如何在PHP中使用HTML表单中的变量？

问题描述：

I just want to receive the input from a HTML form in PHP. I did research for almost 2 hours trying every code I got, but it doesn't work. Here's my code:

HTML:

<!DOCTYPE html>
<html>
<head>
<title>Form site</title>
</head>
<body>
    <table width="300" border="0" align="center" cellpadding="0" cellspacing="1">
        <tr>
           <td>
              <form name="form1" method="post" action="connect.php">
              <table width="100%" border="0" cellspacing="1" cellpadding="3">
                 <tr>
                    <td colspan="3">Insert Data Into mySQL Database </td>
                 </tr>
                 <tr>
                    <td width="71">Name</td>
                    <td width="6">:</td>
                    <td width="301"><input name="name" type="text" id="name"></td>
                 </tr>
                 <tr>
                    <td>Lastname</td>
                    <td>:</td>
                    <td><input name="lastname" type="text" id="lastname"></td>
                 </tr>
                 <tr>
                    <td>Email</td>
                    <td>:</td>
                    <td><input name="email" type="text" id="email"></td>
                 </tr>
                 <tr>
                    <td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
                 </tr>
              </table>
              </form>
           </td>
        </tr>
     </table></body>
</html>

PHP:

<?php
$mysqli = new mysqli("localhost", "root", "", "test");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s
", $mysqli->connect_error);
    exit();
}
$tbl_name="employees"; // Table name

// Get values from form
$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];


// if successfully insert data into database, displays message "Successful".
$sql = "INSERT INTO $tbl_name (name, lastname, email) VALUES ('$name', '$lastname', '$email')";
$mysqli->query($sql);
?>

So, when I run the HTML file in Chrome everything's normal, but if I then fill the form an click the Submit button Chrome says 'This page doesn't work'. The connection to the database works definitely. I read that it may help to set the AllowOverride option in the Apache config file (using XAMPP) to 'ALL'. It didn't. The filename specified in the HTML code matches the filename of the PHP file I'm actually using for this.

PHP is a server-side language, isn't it? So I suspect that I have to configure something to run the PHP file. PHP is already installed on my computer I checked the database connection using PHP in the command-line. Obviously I'm missing something. Does anyone have an idea?

Thanks in advance!

答

Solution: I used Apache as the webserver, but the HTML and PHP files were in another location, so PHP hadn't any webserver to run in.

Here's the working code: PHP:

?php
$mysqli = new mysqli("localhost", "root", "", "test");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s
", $mysqli->connect_error);
    exit();
}
$tbl_name="employees"; // Table name

// Get values from form
$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];

// if successfully insert data into database, displays message "Successful".
$sql = "INSERT INTO $tbl_name (name, lastname, email) VALUES (?, ?, ?)";
$statement = $mysqli->prepare($sql);
$statement->bind_param('sss', $name, $lastname, $email);

// $statement->execute();

if(!$statement->execute()) {
    echo "Query fehlgeschlagen: ".$statement->error;
}
else echo "All done!";

HTML:

<!DOCTYPE html>
<html>
<head>
<title>Form site</title>
</head>
<body>
    <table width="300" border="0" align="center" cellpadding="0" cellspacing="1">
       <tr>
       <td>
          <form name="form1" method="post" action="connect.php">
          <table width="100%" border="0" cellspacing="1" cellpadding="3">
             <tr>
                <td colspan="3">Insert Data Into mySQL Database </td>
             </tr>
             <tr>
                <td width="71">Name</td>
                <td width="6">:</td>
                <td width="301"><input name="name" type="text" id="name"></td>
             </tr>
             <tr>
                <td>Lastname</td>
                <td>:</td>
                <td><input name="lastname" type="text" id="lastname"></td>
             </tr>
             <tr>
                <td>Email</td>
                <td>:</td>
                <td><input name="email" type="text" id="email"></td>
             </tr>
             <tr>
                <td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
             </tr>
           </table>
           </form>
        </td>
        </tr>
     </table></body>
</html>

答

First of all make sure that your html form is sending the data to your php file by adding this line to your php file just after the php tag

    echo '<pre>';
    print_r($_POST);

And you will see that you are receiving three variables i.e

name, lastname and email

now check your php code you're requesting the name as firstname.

如何在PHP中使用HTML表单中的变量？

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