两个 mySQL 数据库到一个 PHP 形式

我正在寻找一种从一个PHP表单向两个mySQL数据库提交数据的方法。

我希望所有文本信息都转到一个数据库，将所有图像转到另一个数据库。

我将如何做到这一点？（请参阅下面的附加代码，它仍然与我想要的地方相去甚远）：

<?php
/*
Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password)
*/
$link = mysqli_connect("****", "****", "****", "****");
// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$title = mysqli_real_escape_string($link, $_POST['title']);
$price = mysqli_real_escape_string($link, $_POST['price']);
$sqm = mysqli_real_escape_string($link, $_POST['sqm']);
$sqm_land = mysqli_real_escape_string($link, $_POST['sqm_land']);
$type = mysqli_real_escape_string($link, $_POST['type']);
$area = mysqli_real_escape_string($link, $_POST['area']);
$location = mysqli_real_escape_string($link, $_POST['location']);
$bedroom = mysqli_real_escape_string($link, $_POST['bedroom']);
$terrace = mysqli_real_escape_string($link, $_POST['terrace']);
$orientation = mysqli_real_escape_string($link, $_POST['orientation']);
$water = mysqli_real_escape_string($link, $_POST['water']);
$seaview = mysqli_real_escape_string($link, $_POST['seaview']);
$pool = mysqli_real_escape_string($link, $_POST['pool']);
$ownerinfo = mysqli_real_escape_string($link, $_POST['ownerinfo']);
$gaddress = mysqli_real_escape_string($link, $_POST['gaddress']);
$description = mysqli_real_escape_string($link, $_POST['description']);


// attempt insert query execution
$sql = "INSERT INTO property (title, price, sqm, sqm_land, type, area, location, bedroom, terrace, orientation, water, seaview, pool, ownerinfo, gaddress, description) VALUES 
('$title', '$price', '$sqm', '$sqm_land', '$type', '$area', '$location', '$bedroom', '$terrace', '$orientation', '$water', '$seaview', '$pool', '$ownerinfo', '$gaddress', '$description' )";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}


// close connection
mysqli_close($link);
?>

任何帮助将不胜感激。