Question

在此,我与公共卫生和社会福利部提出两个问题。是否更简单? 一个问题不是两个问题?

$id = mysql_real_escape_string($_GET["id"]);
$result = mysql_query("SELECT * FROM questionstable WHERE id=$id");
$row = mysql_fetch_assoc($result);
$category = $row[ category ];
$main = mysql_query("SELECT name FROM categorytable WHERE id=$category");

Answer 1

SELECT       questionstable.*, categorytable.name
FROM         questionstable
  INNER JOIN categorytable
  ON         categorytable.id = questionstable.category
WHERE        questionstable.id=$id

除此以外,假定您的<代码>questionstable.id为数字,您可使用$id = (int)$_GET['id],并节省一些写作。 (这也可能是一个更安全的灯塔。) 仅仅因为逃跑,就是指completely。安全的,特别是如果它不是在引文中,[请你提供插图文件]。

Answer 2

请审判:

SELECT name 
FROM categorytable 
WHERE id = (
              SELECT category 
              FROM questionstable 
              WHERE id = $id
           )

Answer 3

$id = mysql_real_escape_string($_GET["id"]);
$main = mysql_query("SELECT c.name FROM categorytable c inner join questionstable q on c.category = q.category WHERE q.id = $id");

Answer 4

如果找不到“<>/strong”类,则不会使用内心加入使用。

SELECT    questionstable.*, categorytable.name
FROM      questionstable
LEFT JOIN categorytable
ON         categorytable.id = questionstable.category
WHERE      id=$id

友情链接