Question

我正在尝试从数据库中获取查询，但是当我回显bindParam和(:name)时，我对$sql print_r($stmtTwo)的值没有受到限制WHERE子句指出WHERE :name而不是来自$wherefinal的字符串。

我的代码是：

$sql= "SELECT Species.Species_ID 
    FROM Species 
    JOIN ( 
        SELECT Species.Species_ID, COUNT(*) AS mynum 
        FROM Species_Opt LEFT JOIN Species ON (Species.Species_ID = Species_Opt.SO_Species_ID) 
        WHERE :name
        GROUP BY SO_Species_ID HAVING mynum = 6 
        ) AS mytable ON Species.Species_ID = mytable.Species_ID";
$stmtTwo = $pdo->prepare($sql);
$stmtTwo->bindParam(':name', $wherefinal);
$stmtTwo->execute();

在sql语句之前定义$wherefinal并定义为：

$where = "";
foreach ($_POST as $k => $v){
    $where .= "(Species_Opt.SO_Option_ID = $v) OR "; 
};
$wherefinal = substr($where, 0, strrpos($where, " OR "));

当回显时，$wherefinal显示：

(Species_Opt.SO_Option_ID = 4) OR (Species_Opt.SO_Option_ID = 12) OR (Species_Opt.SO_Option_ID = 17) OR (Species_Opt.SO_Option_ID = 20) OR (Species_Opt.SO_Option_ID = 21) OR (Species_Opt.SO_Option_ID = 32)

$v来自通过不同的SQL语句生成的表单中的单选按钮的值。

Answer 1

首先，您需要构建占位符部分

$placeholder = str_repeat('?,', count($_POST) - 1) . '?';

然后使用它来构建完整的SQL查询

$sql= "SELECT Species.Species_ID 
       FROM Species 
       JOIN ( 
          SELECT Species.Species_ID, COUNT(*) AS mynum 
          FROM Species_Opt LEFT JOIN Species ON 
               (Species.Species_ID = Species_Opt.SO_Species_ID) 
          WHERE Species_Opt.SO_Option_ID IN ({$placeholder})
       GROUP BY SO_Species_ID HAVING mynum = 6 
       ) AS mytable ON Species.Species_ID = mytable.Species_ID";

之后，您可以将值绑定到每个占位符

$sh = $pdo->prepare($sql);

$i = 1;
foreach($_POST as $value) {
    $sh->bindValue($i, $value); $i++;
}
$sh->execute();

PDO bindParam / Value的问题

1 个答案: