nov 14 2008, 11:05 PM

Ok very hard to explain :

I have a mysql table with 10 fields, i have a form with 10 fields, i have a display page with 10 fields.
So if i fill the 10 fields in my form, it will send the 10 infos to the database and my display page will show the 10 fields.
Problem is, if i fill only 4 fields in my form it will only send 4 info into the table and my display page will show 10 fields, 4 full 6 empty.

Basically i have a CMS that allows me to upload pics to my site, i can upload 10 pictures so my database has 10 fields, on my display page i have 10 lines repeating

CODE

<img src="pictures/". $row['pic'] .">
<img src="pictures/". $row['pic1'] .">
<img src="pictures/". $row['pic2'] .">
....

Problem is if i only uploaded 4 pics, my display page has the code for 10 pics, but no data can be retrieved for 10 so it is displaying a box with a broken image in it :/, what i want to basically do is only show as many images as i uploaded.

Is there like a code i can put between my IMG tags saying IF data than display the code... if NO data than don't display the IMG code ?

I am trying to find a way to it... but i am failing

I NEED HELP PLEASE !!!
Thanks :'(

Jetteh22

Nov 15 2008, 02:02 AM

Not sure if this is exactly wat you mean or not but...

CODE

<?php

if($row['pic'] != "") // checking to see if the field is empty
{
echo ?>
<img src="pictures/<?php echo $row['pic'];?>"/>
<?php
}
?>

basically... it only displays that field if $row['XXXXX'] is NOT empty... A little sloppy, but that's how I code :\ It all makes sense to me in the end. Just change the $row['XXXX'] to whatever pic2, pic3, pic4 ect

swordz

Nov 15 2008, 05:06 PM

QUOTE(Jetteh22 @ Nov 15 2008, 02:02 AM)

Not sure if this is exactly wat you mean or not but...

CODE

<?php

if($row['pic'] != "") // checking to see if the field is empty
{
echo ?>
<img src="pictures/<?php echo $row['pic'];?>"/>
<?php
}
?>

The if statement could be simply

CODE

if($row['pic'])

or possibly better

CODE

if(isset($row['pic']))

, but Jetteh22's way will work.

swordz

MrTouz

Nov 15 2008, 11:50 PM

Thanks a lot, that is saving me some crazy time.

noxus

Nov 28 2008, 08:21 PM

CODE

$result = mysql_query("select image from this_table where image !="" ");

while( list( $image ) = mysql_fetch_row( $result ) ) {
echo "<img src=\"$image\">";
}

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