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Code: Alles auswählen

Allgemeiner Fehler
SQL ERROR [ mysql4 ]

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT smiley_url FROM phpbb_smilies WHERE smiley_id = 43)' at [1064]

SQL

SELECT smiley_id FROM phpbb_smilies WHERE smiley_id <> 43 AND smiley_url = (SELECT smiley_url FROM phpbb_smilies WHERE smiley_id = 43)

BACKTRACE

FILE: includes/db/mysql.php
LINE: 174
CALL: dbal_mysql->sql_error()

FILE: includes/mods/smilies_categories.php
LINE: 123
CALL: dbal_mysql->sql_query()

FILE: includes/acp/acp_icons.php
LINE: 462
CALL: update_smiley_categories()

FILE: includes/functions_module.php
LINE: 507
CALL: acp_icons->main()

FILE: adm/index.php
LINE: 74
CALL: p_master->load_active()

Code: Alles auswählen

		$sql = 'SELECT smiley_id FROM ' . SMILIES_TABLE . '
			WHERE smiley_id <> ' . $smiley_id . '
			AND smiley_url = (SELECT smiley_url FROM ' . SMILIES_TABLE . ' WHERE smiley_id = ' . $smiley_id . ')';
		$result = $db->sql_query($sql);

Code: Alles auswählen

		$sql = 'SELECT smiley_url FROM ' . SMILIES_TABLE . ' WHERE smiley_id = ' . $smiley_id;
		$result = $db->sql_query($sql);
		$row = $db->sql_fetchrow($result);
		$smiley_url = $row['smiley_url'];
		$db->sql_freeresult($result);

		$sql = 'SELECT smiley_id FROM ' . SMILIES_TABLE . '
			WHERE smiley_id <> ' . $smiley_id . "
			AND smiley_url = '$smiley_url'";