2

所有查询都成功执行,当我检查MySQL行中的表时,插入成功且没有任何错误,但lastInsertId() 返回 0。为什么?

我的代码:

// queries executes successfully, but lastInsetId() returns 0
// the menus table has `id` column with primary auto_increment index
// why lastInsertId return 0 and doesn't return actual id?


$insertMenuQuery = " 
 SELECT @rght:=`rght`+2,@lft:=`rght`+1 FROM `menus` ORDER BY `rght` DESC limit 1; 
 INSERT INTO `menus`(`parent_id`, `title`, `options`, `lang`, `lft`, `rght`) 
      values 
  (:parent_id, :title, :options, :lang, @lft, @rght);";
     try {
           // menu sql query
           $dbSmt = $db->prepare($insertMenuQuery);

           // execute sql query
           $dbSmt->execute($arrayOfParameterOfMenu);
           // menu id
           $menuId = $db->lastInsertId();

           // return
           return $menuId;

     } catch (Exception $e) {
          throw new ForbiddenException('Database error.' . $e->getMessage());
     }
4

2 回答 2

7

对于 PDO_MySQL,我们必须使用

$DB->setAttribute(PDO::ATTR_EMULATE_PREPARES,TRUE); // there are other ways to set attributes. this is one

这样我们就可以运行多个查询,例如:

$foo = $DB->prepare("SELECT * FROM var_lst;INSERT INTO var_lst (value) VALUES ('durjdn')");

但可悲的是,这样做可以减轻 $DB 返回正确插入 ID 的负担。您必须单独运行它们才能检索插入 ID。这将返回正确的插入 ID:

$DB->setAttribute(PDO::ATTR_EMULATE_PREPARES,TRUE);
$foo = $DB->prepare("INSERT INTO var_lst (value) VALUES ('durjdn')");
$foo->execute();
echo $DB->lastInsertId();

但这不会:

$DB->setAttribute(PDO::ATTR_EMULATE_PREPARES,TRUE);
$foo = $DB->prepare("SELECT * FROM var_lst;INSERT INTO var_lst (value) VALUES ('durjdn')");
$foo->execute();
echo $DB->lastInsertId();

这甚至不会运行两个查询:

$DB->setAttribute(PDO::ATTR_EMULATE_PREPARES,FALSE); // When false, prepare() returns an error
$foo = $DB->prepare("SELECT * FROM var_lst;INSERT INTO var_lst (value) VALUES ('durjdn')");
$foo->execute();
echo $DB->lastInsertId();
于 2014-01-17T16:08:52.663 回答
5

放置$dbh->lastInsertId();前后$dbh->commit()_$stmt->execute();

于 2016-05-15T18:17:48.620 回答