PHP SQL:如何从一个html表单将数据保存到多个数据库或如何自动将数据从一个数据库复制到另一个数据库

2023-12-15

我有一个 html 表单,举个例子

<form action="form.php" method="post">
First name:<br>
<input type="text" id="fname" name="fname">
<br>
Last name:<br>
<input type="text" id="lname" name="lname">
<br><br>
<input type="submit" value="Submit">
</form>

和表单.php

<?php
$servername = "localhost";
$username = "database1";
$password = "xxxxxxxx";
$dbname = "database1";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

//escape variables for security
$fname = mysqli_real_escape_string($conn, $_POST['fname']);
$lname = mysqli_real_escape_string($conn, $_POST['lname']);

$sql = "INSERT INTO mytable (fname,lname)
VALUES ('$fname','$lname')";

if ($conn->query($sql) === TRUE) {
   echo "Successfully Saved";

    } else {
    echo "Error: Go back and Try Again ! " . $sql . "<br>" . $conn->error;
}

$conn->close();

?>

form.php 将该数据保存到database1 .

我希望将该数据保存到另一个数据库database2随着database1.

是否可以 ??如果是,那么应该在代码中进行哪些更改?

如果不可能,那么是否可以从复制数据database1 to database2自动地?每当添加新行时database1那么它应该自动复制到database2.

我希望相同的数据位于两个不同的数据库中。我怎样才能实现上述任何一个?


从 php 你只需要创建到数据库的新连接。

<?php
   $servername = "localhost";
   $username = "database1";
   $password = "xxxxxxxx";
   $dbname = "database1";


   $servernameS = "localhost";
   $usernameS = "database2";
   $passwordS = "xxxxxxxx";
   $dbnameS = "database2";

   // Create connection
   $conn = new mysqli($servername, $username, $password, $dbname);
   $connS = new mysqli($servernameS, $usernameS, $passwordS, $dbnameS);

   // Check connection
  if ($conn->connect_error) {
       die("Connection failed: " . $conn->connect_error);
   }
   if ($connS->connect_error) {
       die("Connection failed: " . $connS->connect_error);
   }

   //escape variables for security
   $fname = mysqli_real_escape_string($conn, $_POST['fname']);
   $lname = mysqli_real_escape_string($conn, $_POST['lname']);

   $sql = "INSERT INTO mytable (fname,lname) 
   VALUES ('$fname','$lname')";

    if ($conn->query($sql) === TRUE) {
     echo "Successfully Saved";

    } else {
      echo "Error: Go back and Try Again ! " . $sql . "<br>" . $conn->error;
    }

    if ($connS->query($sql) === TRUE) {
     echo "Successfully Saved";

    } else {
      echo "Error: Go back and Try Again ! " . $sql . "<br>" . $connS->error;
    }
    $conn->close();
    $connS->close();

   ?>
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