Question

我的代码如下：

    $sql="SHOW DATABASES WHERE `Database` NOT LIKE '%backup%' AND `Database` NOT LIKE 'mysql' AND `Database` NOT LIKE '%schema%'";
 $query=mysql_query($sql,$connect); 
 while ($row = mysql_fetch_assoc($query)) {
     $sql1="SHOW TABLES FROM `".$row['Database']."`"; 
         $query1=mysql_query($sql1, $connect);
         while( $row2 = mysql_fetch_assoc($query1) ) {
         $row2=implode(" ",$row2);
          $sql1="SELECT COUNT(*) FROM `".$row['Database']."`.`".$row2."`";
          $query1=mysql_query($sql1,$connect);  
          echo implode(" ",mysql_fetch_assoc($query1)).'<br>';
         }
 }

我有两个数据库和两个表，有两个记录上面的代码统计了每个数据库中每个表的每个记录，它将获取两个数据，这些数据包含每个表的记录号。

所以它会输出：

2
2

我需要的是总结它们，结果将是4.Help please？

Answer 1

只需计算PHP中的总和：

$sql="SHOW DATABASES WHERE `Database` NOT LIKE '%backup%' AND `Database` NOT LIKE 'mysql' AND `Database` NOT LIKE '%schema%'";
$query=mysql_query($sql,$connect); 
$total = 0;

while ($row = mysql_fetch_assoc($query)) {
    $sql1="SHOW TABLES FROM `".$row['Database']."`"; 
    $query1=mysql_query($sql1, $connect);

    while($row2 = mysql_fetch_assoc($query1) ) {
        $row2=reset($row2);
        $sql2="SELECT COUNT(*) FROM `".$row['Database']."`.`".$row2."`";
        $query2=mysql_query($sql2,$connect);  
        $total += mysql_result($query2, 0);
    }
}

echo $total;

并且还停止使用 mysql _ * 函数，它们已被弃用...

加入中添加？

1 个答案: