mysqli_error() expects exactly 1 parameter

Hey guys - very new to writing php, but trying…

Here’s the code I got:

$bldg_current = $_GET['bldg'];

$cxn = mysqli_connect($host, $user, $passwd, $dbname) or die(mysqli_error());

$facilitycode = "SELECT * FROM 'facilities' WHERE 'facility_code' = '$bldg_current'";

$result_facilitycode = mysqli_query($cxn, $facilitycode) or die(mysqli_error());

$array_facilitycode = mysqli_fetch_assoc($result_facilitycode);

if ( empty($array_facilitycode) )
    die("nothing found in the database");
else {

[ul]I pass the variable through the url and use _GET to grab it and put it in the variable $bldg_current.

Then I use the value from $bldg_current to grab the row that has the value of $bldg_current in the “facility_code” column of the table “facilities”.

I run the query and pass the value to $result_facilitycode.

Then I put the result into an array and finally extract it so that I may use the column names as the actual variables.[/ul]

Well I get this error: Warning: mysqli_error() expects exactly 1 parameter, 0 given

And it references this line: $result_facilitycode = mysqli_query($cxn, $facilitycode) or die(mysqli_error());

Any help would be appreciated - thanks!

mysqli_error() wants a parameter, namely the MySQLi connection handle.

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