Bug 274877. SQL Error on moreinfo.php when category is specified.

This commit is contained in:
psychoticwolf%carolina.rr.com 2004-12-16 11:46:56 +00:00
Родитель 1a35a6a301
Коммит 28ddf99649
2 изменённых файлов: 4 добавлений и 4 удалений

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@ -113,11 +113,11 @@ $sql = "SELECT TM.ID, TM.Name, TM.DateAdded, TM.DateUpdated, TM.Homepage, TM.Des
INNER JOIN os TOS ON TV.OSID = TOS.OSID";
if ($category && $category !=="%") {
$sql .="INNER JOIN categoryxref TCX ON TM.ID = TCX.ID
$sql .=" INNER JOIN categoryxref TCX ON TM.ID = TCX.ID
INNER JOIN categories TC ON TCX.CategoryID = TC.CategoryID ";
}
if ($editorpick=="true") {
$sql .="INNER JOIN reviews TR ON TM.ID = TR.ID ";
$sql .=" INNER JOIN reviews TR ON TM.ID = TR.ID ";
}
$sql .=" WHERE TM.ID = '$id'";

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@ -113,11 +113,11 @@ $sql = "SELECT TM.ID, TM.Name, TM.DateAdded, TM.DateUpdated, TM.Homepage, TM.Des
INNER JOIN os TOS ON TV.OSID = TOS.OSID";
if ($category && $category !=="%") {
$sql .="INNER JOIN categoryxref TCX ON TM.ID = TCX.ID
$sql .=" INNER JOIN categoryxref TCX ON TM.ID = TCX.ID
INNER JOIN categories TC ON TCX.CategoryID = TC.CategoryID ";
}
if ($editorpick=="true") {
$sql .="INNER JOIN reviews TR ON TM.ID = TR.ID ";
$sql .=" INNER JOIN reviews TR ON TM.ID = TR.ID ";
}
$sql .=" WHERE TM.ID = '$id'";