mysqli_fetch_array（）期望参数1为mysqli_result，在某些列中选中

I'm trying to select all of my column and it works perfectly with this

$query = "SELECT * FROM employee";
$result = mysqli_query($con, $query);

while ($employee = mysqli_fetch_array($result)) {
    # code...
    print (json_encode($employee));
}

and I also succeed select some of my column using this code

$query = "SELECT ID, NAME, LAT FROM employee";
$result = mysqli_query($con, $query);

but I failed when I try to select LONG column, it said that

mysqli_fetch_array() expects parameter 1 to be mysqli_result

It just show when I select LONG column, but it works well if I select another column.

"But i failed when i try to select LONG column"

That's because LONG is a MySQL reserved word.

• http://dev.mysql.com/doc/refman/5.5/en/reserved-words.html

Either wrap it in backticks, or choose another word for it.

I.e.:

$query = "SELECT ID, NAME, `LONG` FROM employee";

or

$query = "SELECT ID, NAME, LAT, `LONG` FROM employee";

LONG is a MySQL reserved word. So, your query is resulting in syntax errors when you run mysqli_query(). To get around that problem, put it in backticks, like this:

$query = "SELECT ID, NAME, LAT, `LONG` FROM employee";
$result = mysqli_query($con, $query);