Dynamic Form to Database - php

Question

I'm trying to make a log of cars going in and out of a parking lot.
The car info is retrieved from a database and is working fine. The problem I'm having is getting the in/out times to store into the database. In a previous page I had done it so that the form was separate from the table and the input info would be updated but for this page I need to have a dynamic amount of fields varying on the cars in the database. I am not sure what I am doing wrong but here is my code, the data is not being sent or stored in the data base.
<h3>Update Car</h3>
<form action="carLog.php" method="post">
<fieldset>
<legend>Car Log</legend>
<?php //This prints out the car log data
$sql = "SELECT * FROM carLog";
$result = $databaseConnection->query($sql);
echo "<table class='TFtable' border='1' style='width':100%>"; //starts the table tag
echo "<tr>
<td>Name</td>
<td>Vehicle</td>
<td>Licence Plate</td>
<td>In</td>
<td>Out</td>
<td>In</td>
<td>Out</td>
<td>Comments</td>
</tr>"; //sets headings
while($row = $result->fetch_assoc()) { //loops for each result
echo "<tr>
<td>".$row['name']."</td>
<td>".$row['vehicle']."</td>
<td>".$row['plate']. "</td>
<td><input type='text' size='5' maxlength='5' name='inTime' value='".$row['inTime']."' id='inTime' /></td>
<td><input type='text' name='outTime' value='".$row['outTime']."' id='outTime' /></td>
<td><input type='text' name='inTime2' value='".$row['inTime2']."' id='inTime2' /></td>
<td><input type='text' name='outTime2' value='".$row['outTime2']."' id='outTime2' /></td>
<td><input type='text' name='comments' value='".$row['comments']."' id='comments' /></td>
</tr>";
}
echo "</table>"; //closes the table
?>
<input type="submit" name="Save" value="Save" />
</fieldset>
</form>
The database connection is fine and working. Here is the php that handles the post:
if (isset($_POST['Save'])){
$name = $_POST['name'];
$vehicle = $_POST['car'];
$plate = $_POST['plate'];
$inTime = $_POST['inTime'];
$outTime = $_POST['outTime'];
$inTime2 = $_POST['inTime2'];
$outTime2 = $_POST['outTime2'];
$comments = $_POST['comments'];
$query = "UPDATE carLog SET inTime = '$inTime', outTime = '$outTime', inTime2 = '$inTime2', outTime2 = '$outTime2' WHERE plate = '$plate'";
$databaseConnection->query($query);

Answer 1

Note:
You only have five (5) input fields inside your while() loop, but you are trying to process eight (8) input fields in your carLog.php. So it will return undefine variables error.
Pass the input fields in array.
Inside the loop, hide the primary id of each car/vehicle in a hidden input (also in array).
Add this inside your while() loop:
/* ASSUMING vehicle_id IS THE PRIMARY ID OF YOUR carLog TABLE; JUST REPLACE IT WITH THE RIGHT COLUMN NAME */
echo '<input type="hidden" name="hidden_id[]" value="'.$row["vehicle_id"].'">';
You have to add [] in your input's name tags.
<td><input type='text' name='outTime[]' .....
Do it to the rest of your inputs.
Then on your carLog.php file, which process the input (at least use
*_real_escape_string to prevent SQL injections). We will be checking each input using for() loop:
if (isset($_POST['Save'])){
for($x = 0; $x< count($_POST["hidden_id"]); $x++){
$vehicleid = $databaseConnection->real_escape_string($_POST['hidden_id'][$x]);
$inTime = $databaseConnection->real_escape_string($_POST['inTime'][$x]);
$outTime = $databaseConnection->real_escape_string($_POST['outTime'][$x]);
$inTime2 = $databaseConnection->real_escape_string($_POST['inTime2'][$x]);
$outTime2 = $databaseConnection->real_escape_string($_POST['outTime2'][$x]);
$comments = $databaseConnection->real_escape_string($_POST['comments'][$x]);
$query = "UPDATE carLog SET inTime = '$inTime', outTime = '$outTime', inTime2 = '$inTime2', outTime2 = '$outTime2' WHERE vehicle_id = '$vehicleid'";
$databaseConnection->query($query);
} /* END OF FOR LOOP */
} /* END OF ISSET Save */
Since you are using mysqli_* already, consider using the prepared statement approach.