Ich denke, dass BookingID eine ganze Zahl ist, also muss Ihre Aktualisierungszeile sein:
$updatequery = mysqli_query($con, "UPDATE booking SET FirstName='" . $_POST['txtfirstname'] . "' WHERE BookingID=" . $_POST['txtid'] . ""); //excute UpDate Query
BEARBEITEN: Ich habe Ihr Skript getestet und das Problem war, dass Sie das Formular außerhalb der While-Schleife geschlossen haben. Jetzt funktioniert es
<!DOCTYPE html>
<head>
<title>Edit Students</title>
</head>
<?php
$user = 'root'; //Database username ("Root for xampp")
$pass = ''; //Database password ("empty for exampp")
$db = 'all_tests'; //Name of database
$con = new mysqli('localhost', $user, $pass, $db) or die("Unable to connect"); //Create new data connection ('name of host/server', user, password, database name)
if (isset($_POST['btnUpdate'])) { //Once Update button pressed perform this code
$updatequery = mysqli_query($con, "UPDATE test_1 SET FirstName='" . $_POST['txtfirstname'] . "' WHERE BookingID='" . $_POST['txtid'] . "'"); //excute UpDate Query
};
$sql = mysqli_query($con, "SELECT *FROM test_1"); //Select All from Booking
//Create Headers for table
echo "<table border='1'>
<tr>
<th></th>
<th>Booking ID</th>
<th>First Name</th>
</tr>";
//Show Edit Form///////////////////////////////////////////////////////////////////////////////////////////////////
while($row = mysqli_fetch_array($sql)) { //Run sql code till there are no more rows to import
echo "<form method=post>"; //Run update code at top of this page
//Populate table with query (sql)
echo "<tr>";
echo "<td> <input name='btnUpdate' type='submit' value='update' /> </td>"; //once press update row this button is apart of
echo "<td> <input type='text' value=" . $row['BookingID'] . " name='txtid' /> </td>";
echo "<td> <input type='text' value=" . $row['FirstName'] . " name='txtfirstname' /> </td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysqli_close($con); //Close connection
?>
