PHP/Mysqli: Why does this code doubles rows insert? - php

Question

I need a help with my code; somehow my code creates two rooms (it inserts two rows into a table at once), I don't know why.
(I need to require an id for every insert to know in which house we create a new room. My database contains table 'house' and table 'room'. Table 'room' has a field 'house_id' which is a foreign key with a field 'id' in table 'house'.)
That is my php page:
<?php
// turn autocommit off
mysqli_autocommit($con, FALSE);
// fetch the houses so that we have access to their names and id
$query = "SELECT name, id
FROM house";
$result = mysqli_query($con, $query);
// check query returned a result
if ($result === false) {
echo mysqli_error($con);
} else {
$options = "";
// create an option
while ($row = mysqli_fetch_assoc($result)) {
// $options .= "".$row['name']."";
$options .= "<option value='".$row['id']."'>";
$options .= $row['name'];
$options .= "</option>";
}
}
include('templates/add_room.html');
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$price = mysqli_real_escape_string($con, $_POST["price"]);
$house = mysqli_real_escape_string($con, $_POST["house_id"]);
$query = "INSERT INTO room (price, house_id)
VALUES ('$price', '$house')";
// run the query to insert the data
$result = mysqli_query($con, $query);
// check if the query went ok
if ( $con->query($query) ) {
echo "<script type= 'text/javascript'>alert('New room created successfully with the id of {$con->insert_id}');</script>";
mysqli_commit($con);
} else {
echo "There was a problem:<br />$query<br />{$con->error}";
mysqli_rollback($con);
}
}
//free result set
mysqli_free_result($result);
?>
and that is my html template with form:
<h2>Add new room</h2>
<form action='' method='POST'>
<fieldset>
<label for='price'>Price:</label>
<input type='number' name='price'>
</fieldset>
<fieldset>
<label for='house_id'>House:</label>
<select name='house_id' required>
<option value='' disabled selected>Select house</options>
<?php echo $options; ?>
</select>
</fieldset>
<button type='submit'>Add</button>
</form>

Answer 1

It inserts 2 rows because of your using the query function twice:
$result = mysqli_query($con, $query);
// check if the query went ok
if ( $con->query($query) ) {
So you'll need to change that conditional statement to:
if ($result) {
By the way, use a prepared statement, it's safer than real_escape_string():
https://en.wikipedia.org/wiki/Prepared_statement

Answer 2

You are inserting it twice
first here:
// run the query to insert the data
$result = mysqli_query($con, $query);
then here:
// check if the query went ok
if ( $con->query($query) ) {
Remove the first one and you should be fine, or check on the result of the first one and remove the second one.

Answer 3

Not 100% certain, but it looks like you run INSERT query twice. Once here:
$result = mysqli_query($con, $query);
and once a moment later when you try to check for something. you inadvertently use the OOP style when you are apparently trying to check for something
if ( $con->query($query) ) {