MySQL Database not Updating when Form Submitted using PHP - php

Question

This is a webpage that I have:
// Info to connect to the Wishlist database
$servername = "em";
$dbusername = "";
$password = "!19";
$dbname = "";
// To connect to the database please
$conn = new mysqli($servername, $dbusername, $password, $dbname);
// If unable to connect to the database display this error
if ($conn->connect_error) {
echo "Connection to wishlist failed";
die("Connection failed: " . $conn->connect_error);
}
echo "Once you have added creatures to your wishlist, click " .
"<strong><a href='http://eggcavity.com/edit-wishlist'>here</a></strong> to edit your wishlist.";
// Get current user's username
$current_user = wp_get_current_user();
$username = $current_user->user_login;
// Retrieve data from the database
$sql = "SELECT Name, Stage1 FROM Creatures";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// Display all of the data from the database
echo '<form method="POST">';
while($row = $result->fetch_assoc()) {
echo '<div style="display: inline-block; width: 30%;">' .
'<img src="' . $row["Stage1"] . '"><br>'.
$row["Name"] .
'<br><input type="checkbox" name="creautres[]" value="' .
$row["Name"] .
'"></div>';
}
echo '<br><br><input type="submit" value="Submit"></form>';
} else {
echo "Creatures not found";
}
if(isset($_POST['submit'])){
foreach($_POST['creatures'] as $selected){
$sql = "INSERT INTO " . $username .
" (Creature, Picture, Stage, Gender, Frozen, Notes) VALUES ('" .
$selected . "', 'http://static.eggcave.com/90x90/" . $selected .
"_1', 'Stage1', 'Unspecified', 'Unspecified', 'Unspecified', '')";
if ($conn->query($sql) === FALSE) {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
}
// Close the connection to the database
$conn->close();
It displays like I want it to:
But it doesn't update the database when I click the submit button.
I've tried echoing $stmt, it seemed to be written as it should be.
When I try echoing $selected, within the loop it doesn't seem to output anything.
Can you help me?
I have updated the code to use one database. Please help me. It still isn't adding.

Answer 1

You had a typo in your checkbox name. I re-did the submit line in the form and the isset() I believe.
The below includes activating error reporting, a try/catch, binding for safety against sql injection. The data saves. You will need to deal with what should be unique data getting saved more than once of course. For instance, a unique key on (Creature,Username). And I would re-think the columns for Id's, but this was your table design. Thanks for allowing us to show you a re-use of a table. Good luck.
schema (from you):
drop table if exists Wishlists;
CREATE TABLE `Wishlists` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`Creature` varchar(100) DEFAULT NULL,
`Picture` varchar(200) DEFAULT NULL,
`Stage` varchar(100) DEFAULT NULL,
`Gender` varchar(100) DEFAULT NULL,
`Frozen` varchar(100) DEFAULT NULL,
`Notes` varchar(500) DEFAULT NULL,
`Username` varchar(100) DEFAULT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB; -- <------------------------ went with InnoDB
-- truncate table Wishlists; -- used during early testing
PHP (eggs01.php):
// Info to connect to the Wishlist database
$servername = "serve it up";
$dbusername = "dbu dbu dbu";
$password = "OpenSesame";
$dbname = "my db name";
try {
// To connect to the database please
$conn = new mysqli($servername, $dbusername, $password, $dbname);
if ($conn->connect_error) {
die('Connect Error (' . $conn->connect_errno . ') '
. $conn->connect_error);
}
echo "I am connected and feel happy.<br/>";
if(isset($_POST['submit'])){
// Postback - submit
// Get current user's username
//$current_user = wp_get_current_user(); // remmed out, I don't have your system
//$username = $current_user->user_login; // remmed out, I don't have your system
$theCount=0;
foreach($_POST['creatures'] as $selected){
$Creature=$selected;
$Picture="http://static.eggcave.com/90x90/" . $selected . "_1";
$Stage="Stage1";
$Gender="Unspecified";
$Frozen="Unspecified";
$Notes="Unspecified";
$Username="Stackoverflow123";
$sql = "INSERT Wishlists (Creature, Picture, Stage, Gender, Frozen, Notes, Username) " .
" VALUES (?,?,?,?,?,?,?)";
$stmt = $conn->prepare($sql); // SQL Injection - safe prepare / bind / execute
// 7 s's means 7 strings:
$stmt->bind_param('sssssss', $Creature, $Picture, $Stage, $Gender, $Frozen, $Notes, $Username);
$stmt->execute();
$theCount++;
}
echo "<br>Santa has been notified, count = ".$theCount."<br>";
}
else {
// Just display the form
// Retrieve data from the database
$result = $conn->query("SELECT Name, Stage1 FROM Creatures");
if ($result->num_rows > 0) {
// Display all of the data from the database
echo '<form method="POST">';
while($row = $result->fetch_array(MYSQLI_ASSOC)) {
echo '<div style="display: inline-block; width: 30%;">' .
'<img src="' . $row["Stage1"] . '"><br>'.
$row["Name"] .
'<br><input type="checkbox" name="creatures[]" value="' .
$row["Name"] .
'"></div>';
}
echo '<br><br><button type="submit" name="submit">Submit</button></form>';
$result->close();
} else {
echo "Creatures not found";
}
}
} catch (mysqli_sql_exception $e) {
throw $e;
}
After the submit having selected 3 eggs:
Database Image:

Answer 2

First of all, look at code
$dbname1 = *****";
Here you are missing " :
So replace this with
$dbname1 = "*****";
and try again

Answer 3

First issue is you should never run a SQL statement in a Loop ALWAYS AVOID THIS - as it will put your server through a lot of strain.
And 2nd try this and tell me the out put
foreach($_POST['creatures'] as $selected){
$stmt = "INSERT INTO " . $username . " (Creature, Picture, Stage, Gender, Frozen, Notes) VALUES ('" . $selected . "', 'http://static.eggcave.com/90x90/" . $selected . "_1', 'Stage1', 'Unspecified', 'Unspecified', 'Unspecified', '')";
if ($conn->query($stmt) === TRUE) {
}
}
TO
foreach($_POST['creatures'] as $selected){
$stmt = "INSERT INTO " . $username . " (Creature, Picture, Stage, Gender, Frozen, Notes) VALUES ('" . $selected . "', 'http://static.eggcave.com/90x90/" . $selected . "_1', 'Stage1', 'Unspecified', 'Unspecified', 'Unspecified', '')";
$result = $conn->query($stmt) OR die(var_dump($conn));
var_dump($result->fetch_array(MYSQLI_ASSOC));
die;
}