This is my block of code for doing that. It works fine until it reaches the last if statement. I cannot get it to find the Graphics column using the Department_ID. I am trying to check if the user input is equal to a id within the table. Then check if that id requires graphic work done. To do that, I need to find out that for that specific project graphic is a 1 in the database.
if($graphics_id != Null)
{
$query = mysqli_query($connect,"SELECT * FROM Project_Overview WHERE Project_ID='".$graphics_id."'");
$row = mysqli_fetch_assoc($query);
//echo $row['Project_ID'];
if($graphics_id == $row['Project_ID']) //if the graphics_id matches a project_id in the table
{
$result = mysqli_query($connect, "SELECT Graphics FROM Department WHERE Department_ID ='".$graphics_id."'")
$row = mysqli_fetch_assoc($result);
if($result)
{
echo $row['Department_ID'];
} else {
echo "This Project does not require graphics!"
}
} else {
echo "Project_ID ".$graphics_id." does not exist!";
}
}
A few thoughts:
Graphics column, but later you are echoing $row['Department_ID']; which should be empty as the only key in $row would be Graphicsif($result). Don't you mean if($row)? If $result is false (and hence "This Project does not require graphics!" is printed out), this would indicate, that mysqli_query has failed, possibly because of an error in your second SQL statement.SELECT * is not wrong but returns (probably) more than you need.$graphics_id comes from (a user input?) you should consider escaping it for security reasons ($graphics_id_escaped = mysqli_real_escape_string($connect, $graphics_id); - just in case ;)