Reputation: 2881
I have a php script to insert form input into a database called XXXX_comments into the Table called Comments that has Name and Comment as columns.When a user hits Save button the form should be inserted into the DB. I connect to the database using connect.php :
<?php
$servername = "localhost";
$username = "XXXXX";
$password = "XXXXX";
$Dbconnect = "XXX_comments";
// Create connection
$conn = new mysqli($servername, $username, $password);
mysqli_select_db($conn,$Dbconnect);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>
in my index.php file i have the following form:
<?php
include ('connect.php');
?>
<?php
if(isset($_POST['Save'])){
$sql = $conn->prepare("INSERT INTO Comments (Name, Comment) VALUES (?, ?)");
$stmt->bind_param("ss", $firstname, $comment);
$firstname = $_POST['firstname'];
$comment = $POST['comment'];
$stmt->execute();
Echo "Succefully inserted to table";
}
?>
<div class="container">
<div class="row" id="post-review-box" style="display:none;">
<div class="col-md-12">
<form id="form" accept-charset="UTF-8" action="index.php" method="post">
<input type="text" class="form-control animated" id="firstname" type="hidden" placeholder="Enter your Name">
<br>
<input id="ratings-hidden" name="rating" type="hidden">
<textarea class="form-control animated" cols="50" id="comment" placeholder="Enter your review here..." rows="5"></textarea>
<br>
<div class="text-right">
<div class="stars starrr" data-rating="0"></div>
<a class="btn btn-danger btn-sm" href="#" id="close-review-box" style="display:none; margin-right: 10px;">
<span class="glyphicon glyphicon-remove"></span>Cancel</a>
<button class="btn btn-success btn-lg" type="submit" name="Save">Save</button>
</div>
</form>
</div>
</div>
</div>
<div class="display"></div>
</div>
</div>
</div>
The connection is successful but the data is not being inserted.
Upvotes: 0
Views: 123
Reputation: 2881
For any one intrested I have solved the issue, It seemed like the submit button wasn't responding So i added a javascript onclick function to force submit the form.
<input class="btn btn-default" onclick="myFunction()" value="Submit">
<script>
function myFunction() {
document.getElementById("form").submit();
}
</script>
Thanks for all the answers.
Upvotes: 0
Reputation: 1771
Few remarks:
First of all, in the connect.php file you're mixing OOP and Procedural styles, you can gain the same result by using the following code:
$conn = new mysqli($servername, $username, $password, $Dbconnect);
Next, in your index.php, you've created a prepared statement, but you assigned that to $sql variable, and later you tried to use the undefined $stmt variable, so in order to fix that just change the first 2 lines
$stmt = $conn->prepare("INSERT INTO Comments (Name, Comment) VALUES (?, ?)");
$stmt->bind_param("ss", $firstname, $comment);
Hope it will solve your little issue!
Upvotes: 1
Reputation: 6004
You should give name to your inputs and textarea inside the form as
<input type="text" name="firstname" class="form-control animated" id="firstname" type="hidden" placeholder="Enter your Name">
<input id="ratings-hidden" name="rating" type="hidden">
<textarea class="form-control animated" cols="50" name="comment" id="comment" placeholder="Enter your review here..." rows="5"></textarea
and you are assigning values after the binding it to the query, you should assign values for $firstname
and $comment
before binding it to the query as
if(isset($_POST['Save'])){
$firstname = $_POST['firstname'];
$comment = $_POST['comment'];
$sql = $conn->prepare("INSERT INTO Comments (Name, Comment) VALUES (?, ?)");
$stmt->bind_param("ss", $firstname, $comment);
$stmt->execute();
echo "Succefully inserted to table";
}
Edit: The code was using $POST instead of $_POST.
Upvotes: 1
Reputation: 740
You need to assign values before you can use them as below code:
$firstname = $_POST['firstname'];
$comment = $_POST['comment'];
$sql = $conn->prepare("INSERT INTO Comments (Name, Comment) VALUES (?, ?)");
$stmt->bind_param("sss", $firstname, $comment);
Upvotes: 1