我有三个具有相同结构（相同表）的 mysql 数据库。然后我有一个查询，它从每个数据库返回不同的结果。
我想要一个 php 页面，其中有单选按钮、列表框等（没有提交按钮），我将在其中选择数据库（DB1/DB2/DB3），然后根据所选数据库查看结果（我想要它是实时的，没有提交按钮）。
我有的：
索引.php
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
</head>
<?php
include_once ('connection_db_1.php');
?>
<body>
<form action="">
<select name="database">
<option value="DB1">DB1</option>
<option value="DB2">DB2</option>
<option value="DB3">DB3</option>
</select>
</form>
<?php
include ('queries.php');
$test_1 = mysqli_query($mysqli_db, $test);
echo "
<table>
<thead>
<tr>
<th>Column_1</th>
</tr>
</thead>";
while ($row = mysqli_fetch_array($test_1)) {
echo "<form method=\"post\"><tr>";
echo "<td>" . $row['Column_1'] . "</td>";
echo "</tr></form>";
}
echo "</table><br>";
mysqli_close($mysqli_db);
?>
</body>
</html>
查询.php
<?php
$test = "select Column_1 from TEST; ";
?>
connection_db_1.php
<?php
// Connection data
$servername = "servename";
$username = "username";
$password = "pasword";
$dbname = "dbname_1";
// Create connection
$mysqli_db = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($mysqli_db->connect_error) {
die("Connection failed: " . $mysqli_db->connect_error);
}
?>
connection_db_2.php
<?php
// Connection data
$servername = "servename";
$username = "username";
$password = "pasword";
$dbname = "dbname_2";
// Create connection
$mysqli_db = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($mysqli_db->connect_error) {
die("Connection failed: " . $mysqli_db->connect_error);
}
?>
我认为，我需要一些 javascript/ajax 解决方案，但我不知道如何有效地使用它。
红糖糍粑
aluckdog