getting some errors

[assassingod]

You don't need to use a variable, just use:

PHP Code:

UPDATE gallery_index SET gallery_views='gallery_views+1' WHERE gallery_id='$id' 


[AN-net]

hmm it just changes it to 0

[assassingod]

1. I wouldn't query in a loop if I were you.
2. I'm not sure why it's not updating - I used the exact same method for my Tutorial Hack and it works perfectly.

[AN-net]

oh i removed, the update query from the loop
heres my current code:

PHP Code:

if($view==gal){
$fgall= $DB_site->query("SELECT gallery_index.gallery_views,gallery.gallery_id,gallery.image_id,gallery.image_name,gallery.image_url,gallery.image_des,gallery.image_views,gallery.image_rating,gallery.image_date FROM gallery,gallery_index WHERE gallery_index.gallery_id='$id' AND gallery.gallery_id='$id'");
while($gallery= $DB_site->fetch_array($fgall))
{
    eval('$gallerybits .= "' . fetch_template('gallery_gal_lvl1') . '";');    
}
$fv= $DB_site->query("SELECT gallery_views FROM gallery_index WHERE gallery_id='$id'");
$views=$DB_site->fetch_array($fv);
$DB_site->query("UPDATE gallery_index SET gallery_views='gallery_views+1' WHERE gallery_id='$id'");
eval('$navbar .= "' . fetch_template('navbar') . '";');
eval('print_output("' . fetch_template('gallery_gal') . '");');
} 


[AN-net]

hmmm i can get this to insert stuff in the table
heres the code:

PHP Code:

if($do==insert){
$DB_site->query("INSERT INTO gallery (gallery_id,artist_name,artist_id,image_name,image_url,image_des) VALUES('$id','$artname','$artid','$imgname','$imgurl','$imgdes')");
} 


heres the form i used:

<table align="center" class="tborder" style="border: 1px outset;">
<tr>
<td><strong>At this time you must host your own images</strong></td>
</tr>
<tr>
<td class="thead">Required Information</td>
</tr>
<tr>
<td class="gal1"><form action="gallery.php?do=insert" method="get">
<input type="hidden" name="id" value="$id">
<input type="hidden" name="artid" value="$bbuserinfo[userid]">
<input type="hidden" name="artname" value="$bbuserinfo[username]">
Image Name:<input type="text" name="imgname" value="Image Name" maxlength="35" size="20">
</td>
</tr>
<tr>
<td class="gal2">Image Description:<br /><textarea name="imgdes" maxlength="900" rows="3" cols="45">Put your image's description here</textarea>
</td>
</tr>
<tr>
<td class="gal1">Image Url:<input type="text" name="imgurl" value="Image's Url" maxlength="255"></td>
</tr>
<tr>
<td class="galhead"><input type="submit" value="Submit Entry"></td>
</tr>
</table>
</form>

[Zero Tolerance]

PHP Code:

if($view==gal){ 
$fgall= $DB_site->query("SELECT gallery_index.gallery_views,gallery.gallery_id,gallery.image_id,gallery.image_name,gallery.image_url,gallery.image_des,gallery.image_views,gallery.image_rating,gallery.image_date FROM gallery,gallery_index WHERE gallery_index.gallery_id='$id' AND gallery.gallery_id='$id'"); 
while($gallery= $DB_site->fetch_array($fgall)) 
{ 
    eval('$gallerybits .= "' . fetch_template('gallery_gal_lvl1') . '";');     
} 
$fv= $DB_site->query("SELECT gallery_views FROM gallery_index WHERE gallery_id='$id'"); 
$views=$DB_site->fetch_array($fv); 
$DB_site->query("UPDATE gallery_index SET gallery_views=gallery_views+'1' WHERE gallery_id='$id'"); 
eval('$navbar .= "' . fetch_template('navbar') . '";'); 
eval('print_output("' . fetch_template('gallery_gal') . '");'); 
} 


Mysql 'field' doesnt go inside the quotes, atleast, i never do that and it works fine for me

[g-force2k2]

First off

PHP Code:

$fv= $DB_site->query("SELECT gallery_views FROM gallery_index WHERE gallery_id='$id'"); 
$views=$DB_site->fetch_array($fv); 


is a waste of a query as it is getting data that already exists.

Also you don't need any quotes at all

PHP Code:

gallery_views=gallery_views+1 


will suffice, and where in the script are you definning the $id, and does everything display correctly just the views will note update?

Only reason I can see that the data doesn't update correctly is that the url is not yourscript.php?view=gal, other then that it should update, unless you also defined the templates to eval elsewhere. You can free the memory that is used by the query after the while loop

PHP Code:

$DB_site->free_result ( $fgall ) ; 


Cheers,
g-force2k2

[AN-net]

thanks guys ill try it out when i get home
gforce i am defining $id through the link to the gallery from the gallery index.
example:
gallery.php?view=gall&id=1
where should i put that free result
im a tid bit new to php but i think im doing ok, also can someone help me out with my post on inserting stuff into the db, i have clue y it doesnt work

edit:
thanks to both of you, its updating views now now i gotta sort out that inserting>_<

edit2:
ok i changed the insert query to this but it still aint working:

PHP Code:

if($do==insert){
$image_date= time();
$DB_site->query("INSERT INTO gallery (gallery_id,artist_name,artist_id,image_name,image_url,image_des,image_date) VALUES('$gallery_id','$artist_name','$artist_id','$image_name','$image_url','$image_des','$image_date')");
} 


<form action="gallery.php?do=insert" method="get">
<input type="hidden" name="gallery_id" value="$id">
<input type="hidden" name="artist_name" value="$bbuserinfo[username]">
<input type="hidden" name="artist_id" value="$bbuserinfo[userid]">
Image Name:<input type="text" name="image_name" value="Image Name" maxlength="35" size="20">
</td>
</tr>
<tr>
<td class="gal1">Image Url:<input type="text" name="image_url" value="Image's Url" maxlength="255"></td>
</tr>
<tr>
<td class="gal2">Image Description:<br /><textarea name="image_des" maxlength="900" rows="3" cols="45">Put your image's description here</textarea>
</td>
</tr>
<tr>
<td class="galhead"><input type="submit" value="Submit Entry"></td>
</tr>
</table>
</form>

[AN-net]

anyone know how to solve my insert query problem?

[g-force2k2]

Originally Posted by AN-net

anyone know how to solve my insert query problem?

Try changing the method to POST and then make it $_POST['do'] == "insert" and also make the form action="gallery.php"

Hope that helps somewhat.

Cheers,
g-force2k2