Question

我有一个非常简单的查询，但我需要在其中进行另一个查询。我对SQL和PHP知之甚少，所以我想请你帮忙。

$result = mysql_query("SELECT * FROM img");
while ($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

这看起来很简单JOIN来获取类别名称。您可以使用单个查询：

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

用上面的代码替换你的初始查询，它将消除对内部查询的需要。

Answer 2

这应该有效。您将看到每行中每个表的所有img属性以及cat.name值。

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

Answer 3

你可以在这个上做一个简单的子查询：

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

PHP - 查询内部查询

3 个答案: