PHPの3つのデータベースに関する4つのクエリを実行

Question

私はPHPとMySQLにはかなり新しく、さまざまな場所で読んだことがありますが、このスクリプトを一緒にチョップしてしまいました。

Buffer Status 2H: Notice: Undefined variable: row in /var/www/secretbit/allbuff.php on line 53 0 records. <br /> 
Buffer Status TC: Notice: Undefined variable: row in /var/www/secretbit/allbuff.php on line 54 0 records. <br /> 
Buffer Status ST: Notice: Undefined variable: row in /var/www/secretbit/allbuff.php on line 55 0 records. <br /> 
Buffer Status 3C: Notice: Undefined variable: row in /var/www/secretbit/allbuff.php on line 56 0 records <br /> 

そして、ここで私はこのモンスターをコンパイルするために使用したコードは次のとおりです：期待、私はの多くを得る

<?php 

error_reporting(-1); 
ini_set('display_errors', true); 

$con = mysqli_connect(
    '192.168.20.10', 
    'root2', 
    'password', 
    'contaque' 
); 

$con2 = mysqli_connect(
    '192.168.20.222', 
    'root2', 
    'password', 
    'contaque' 
); 

$con3 = mysqli_connect(
    '192.168.20.8', 
    'root2', 
    'password', 
    'contaque' 
); 

$res = mysqli_query(
$con, 
    "SELECT COUNT(*) AS 'total1' FROM contaque_hopper WHERE campaign_id = 'CLOSERUK';" 
); 

$res = mysqli_query(
$con2, 
    "SELECT COUNT(*) AS 'total2' FROM contaque_hopper WHERE campaign_id = 'CLOSERTC';" 
); 

$res = mysqli_query(
$con2, 
    "SELECT COUNT(*) AS 'total3' FROM contaque_hopper WHERE campaign_id = 'CLOSERST';" 
); 

$res = mysqli_query(
$con3, 
    "SELECT COUNT(*) AS 'total4' FROM contaque_hopper WHERE campaign_id = 'CLOSER3C';" 
); 

echo 'Buffer Status 2H: ', (int)$row['total1'], ' records. <br />'; 
echo 'Buffer Status TC: ', (int)$row['total2'], ' records. <br />'; 
echo 'Buffer Status ST: ', (int)$row['total3'], ' records. <br />'; 
echo 'Buffer Status 3C: ', (int)$row['total4'], ' records. <br />'; 

?> 

Answer 1

$rowが定義されていないようにまあ、それは見えます。クエリの結果を得るには、おそらくメソッドmysqli_fetch_assoc()を使用する必要があると思います。