0

可能重复：
警告：mysql_fetch_array()：提供的参数不是有效的 MySQL 结果

这是代码：

$result=mysql_query("
SELECT items.items_id, 
COUNT(ratings.item_id) AS TotalRating,
AVG(ratings.rating) AS AverageRating
FROM 'items'
LEFT JOIN ratings ON (ratings.item_id = items.items_id)
WHERE ratings.item_id = '{$item_id}' ;");

echo "Error message = ".mysql_error();

while($row=mysql_fetch_assoc($result)) {
      $output[]=$row;
}

这是错误：

Error message = 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 ''items' LEFT JOIN ratings ON (ratings.item_id = items.items_id) WHERE ' at line 4
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/k/i/c/kickinglettuce/html/Kickinglettuce/ratethis/get_ratings.php on line 38
null

我已经确认 $item_id 是基于 echo 语句的正确响应。

4

1 回答 1

2

您items在 SINGLE QUOTES (') 而不是 BACKTICKS(`) 中有表。

于 2012-08-01T03:14:28.330 回答