What is causing this error in my PHP SQL query?

Your mysqli_query command will return false. Use mysqli_error to diagnose the problem.

if (!mysqli_query($link, $query)) {
    printf("Errormessage: %s\n", mysqli_error($link));
}

      

        
        
        
      

    

You will need to do the above check to determine for sure, but the problem with your query may be related to this section, which does not indicate what appears to be a string value:

if ($form_member_name !=0) {
    $query .= "AND members.member_name = $form_member_name ";
}

      

        
        
        
      

    

$ form_member_name should be surrounded by single quotes at least, but you should definitely use parameterized operators for this, rather than embedding unparsed variables in your queries, as you leave yourself wide open to SQL injection . Here's a revised version, but remember that I'm a little rusty with mysqli and can't test it without your database:

$query = "
    SELECT
    photos.photo_id, members.member_name, photos.photo_title, photos.photo_film,
        photos.photo_height, photos.photo_width
    FROM members, photos
    WHERE members.member_id = photos.member_id
";

$types = "";
$params = array();

if ($form_photo_title !="") {
    $query.= "AND photos.photo_title = ? ";
    $types .= "s";
    $params[] = $form_photo_title;
}

if ($form_member_name !=0) {
    $query .= "AND members.member_name = ? ";
    $types .= "s";
    $params[] = $form_member_name;
}

if ($form_type !="") {
    $query .= "AND photo.photo_film = ? ";
    $types .= "s";
    $params[] = $form_type;
}

if (!($statement = mysqli_prepare($link, $query)))
    throw new Exception(mysqli_error($link));

// this tells the statement to substitute those question marks with each of 
// the values in the $params array. this is done positionally, so the first 
// question mark corresponds to the first element of the array, and so on. 
// the $types array is just a string with an indication of the type of the 
// value stored at each position in the array. if all three of the above 
// clauses are applied, then $types will equal "sss", indicating that the 
// first, second and third elements in $params are string types. 
// worse still, because the parameters to the query are dynamic, we can't 
// call mysqli_stmt_bind_param directly as it does not allow an array to be 
// passed, so we have to call it dynamically using call_user_func_array!
// i really hate this about mysqli.
// if all three of your above query clauses are applied, this call translates to
//     mysqli_stmt_bind_param(
//         $stmt, $types, 
//         $form_photo_title, $form_member_name, $form_type
//     );
array_unshift($values, $stmt, $types);
call_user_func_array("mysqli_stmt_bind_param", $values);

mysqli_stmt_execute($stmt);

// this instructs mysqli to assign each field in your query to each of 
// these variables for each row that is returned by mysqli_stmt_fetch(). 
// this is also positional - if you change the order or number of fields 
// in your query, you will need to update this.
mysqli_stmt_bind_result($photo_id, $member_name, $photo_title, $photo_film, $photo_height, $photo_width);

while (mysqli_stmt_fetch($stmt)) {
    // $photo_id will be reassigned to the value from the row on each 
    // loop iteration
    echo $photo_id."<br />";
}

      

        
        
        
      

    

I forgot what a terrible beast the mysqli extension is - if you have access to the PDO extension , I can't recommend more emphatically that you find out your way around it and use it instead.

I think there is something wrong with the SQL you are trying to execute. Print the SQL on the output before the mysqli_query () command. Then take that line and run it from the mysql console to see if the query can be executed.