这是我创造我把手sqlite3的和last_insert_rowid
$this->_handle = new SQLite3($this->_dbname);
这是如何使我的数据库查询(缩短):
$stmt = $this->_handle->prepare($sql);
// execute query
$result = $stmt->execute();
// get all results
while($row = $result->fetchArray(SQLITE3_ASSOC)){
$res[] = $row;
}
现在我想获得最后插入的ID 。我试着这样说:
public function last_insert_rowid()
{
$result = sqlite_last_insert_rowid($this->_handle);
var_dump($result);
echo '<pre>';
print_r($result);
echo '</pre>';
return $result;
}
我得到这个错误:
Warning: sqlite_last_insert_rowid() expects parameter 1 to be resource, object given in ... on line 444 NULL
编辑:这可不行,因为我必须有它sqlite_open()
打开。
我也与此代码试了一下:
function last_insert_rowid(){
global $db;
$sql = "SELECT last_insert_rowid();";
$result = $db->ExecuteQuery($sql);
var_dump($result);
return $result;
//return ($db->last_insert_rowid());
}
我收到以下错误信息:
NULL Warning: SQLite3Stmt::execute() [sqlite3stmt.execute]: Unable to execute statement: constraint failed in ..
错误指向此行$stmt->execute();
如何让我的最后插入ID?