php - MySQL 尝试执行 JOINed 查询时抛出错误消息

Question

这个问题已经有答案了:

已关闭 10 年前。

Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

这里是代码:

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

我已根据 echo 语句确认 $item_id 是正确的响应。

Answer 1

表 items 位于单引号 (') 中，而不是反引号 (`) 中。