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私はこれを持っていて、すべてうまくいきました（私は一般的なテーブルビルダーを持っていましたが、今はそれから離れなければなりません）：

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

今、私はこれに似たものを持っています:

 $result = mysqli_query($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'];
    }

$fields[] = $x['Field'];という単語が原因で、行のエラーが発生していますField。なんで？完全なクエリができたからですか？各フィールド名を参照せずにこれを修正するにはどうすればよいですか?

4

1 に答える 1

1

Fieldクエリ結果に名前が付けられたフィールドがないため:

'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'

IDクエリ結果には、Name、 、 などのフィールドがいくつかありますApple。以下のようにこれらのフィールドをフェッチするか、クエリ コマンドを変更してみてください。

while ($x = mysqli_fetch_assoc($result))
{
    $fields[] = $x['ID'];
}
于 2013-10-08T02:42:02.733 に答える