如果问题在于你不知道用户将选择哪一种标准,但希望将结果退回到“空白”标准上,那么你可以采取以下措施:

$criteria_1 = $_POST[ criteria_1 ];
$criteria_2 = $_POST[ criteria_2 ];
$criteria_3 = $_POST[ criteria_3 ];

if(!$criteria_1 && !$criteria_2 && !$criteria_1) {

     echo "You must select at least one criteria!";

} else {

   // Run query mentioned below and return results.       

} 

接下来,他会问:

SELECT * from mytable
WHERE
(criteria1 =  $criteria_1  OR  $criteria_1  =   ) AND
(criteria2 =  $criteria_2  OR  $criteria_2  =   ) AND
(criteria3 =  $criteria_3  OR  $criteria_3  =   )

这将把任何空白(非选用)参数作为空白处理,忽视这些参数。 了解到,如果没有规定标准,它将恢复所有成果。

Another way to write the above is:

SELECT * from mytable
WHERE
criteria1 IN ( $criteria_1 ,   ) AND
criteria2 IN ( $criteria_2 ,   ) AND
criteria3 IN ( $criteria_3 ,   )

同样,允许完全不加入所有标准1 的结果。

下面是你再次要求的例子:

$query = "SELECT * FROM mytable";

if ($_POST[ name ] == "Jack") {
   $query .= " WHERE name =  Jack ";
}
if ($_POST[ name ] == "Bob") {
   $query .= " WHERE name =  Bob ";
}

if ($_POST[ state ] != "") {
   $query .= " AND state =  " . mysql_real_escape_string($state) . " ";
}

//So now, in total, your query might look like this
//"SELECT * FROM mytable WHERE name =  Bob  AND state =  $state "    

$result = mysql_query($query);

页: 1 随附if的报表,在您对所有<代码>_POST变量进行抽查后即执行查询。

我看到了这样的问题,这样,如果你不想把某一栏的价值放在一栏中,你就把这栏放在了联合国利比里亚办事处:

SELECT *
FROM users
WHERE param1 = :param1
UNION
SELECT *
FROM users
WHERE param2 = :param2
UNION
SELECT *
FROM users
WHERE param3 = :param3

这假设,每栏中每一栏都有索引,你重新进行“蓝色”和搜索(并使用《设计书》)。

use your scripting language (php) to loop over the inputs...

这样就有一个结构:

WHERE 1=1

之后添加:

AND paramx =  $px  

目 录

$criteria = array();

//Populate your criteria and parameter arrays with input from the web page here
...
// $criteria should now have stuff in it

$sql = "SELECT * FROM mytable ";//Or whatever your sql query is
$count = 0;
foreach ($criteria as $key => $parameter) {
  if ($count == 0) {
    $sql = $sql."WHERE ".$key." = ".$parameter;
  } else {
    $sql = $sql."AND ".$key." = ".$parameter;
  }
  $count++;
}

尽管如此,这极易受到 s的注射攻击。 Try using