创建视图：Php->PDO->postgres PDOStatement::execute(): SQLSTATE[42P18]: Indeterminate datatype

Question

正常的选择、插入等工作，现在我被困在试图创建一个视图并且占位符似乎存在一些问题，如果我插入一个静态数字而不是占位符，则该行可以工作...我首先尝试没有强制转换，然后是隐式强制转换 (::int)，现在是显式强制转换，但它不想工作....

$sql = "CREATE OR REPLACE VIEW allowedlists AS
         SELECT distinct l.id, l.nom FROM listing as l,acl,user2acl
         WHERE userid=cast(:userid as INTEGER) AND acl.id=aclid 
          and value > (select value from acl where name='saisie');";
$stmt = $csh->handler->pdo->prepare($sql); 
if(!$stmt->bindValue(":userid", $_SESSION["userid"],PDO::PARAM_INT))
             print("failed to bind 'userid = ".$_SESSION["userid"]."' as ".PDO::PARAM_INT." to ".print_r($stmt,true)."<br>\n"); 
else print("bound 'userid' to ".$_SESSION["userid"]." of type '".PDO::PARAM_INT."'<br\n>");
print("before create view<br>\n"); 
if(!($viewcreation = $stmt->execute()))
{   
  print("<p style=\"color:red;\">Error: (".$stmt->errorCode()."): ".print_r($stmt->errorInfo(),true)."</p>\n");
  $stmt->debugDumpParams();  
} 
print("after create view<br>\n");

失败：

before create view(1)<br>
<br />
<b>Warning</b>:  PDOStatement::execute(): SQLSTATE[42P18]: Indeterminate datatype: 7 ERROR:  could not determine data type of parameter $1 in <b>/var/www/html/content/listing.php</b> on line <b>16</b><br />
<p style="color:red;">Error: (42P18): Array
(
    [0] => 42P18
    [1] => 7
    [2] => ERROR:  could not determine data type of parameter $1
)       
</p>    
SQL: [209] CREATE OR REPLACE VIEW allowedlists AS SELECT distinct l.id, l.nom FROM listing as l,acl,user2acl WHERE userid=cast(:userid as INTEGER) AND acl.id=aclid and value > (select value from acl where name='saisie');
Params:  1
Key: Name: [7] :userid
paramno=0
name=[7] ":userid"
is_param=1
param_type=1
after create view<br>

似乎我在这里遗漏了明显的东西，但我现在转了 3 天，但我唯一设法建立的是挫败感？

据我所知，占位符绑定得很好，类型看起来也不错，如上所述，如果我用静态数字替换占位符，它可以工作，但不能用占位符:(

Answer 1

创建 VIEW 是一个 DDL 操作，类似于创建表。因此，它不能被参数化。然后使用“参数化”它，使用参数是select语句的where子句。

create or replace view allowedlists as
   select distinct l.id, l.nom 
     from listing as l,acl,user2acl
         where  acl.id=aclid 
          and value > (select value from acl where name='saisie'); 
... 

select id, nom 
  from allowedlists
 where id = :userid; 
 
 OR       

select id, nom 
  from allowedlists
 where id = cast(:userid as integer); 

由于 where 子句会延迟到从视图中选择，因此可能需要将列添加到创建视图的选择中。由于您没有为所有列加上别名，我​​无法分辨。参数化名称需要类似的方法，但您的选择