我需要使用 PHP 连接到两个数据库,并使用第一个查询的结果从第二个数据库中获取我需要的其余数据.
I need to connect to two databases using PHP and use the results from the first query to get the rest of the data I need out of a second database.
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?
So for the second connection, I need to connect to the second database and Select state and zipcode where the results from connection 1 (client) is equal to the firstname in database 2. How would I do this?
<?php
// 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();
连接到数据库代码类似于下面
Connection To Database Code is similar to the below
<?php
//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);
}
这没有经过测试,但我认为它会像这样.
This isn't tested, but I think it would go something like this.
<?php
$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;
}
?>
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