我现在需要连接到使用PHP两个数据库,并使用结果从第一个查询得到我需要从第二个数据库的数据的其余部分.

因此,对于第二个连接,我需要连接到第二个数据库,并选择状态和邮政编码,其中连接1(客户端)的结果等于数据库2中的名字.我该怎么做？

// check if the 'id' variable is set in URL, and check that it is valid

if (isset($_GET['cd']) && is_numeric($_GET['cd']))

// get id value

$id = intval($_GET['cd']);

$results = $id;

//Open a new connection to the MySQL server

require "calendarconnect.php";

//chained PHP functions

$client = $mysqli->query("SELECT client FROM appointments WHERE ID = $results")->fetch_object()->client;

print $client; //output value

$mysqli->close();

?>

到数据库的连接代码类似于下面的

//Open a new connection to the MySQL server

$mysqli = new mysqli('localhost','some database','some password','some username');

//Output any connection error

if ($mysqli->connect_error) {

die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);

}

?>

解决方法:

这未经测试,但我认为它会像这样.

$dbc1 = new MySQLi()or die('error connecting to database');

$dbc2 = new MySQLi()or die('error connecting to database');

//build query 1

$query1 = "SELECT * FROM Table";

$result1 = $dbc1->query($query) or die("Error in query");

$thing1 = '';

// check result

if($result1->num_rows){

//fetch result as object

$row = $result1->fetch_object();

//set attributes

$thing1 = $row->Name;

}

//build query 2

$query2 = "SELECT * FROM AnotherTable WHERE Id = '$thing1'";

$result2 = $dbc2->query($query) or die("Error in query");

$thing2 = '';

// check result

if($result2->num_rows){

//fetch result as object

$row = $result2->fetch_object();

//set attributes

$thing2 = $row->Name;

}

?>

标签：php,mysqli

来源： https://codeday.me/bug/20191011/1891401.html