Question

我有一个非常简单的问询,但我需要再问一下。我对卡什卡和购买力平价的了解很少,因此我要请你提供帮助。

$result = mysql_query("SELECT * FROM img");
while ($row = mysql_fetch_assoc($result))
{
   $imgName = $row[ name ];
   $catID = $row[ catid ];
   // Problem
   // Need to get category NAME from category ID
   $result2 = mysql_query("SELECT name FROM cat WHERE id = $catID");
   while ($row2 = mysql_fetch_assoc($result2))
   {
      $catName = $row2[ name ];
   }

   echo "Image: $imgName <br />";
   echo "Category: $catName";
}

Answer 1

这看起来是一个简单的<代码>JOIN,以获得这一类别的名称。您可使用单一询问:

SELECT
  img.id AS imgId,
  img.name AS imgName,
  cat.name AS catName
FROM img JOIN cat ON img.catid = cat.id

将最初的询问改为上文,这将消除内部询问的需要。

Answer 2

您可以在此问题上采取简单的形式:

$result = mysql_query("
SELECT name, catid, (SELECT name FROM cat WHERE id = img.catid) AS cat_name
FROM img
");

Answer 3

这应当行之有效。参看每一行各表的所有成像特性,以及目录价值。

SELECT *, cat.name as catname FROM img
INNER JOIN cat
ON cat.id = img.catid

友情链接