Home/ Questions/Q 6096395
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-23T12:55:20+00:00

I have been struggling through PHP and sqlite for a bit now and I’m

  • 0

I have been struggling through PHP and sqlite for a bit now and I’m just confusing myself.
I have an html form that accessess a php script called processFeedback.php.My html code looks like this..

<html>
<head>
</head>
<body>
<form action="processFeedback.php" method="POST">
    <table>
        <tr>
            <td>Name:</td><td><input name="name"/></td>
        </tr>
        <tr>
            <td>Email:</td><td><input name="email"/></td>
        </tr>
        <tr>
            <td>Comments:</td><td><textarea name="comments"></textarea></td>
        </tr>
        <tr>
            <td></td><td><input type="submit" value="Submit"/></td>
        </tr>
    </table>
</form>
</body>
</html>

and my php file looks like this…

<?php
try
{
//open the database
$db = new PDO('sqlite:feedback.db');

$name = $_POST["name"]; 
$email = $_POST["email"]; 
$comments = $_POST["comments"]; 

//Insert record  
$db->exec("INSERT INTO feedback (name, email,comments) VALUES ('&name', '&email','&comments');");

//now output the data to a simple html table...
print "<table border=1>";
print "<tr><td>Id</td><td>Name</td><td>Email</td><td>Comments</td></tr>";
$result = $db->query('SELECT * FROM feedback');
foreach($result as $row)
{
print "<tr><td>".$row['feedbackid']."</td>";
print "<td>".$row['name']."</td>";
print "<td>".$row['email']."</td>";
print "<td>".$row['comments']."</td>";
}

print "</table>";

$db = NULL;
}
catch(PDOException $e)
{
print 'Exception : ' .$e->getMessage();
}

?>

And here is my table creation method…

CREATE TABLE feedback (feedbackid INTEGER PRIMARY KEY,name TEXT,email TEXT,comments TEXT);

The form is outputting the table headers and also a record that I manually entered using the Terminal but it won’t insert a record in??? Can anyone see an easy mistake?

Disco

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    //Insert record
    $db->exec(“INSERT INTO feedback (name, email,comments)
    VALUES (‘&name’, ‘&email’,’&comments’);”);

    One obvious issue is that you’re doing no error checking:

    if (1!=$db->exec(“INSERT…)) {
    print “Error: ” . implode(‘/’,$db->errorInfo()) . “
    \n”;
    }

    While I’m not overly familiar with PDO, I don’t see how your placeholders are getting mapped to the corresponding PHP variables – which is probably the cause of the error:

    $bound=array(
     $_POST["name"], $_POST["email"], $_POST["comments"] 
);
$stm=$db->prepare("INSERT INTO feedback (name, email,comments) 
   VALUES (?, ?,?));"

if (!$stm || !$stm->execute($bound)) {
     print "Error: " . implode('/',$db->errorInfo()) . "<br />\n";
}
    • 0