Question

我尝试使用POST方法将Value从一个页面发送到另一个页面，其中数据库的名称是“firstdb”，表格名为“TAB1”。这是我的代码：

名为“addStud.php”的第一页。代码是：

<html>
<body>

<form action= "addStud_D.php" method="POST">

<center>
<fieldset>
Last Name : <input type = 'text' name ='LName'>
</br> </br>

First Name : <input type = 'text' name = 'FName'>
</br></br>

 Date of birth : <input type = 'text' name = 'dat'>
</br></br>
<input type='submit' value='OK'>

</fieldset>
</center>
</form>

</body >
</html>

接收名为“addStud_D.php”的数据的第二页。代码是：

<html>
<body>

<?php

$V1 = $_POST['LName'];
$V2 = $_POST['FName'];
$V3 = $_POST['dat'];

$db=mysql_connect('127.0.0.1','root','','firstdb')
  or die('error connecting to MySQL server.');

 mysql_select_db('firstdb');
mysql_query("INSERT INTO `firstdb`.`TAB1` (`ID`, `Last_Name`, `First_Name`,`date`)
  VALUES (NULL,'$V1','$V2', '$V3')");

?>

</body>
</html>

静止。这个方法不起作用，这是错误：

Notice: Undefined index: LName in D:\Computer Scince\Web\EasyPHP-5.3.3.1\www\APP04\addStud_D.php on line 6

Notice: Undefined index: FName in D:\Computer Scince\Web\EasyPHP-5.3.3.1\www\APP04\addStud_D.php on line 7

Notice: Undefined index: dat in D:\Computer Scince\Web\EasyPHP-5.3.3.1\www\APP04\addStud_D.php on line 8

Answer 1

更改方法的拼写=＆＃34; POST＆＃34; to method =＆＃34; POST＆＃34;，你的代码的其余部分也存在一些问题，因为你应该阅读有关准备好的语句，这些语句在处理用户输入时是有效的。

参考：tutorial

Answer 2

在第一页中，您应将'methode'更改为'method'。

第二页：

<html>
<body>

<?php

$servername    = "localhost";
$username = "root";
$password = "";
$dbname   = "firstdb";
$conn     = new mysqli($servername, $username, $password, $dbname);

$V1 = isset($_POST['LName']) ? mysqli_real_escape_string($conn, $_POST['LName']) : '';
$V2 = isset($_POST['FName']) ? mysqli_real_escape_string($conn, $_POST['FName']) : '';
$V3 = isset($_POST['dat']) ? mysqli_real_escape_string($conn, $_POST['dat']) : '';


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

$sql
    = "INSERT INTO firstdb (id, last_name, first_name, date)
VALUES ('', '" . $V1 . "', '" . $V2 . "', '" . $V3 . "')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

Answer 3

修复addStud_D.php档案

<html>
<body>

<?php

$V1 = (isset($_POST['LName'])?$_POST['LName']:'';
$V2 = (isset($_POST['FName'])?$_POST['FName']:'';
$V3 = (isset($_POST['dat'])?$_POST['dat']:'';

$db=mysqli_connect('127.0.0.1','root','','firstdb')
  or die('error connecting to MySQL server.');

 mysql_select_db('firstdb');
mysql_query("INSERT INTO `firstdb`.`TAB1` (`ID`, `Last_Name`, `First_Name`,`date`)
  VALUES (NULL,'$V1','$V2', '$V3')");

?>

</body>
</html>

您可以验证表格

<html>
<body>
<?php
$V1 = (isset($_POST['LName'])?$_POST['LName']:'';
$V2 = (isset($_POST['FName'])?$_POST['FName']:'';
$V3 = (isset($_POST['dat'])?$_POST['dat']:'';
if(!empty($V1) && !empty($V2) && !empty($V3)){
    $db=mysqli_connect('127.0.0.1','root','','firstdb')  or die('error connecting to MySQL server.');
    mysqli_query("INSERT INTO `firstdb`.`TAB1` (`ID`, `Last_Name`, `First_Name`,`date`) VALUES (NULL,'$V1','$V2', '$V3')");
}else{
    echo "Error";
}
?>
</body>
</html>

使用POST方法HTML在页面上传递值

3 个答案: