Question

我想从两个不同的数据库访问数据。

我想将两个数据库中的记录打印到同一视图中。

现在我的第一个查询来自另一个数据库，然后第二个查询来自默认数据库。所以我得到一个错误，SecondDatabase.tablenmae不存在，实际上该表在默认数据库中

所以我这样定义了它们：

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

现在在我的视图中，我有一行从默认数据库中获取数据。

然后在第二行我想要来自另一个数据库的数据，所以我加载了该数据库：

$this->db = $this->load->database('bizlms', TRUE);

但是我在第一行从默认数据库中获取数据时出现错误：

表bizlms.tablname不存在

如何在同一视图上使用这两个数据库？

Answer 1

是的，您可以使用2个或更多数据库获取信息，然后在视图中显示

应用/配置/ database.php中

$active_group = 'default';

$db['default']['hostname'] = "host";
$db['default']['username'] = "username";
$db['default']['password'] = "password";
$db['default']['database'] = "database1";
$db['default']['dbdriver'] = "mysqli";
$db['default']['dbprefix'] = "";
$db['default']['pconnect'] = TRUE;
$db['default']['db_debug'] = TRUE;
$db['default']['cache_on'] = FALSE;
$db['default']['cachedir'] = "";
$db['default']['char_set'] = "utf8";
$db['default']['dbcollat'] = "utf8_general_ci";

$db['stats']['hostname'] = "host2";
$db['stats']['username'] = "username";
$db['stats']['password'] = "password";
$db['stats']['database'] = "database2";
$db['stats']['dbdriver'] = "mysqli";
$db['stats']['dbprefix'] = "";
$db['stats']['pconnect'] = TRUE;
$db['stats']['db_debug'] = TRUE;
$db['stats']['cache_on'] = FALSE;
$db['stats']['cachedir'] = "";
$db['stats']['char_set'] = "utf8";
$db['stats']['dbcollat'] = "utf8_general_ci";

Controller.php这样

function do_somthing(){
    $this->load->model(array('model_db1', 'model_db2'));

    $info_db1  = $this->model_db1->get_info1();
    $info_db2  = $this->model_db2->get_info2();

    $data      = array(
                       'info1'  =>  $info_db1,
                       'ifno2'  =>  $info_db2
                      );

    $this->load->view('view', $data);
}

model_db1.php

public function __construct(){
    parent::__construct();

    $database1 = $this->load->database('default', TRUE);
}

function get_ifno1(){
    return $database1->db->get('table')->result();
}

model_db2.php

public function __construct(){
    parent::__construct();

    $database2 = $this->load->database('stats', TRUE);
}

function get_ifno2(){
    return $database2->db->get('table')->result();
}

view.php

<div>
    <?php var_dump($info1)?> 
</div> 
<hr>
<div>
    <?php var_dump($info2)?> 
</div>

Answer 2

尝试这样，将db name保留在一个变量中，并使用该变量来获取数据：

[TestMethod]
public void Factory_Should_Get_Claims_Principal() {
    //Arrange
    var fakeIdentity = new ClaimsIdentity();
    var mockUserClaimsPrincipal = new Mock<IClaimsPrincipalFactory>();
    mockUserClaimsPrincipal.Setup(ucp => ucp.GetClaimsPrincipal().Identity).Returns(fakeIdentity);

    IClaimsPrincipalFactory userClaimsPrincipal = mockUserClaimsPrincipal.Object;
    System.Security.Claims.ClaimsIdentity claimIdentity = userClaimsPrincipal.GetClaimsPrincipal().Identity as System.Security.Claims.ClaimsIdentity;

    //Act
    //...other code that uses the claimIdentity


    //Assert
    //...other assertions
}

我们可以在Codeigniter中的同一视图中使用两个数据库吗？

2 个答案: