我有三個具有相同結構（相同表）的 mysql 數據庫。然后我有一個查詢，它從每個數據庫返回不同的結果。我想要一個 php 頁面，其中有單選按鈕、列表框等（沒有提交按鈕），我將在其中選擇數據庫（DB1/DB2/DB3），然后根據所選數據庫查看結果（我想要它是實時的，沒有提交按鈕）。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>查詢.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, $dbname);// Check connectionif ($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, $dbname);// Check connectionif ($mysqli_db->connect_error) { die("Connection failed: " . $mysqli_db->connect_error);}?>我認為，我需要一些 javascript/ajax 解決方案，但我不知道如何有效地使用它。
2 回答
紅糖糍粑
TA貢獻1815條經驗 獲得超6個贊
現在，我看不到您展示任何使用或已實現任何 ajax 處理程序的示例。我可以建議您使用純 Javascript 或 jQuery（通過包含 javascript 庫，請參閱她
在考慮實施 AJAX 之前，您可以做以下事情（它將包括第一次選擇和表單提交后的數據庫）：
<?php
if (isset($_POST["database"])) {
$db = $_POST["database"];
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8"/>
</head>
<?php
include_once("connection_db_" . $db . ".php"); /* This will include the selected connection */
?>
<body>
<form action="" method="post">
<select name="database">
<option value="1">DB1</option>
<option value="2">DB2</option>
<option value="3">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>
aluckdog
TA貢獻1847條經驗 獲得超7個贊
數據庫是否在同一臺服務器上？如果是這樣，您只需要一個與服務器的連接。您可以通過在查詢中添加數據庫名稱來進行查詢：
select Column_1 from dbname_1.TEST
或者
select Column_1 from dbname_2.TEST
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