Nieuwste plaatje weergeven in Portal: SQL foutmelding

[_Ron_]

Ik heb vandaag de Foto album MOD geïnstalleerd: Werkt perfect!
Nu heb ik tevens de add-on "laatste plaatje weergeven" voor de Portal geïnstalleerd, maar krijg ik een SQL-error.

Code: Selecteer alles

Could not query album information

DEBUG MODE

SQL Error : 1054 Unknown column 'id' in 'field list'

SELECT id, title, username, timestamp FROM phpbb_album ORDER BY timestamp DESC LIMIT 0,1

Line : 139
File : /home/virtual/site109/fst/var/www/html/forum/portal.php

Ik ben geen SQL-held, dus beroep ik me maar op de experts.

Dit is wat er is gewijzigd in portal.php:

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#-----[ VIND ]----------------------------------------
#
else
{
	$l_total_user_s = $lang['Registered_users_total'];
}

#
#-----[ VOEG DAARNA TOE ]-----------------------------------
#
//
// Haal nieuwste plaatje
//
$sql = "SELECT id, title, username, timestamp FROM phpbb_album ORDER BY timestamp DESC LIMIT 0,1";

if (!$result = $db->sql_query($sql))
{
	message_die(GENERAL_ERROR, 'Could not query album information', '', __LINE__, __FILE__, $sql);
}

$picrow = $db->sql_fetchrow($result);
//
// BEEINDIG - Haal nieuwste plaatje
//


#

Rara waar zit de vaudt?

-edit- linkje: http://www.hanff.nl -edit-

[Carlo Claessen]

welke versie foto album heb je

[_Ron_]

Oja.... wel handig als ik dat erbij vermeld
Het is versie 2.0.51
Ik heb ook even gekeken op de Smartor site. Daar wordt trouwens dezelfde vraag gesteld: http://smartor.is-root.com/viewtopic.ph ... ght=#20556

[WebSiteNet]

ik heb de orginele sql query die de tabel aanmaakt:

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CREATE TABLE phpbb_album (
	pic_id int(11) UNSIGNED NOT NULL auto_increment,
	pic_filename varchar(255) NOT NULL,
	pic_thumbnail varchar(255),
	pic_title varchar(255) NOT NULL,
	pic_desc text,
	pic_user_id mediumint(8) NOT NULL,
	pic_username varchar(32),
	pic_user_ip char(8) NOT NULL DEFAULT '0',
	pic_time int(11) UNSIGNED NOT NULL,
	pic_cat_id mediumint(8) UNSIGNED NOT NULL DEFAULT '1',
	pic_view_count int(11) UNSIGNED NOT NULL DEFAULT '0',
	pic_lock tinyint(3) NOT NULL DEFAULT '0',
	pic_approval tinyint(3) NOT NULL DEFAULT '1',
	PRIMARY KEY (pic_id),
	KEY pic_cat_id (pic_cat_id),
	KEY pic_user_id (pic_user_id),
	KEY pic_time (pic_time));

Daarvan hoef je niet denderend veel van te begrijpen. elke kolomnaam begint met pic_ en in die foute query vraagt hij veldnamen uit die tabel zonder pic_
Dit moet die query een beetje zijn.

Code: Selecteer alles

SELECT pic_id, pic_title, pic_username, pic_timestamp FROM phpbb_album ORDER BY pic_timestamp DESC LIMIT 0,1

Rara waar zit de vaudt?

in je spelling.

[993ti]

In portal.php

Code: Selecteer alles

Vind-----
// Exceptional Forums for Recent Topics, eg. '2,4,10' (note: my Recent Topics script has its own permission checking, so you can leave this variable blank)
$CFG['exceptional_forums'] = '';

Voeg daarna toe:

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// Recent Pics please select sort by 
$sortby = 'last_pic' ; // Sort By Last Pic 
//$sortby = 'good_rate' ; // Sort by Good Average Vote 
//$sortby = 'time_rate' ; // Sort By rating et Last Time 

Vind

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$l_total_user_s = $lang['Registered_users_total'];
}

Voeg daarna toe

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// Get Newest Pic 
// 
$sql = "SELECT * 
FROM ". ALBUM_CONFIG_TABLE; 
if(!$result = $db->sql_query($sql)) 
{ 
message_die(GENERAL_ERROR, "Could not query Album config information", "", __LINE__, __FILE__, $sql); 
} 
while( $row = $db->sql_fetchrow($result) ) 
{ 
$album_config_name = $row['config_name']; 
$album_config_value = $row['config_value']; 
$album_config[$album_config_name] = $album_config_value; 
} 

$sql = "SELECT p.pic_id, p.pic_title, p.pic_desc, p.pic_user_id, p.pic_user_ip, p.pic_username, p.pic_time, p.pic_cat_id, p.pic_view_count, u.user_id, u.username, r.rate_pic_id, AVG(r.rate_point) AS rating, COUNT(DISTINCT c.comment_id) AS comments 
FROM " . phpbb_album . " AS p 
LEFT JOIN ". USERS_TABLE ." AS u ON p.pic_user_id = u.user_id 
LEFT JOIN ". ALBUM_RATE_TABLE ." AS r ON p.pic_id = r.rate_pic_id 
LEFT JOIN ". ALBUM_COMMENT_TABLE ." AS c ON p.pic_id = c.comment_pic_id 
WHERE pic_approval = 1 
GROUP BY p.pic_id "; 

if ($sortby == "last_pic") { 
$sql .= " ORDER BY pic_time DESC LIMIT 0,1"; 
} 
if ($sortby == "good_rate") { 
$sql .= " ORDER BY rating DESC LIMIT 0,1"; 
} 
if ($sortby == "time_rate") { 
$sql .= " ORDER BY rating DESC,pic_time DESC LIMIT 0,1"; 
} 

if (!$result = $db->sql_query($sql)) 
{ 
message_die(GENERAL_ERROR, 'Could not query album information', '', __LINE__, __FILE__, $sql); 
} 

$picrow = $db->sql_fetchrow($result); 

if(!$picrow['rating']) 
{ 
$picrow['rating'] = $lang['Not_rated']; 
} 
else 
{ 
$picrow['rating'] = round($picrow['rating'], 2); 
} 


// 
// END - Get Newest Pic 

In portal_body.tpl
Vind:

Code: Selecteer alles

<table width="100%" cellpadding="2" cellspacing="1" border="0" class="forumline">
		   <tr>
			<td class="catHead" height="25"><span class="genmed"><b>{L_STATISTICS}</b></span></td>
		   </tr>

Voeg daarvoor toe:

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<table width="100%" cellpadding="2" cellspacing="1" border="0" class="forumline"> 
              <tr> 
<td class="catHead" height="25"><span class="mcattitle"><b>{L_NEWEST_PIC}</b></span></td> 
</tr> 
<tr> 
<td class="row1" align="left"><span class="gensmall" style="line-height:150%"><center><a href="{U_PIC_LINK}"><img src="{PIC_IMAGE}" border="0"></a><br /></span></td> 
</tr> 
<tr> 
<td align="left" valign="top" class="row2"><span class="gensmall"> 
{L_PIC_TITLE}: {TITLE}<br /> 
{L_POSTER}: {POSTER}<br /> 
{L_POSTED}: {TIME}<br /> 
{L_VIEW}: {VIEW}<br /> 
{RATING} 
{COMMENTS} 
<div align="right"></a></div></span></td> 
</tr> 
<tr><td height="15"></td></tr>

              </table> 

              <br />

Heb zelf ook 2.0.51 en werkt perfect.

Grtzzzzzz

[_Ron_]

Ik ga 't dit weekend proberen. Thnx voor het reageren!

[_Ron_]

Helaas......... Ik heb zomaar het idee dat een andere mod/hack conflicteert i.c.m. met het gebruik van het Aeolus-Template.

Dit gedeelte geef ik maar op..... Bedankt voor jullie hulp allemaal!
:thumb: