Question

我想显示数据，但数据库与config / database中的设置数据库不同。但是，当我运行此代码时，它会给我错误。

这是我的代码：

import numpy as np
from libc.stdlib cimport malloc, calloc, realloc, free

def calculate_signed_areas(double[:,:,::1] triangles):
    cdef:
        int i, n
        double area
        double[::1] areas
        double x0, x1, x2
        double y0, y1, y2

    n = triangles.shape[2]
    areas = <double[:n]>malloc(n * sizeof(double))
    for i in range(n):
        x0 = triangles[0,0,i]
        y0 = triangles[1,0,i]
        x1 = triangles[0,1,i]
        y1 = triangles[1,1,i]
        x2 = triangles[0,2,i]
        y2 = triangles[1,2,i]
        area = (
            + x2 * (y0 - y1)
            + x0 * (y1 - y2)
            + x1 * (y2 - y0)
        )/2.0
        areas[i] = area
    return np.asarray(areas)

我希望你能帮助我

Answer 1

您可以在codeigniter中拥有多个具有不同组名的数据库连接组。

    $db['default']['hostname'] = "localhost";
    $db['default']['username'] = "root";
    $db['default']['password'] = "";
    $db['default']['database'] = "database_name";
    $db['default']['dbdriver'] = "mysql";
    $db['default']['dbprefix'] = "";
    $db['default']['pconnect'] = TRUE;
    $db['default']['db_debug'] = FALSE;
    $db['default']['cache_on'] = FALSE;
    $db['default']['cachedir'] = "";
    $db['default']['char_set'] = "utf8";
    $db['default']['dbcollat'] = "utf8_general_ci";
    $db['default']['swap_pre'] = "";
    $db['default']['autoinit'] = TRUE;
    $db['default']['stricton'] = FALSE;

    $db['otherdb']['hostname'] = "localhost";
    $db['otherdb']['username'] = "root";
    $db['otherdb']['password'] = "";
    $db['otherdb']['database'] = "other_database_name";
    $db['otherdb']['dbdriver'] = "mysql";
    $db['otherdb']['dbprefix'] = "";
    $db['otherdb']['pconnect'] = TRUE;
    $db['otherdb']['db_debug'] = FALSE;
    $db['otherdb']['cache_on'] = FALSE;
    $db['otherdb']['cachedir'] = "";
    $db['otherdb']['char_set'] = "utf8";
    $db['otherdb']['dbcollat'] = "utf8_general_ci";
    $db['otherdb']['swap_pre'] = "";
    $db['otherdb']['autoinit'] = TRUE;
    $db['otherdb']['stricton'] = FALSE;

$CI = & get_instance();
$CI->db = $this->load->database('otherdb', true);

Answer 2

**Use this db config code:**

$db_name = 'test_db_name';
$db_user = 'root';
$db_password = '123456';
$config['dsn'] = "mysql:host=localhost;dbname=$db_name";
$config['hostname'] = "localhost";
$config['username'] = "$db_user";
$config['password'] = "$db_password";
$config['database'] = "$db_name";
$config['dbdriver'] = "pdo";
$config['dbprefix'] = "";
$config['pconnect'] = FALSE;
$config['db_debug'] = TRUE;
$this->db1 = $this->load->database($config, TRUE);

如何使用codeignter在另一个数据库中显示数据

2 个答案: