将数据从一个PHP文件发布到另一个PHP文件会导致错误

Question

我有两个php文件，一个文件将数据提交到第二个文件，以便对mysql数据库进行更新操作。

下面的

是提交数据的文件的代码

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="test_mysql"; // Table name 
$server_name="localhost";


// Create connection
$con = new mysqli($server_name, $username, $password, $db_name , 3306);

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

// get value of id that sent from address bar
$id=$_GET['id'];

// Retrieve data from database 
$sql="SELECT * FROM $tbl_name WHERE id='$id'";

$result = $con->query($sql);

$rows = $result->fetch_assoc();          

?>

<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<form name="form1" method="post" action="update_ac.php">
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>&nbsp;</td>
<td colspan="3"><strong>Update data in mysql</strong> </td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Lastname</strong></td>
<td align="center"><strong>Email</strong></td>
</tr>
<tr>
<td>&nbsp;</td>
<td align="center">
<input name="name" type="text" id="name" value="<?php echo $rows['name']; ?>">
</td>
<td align="center">
<input name="lastname" type="text" id="lastname" value="<?php echo $rows['lastname']; ?>" size="15">
</td>
<td>
<input name="email" type="text" id="email" value="<?php echo $rows['email']; ?>" size="15">
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<input name="id" type="hidden" id="id" value="<?php echo $rows['id']; ?>">
</td>
<td align="center">
<input type="submit" name="Submit" value="Submit">
</td>
<td>&nbsp;</td>
</tr>
</table>
</td>
</form>
</tr>
</table>

<?php
// close connection 
$con->close();
?>

更新操作的第二个文件如下所示

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="test_mysql"; // Table name 
$server_name="localhost";

// Create connection
$con = new mysqli($server_name, $username, $password, $db_name , 3306);

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

// update data in mysql database 
$sql="UPDATE $tbl_name SET name='$name', lastname='$lastname', email='$email' WHERE id='$id'";
$result=$con->query($sql);

// if successfully updated. 
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='list_records.php'>View result</a>";
}

else {
echo "ERROR";
}

?>

错误页面如下所示。

解决问题的建议

Answer 1

嗯，你还没有设置这些值，这就是它出错的原因。 首先，您必须包装第二个文件以检查它是否已提交表单。然后在里面设置这些变量。

<?php
if(isset($_POST['Submit'])) {
  $name = $_POST['name'];
  $lastname = $_POST['lastname'];
  $email = $_POST['email'];
  $id = $_POST['id'];

  // rest of your code goes here
}

Answer 2

将您的第二个文件更改为

 <?php
if(isset($_POST['submit'])) {
  $name = $_POST['name'];
  $lastname = $_POST['lastname'];
  $email = $_POST['email'];
  $id = $_POST['id'];
// add you rest of code
}

Answer 3

你应该在行

之前定义一些变量

$sql="UPDATE $tbl_name SET name='$name', lastname='$lastname', email='$email' WHERE id='$id'";

例如

$tbl_name = 'mytable';
$name = 'ABC';
$lastname = 'XYZ';
$email = 'abc@example.com';
$sql="UPDATE $tbl_name SET name='$name', lastname='$lastname', email='$email' WHERE id='$id'";

Answer 4

在你编写sql查询来更新数据的第二个文件中， 您正在使用： -

取消定义变量 name, lastname, email, id

在sql查询之前，在这些变量中获取发布值： -

 $name= $_POST['name'];
 $lastname= $_POST['lastname'];

Answer 5

在请求页面顶部使用此功能。

 $name=$_POST['name'];
 $lastname=$_POST['lastname'];
 $email=$_POST['email'];
 $id=$_POST['id'];