Fill array from query with mysqli_fetch_assoc()

Question

I had this and everything worked fine (I had a generic table builder, but now I am having to stray from that):

 while ($x = mysqli_fetch_assoc($result))
 {
   $fields[] = $x['Field'];
 }

Now I have something similar to this:

 $result = mysqli_query($con, 'SELECT r.id AS ID, CONCAT(g.fname, g.lname) AS Name, r.apple AS Apple, 
                            r.dog AS Dog, DATEDIFF(r.Dog, r.Apple) AS Days, 
                            r.total_price AS "Total Price", u.name AS Name, r.in AS "In",
                            r.out AS "Out", r.time_in AS "Time In", r.time_out AS "Time Out", 
                            CONCAT(c.fname,c.lname) AS Charlie, r.here AS "Apple",
                            r.leave AS "Dog"
                            FROM really r, georgia g, unit u, charlie c 
                            WHERE g.id = r.georgia AND r.unit = u.id AND r.charlie = c.id
                            HAVING r.in = TRUE AND r.out = FALSE');

    //fill fields array with fields from table in database
    while ($x = mysqli_fetch_assoc($result))
    {
        $fields[] = $x['Field'];
    }

I am now getting an error for the line $fields[] = $x['Field']; because of the word Field. Why? Because I now have a full query? How can I fix this without referencing each field name?

why not use mysqli_fetch_array?

Jacques
– Jacques

2013-10-08 02:34:26 +00:00
Commented Oct 8, 2013 at 2:34 — Jacques
– Jacques, Commented Oct 8, 2013 at 2:34

srain · Accepted Answer · 2013-10-08 02:42:02Z

Because there is not a field named Field in you query result:

'SELECT r.id AS ID, CONCAT(g.fname, g.lname) AS Name, r.apple AS Apple, 
                        r.dog AS Dog, DATEDIFF(r.Dog, r.Apple) AS Days, 
                        r.total_price AS "Total Price", u.name AS Name, r.in AS "In",
                        r.out AS "Out", r.time_in AS "Time In", r.time_out AS "Time Out", 
                        CONCAT(c.fname,c.lname) AS Charlie, r.here AS "Apple",
                        r.leave AS "Dog"
                        FROM really r, georgia g, unit u, charlie c 
                        WHERE g.id = r.georgia AND r.unit = u.id AND r.charlie = c.id
                        HAVING r.in = TRUE AND r.out = FALSE'

There are some fields in your query result: ID, Name, Apple, etc.. You can try to fetch the these field as below, or change your query command.

while ($x = mysqli_fetch_assoc($result))
{
    $fields[] = $x['ID'];
}

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