How to Transfer Data in Web Form to a Database

Question

I'm sure this has been asked 1000 times, but I have looked all over and can't seem to get this to work.

form:

<form action="sendinfo.php" method="post">               
    <h4>ID:</h4>
    <input type="text" name="CustomerID">
<h4>First name:</h4>
    <input type="text" name="FirstName">
    <h4>Last Name:</h4>
    <input type="text" name="LastName">
    <h4>Street:</h4>
    <input type="text" name="Street">
    <h4>City:</h4>
    <input type="text" name="City">
    <h4>Zip:</h4>
    <input type="text" name="Zip">
    <h4>State:</h4>
    <input type="text" name="State">
    <h4>Phone:</h4>
    <input type="text" name="Phone">
    <h4>Email:</h4>
    <input type="text" name="Email">
    <input type="submit">
</form>

sendinfo.php

    <?php
    include('connection.php');
    $dbh = con();

    $dbh->query = "INSERT INTO Customer (CustomerID, FirstName, LastName, Street, City, State, Zip, Phone, Email)
    VALUES  ('$_POST[CustomerID]', ('$_POST[FirstName]', ('$_POST[LastName]', ('$_POST[Street]', ('$_POST[City]', ('$_POST[State]', ('$_POST[Zip]', ('$_POST[Phone]', ('$_POST[Email]')";
    if (!mysql_query($user_info, $connect)) { die('Error: ' . mysql_error()); } echo “Your information was added to the database.”; mysql_close($connect); 

    ?>

connection.php

<?php
define("DB_HOST", "localhost");
define("DB_NAME", "Impact_Technologies");
define("DB_USER", "root");
define("DB_PASS", "password");

function con(){
    try {
        $db_connection = new PDO('mysql:host='. DB_HOST .';dbname='. DB_NAME . ';charset=utf8', DB_USER, DB_PASS);
        return $db_connection;
    } catch (PDOException $e) {
        echo "Sorry, there was a problem connecting to the database." . $e->getMessage();
    }
}
?>

When submit is clicked, no messages displayed and no information entered into the db

Daniel Arechiga · Accepted Answer · 2016-05-03 22:32:22Z

2

I see what's the problem, ('$_POST[CustomerID]', ('$_POST[FirstName]', ('$_POST[LastName]', ('$_POST[Street]', ('$_POST[City]', ('$_POST[State]', ('$_POST[Zip]', ('$_POST[Phone]', ('$_POST[Email]')

You're opening a parenthesis before every value, it should be like this:

('$_POST[CustomerID]', '$_POST[FirstName]', '$_POST[LastName]', '$_POST[Street]', '$_POST[City]', '$_POST[State]', '$_POST[Zip]', '$_POST[Phone]', '$_POST[Email]')

answered May 3, 2016 at 22:32

Daniel Arechiga

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2 Comments

Randy Over a year ago

Damn, copy and paste haha. I found one other thing plus what you said and its working! thank you so much!

Daniel Arechiga Over a year ago

you're very welcome, please mark the answer as correct :) !!

Ukor · Accepted Answer · 2016-05-03 22:55:04Z

First, Simplify your code by doing this

    $id = $_POST['custormerID'];
    $firstName = $_POST['FirstName'];
//and so on

Secondly, Remove all the opening parenthesis before every value

    $dbh->query = "INSERT INTO Customer (CustomerID, FirstName, 
LastName, Street, City, 
State, Zip, Phone, Email)
    VALUES  ('$id', '$FirstName', 
'$LastName', '$Street', '$City', 
'$State', '$Zip', '$Phone', '$Email')";
    if (!mysqli_query($user_info, $connect)) { 
die('Error: ' . mysqli_error()); 
} 
echo “Your information was added to the database.”; mysql_close($connect);

Thirdly, mysql is depreciated, use mysqli or PDO instead

    if (!mysql_query($user_info, $connect)) { 
die('Error: ' . mysql_error()); } 
echo “Your information was added to the database.”; mysql_close($connect);

do this instead:

 if (!mysqli_query($user_info, $connect)) {
 die('Error: ' . mysqli_error()); 
} 
echo “Your information was added to the database.”; mysqli_close($connect);

Side Note: Is either you use mysqli or PDO, don't use the both. You can learn about PDO http://wiki.hashphp.org/PDO_Tutorial_for_MySQL_Developers and http://www.w3schools.com/php/php_mysql_intro.asp

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