Question

我想知道为什么我这样做有一些错误：

 $query = "SELECT * FROM servers ";
      while($rows = mysql_fetch_array(mysql_query($query)))
    {
    $database1 = array();

 $database1['host']     = $rows['db_host'];
 $database1['user']     = $rows['db_user'];
 $database1['password'] = $rows['db_pass'];
 $database1['table']    = $rows['db_user'];

 $con1 = @mysql_connect($database1['host'], $database1['user'], $database1['password']) or die($connect_error);
 if(!$con1) {
       exit;
 }


$connect_error = 'Sorry, there are some connection problems.';

mysql_select_db($database1['table']) or die($connect_error);


$info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"));
 echo "No servidor ".$rows['name']." existem um total de ".$info['total']. " contas      registradas, alem disso";

    }

确定。以免解释：我有两个不同的数据库，在我编写此查询的fisrt中：

   query = "SELECT * FROM servers ";

我在表服务器中有一个mysql连接。所以我必须连接不同的mysql来获取mysql主机，用户和密码。当我得到这些东西时，我必须获得每个连接的值并回显此值当我这样做时，使用COUNT作为变量：

 $info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"));
echo "No servidor ".$rows['name']." existem um total de ".$info['total']. " contas      registradas, alem disso";

我希望你们不会因为我不会说英语而感到满意）：感谢你的关注......

Answer 1

你应该{@ 1}}一次不在循环中运行。您的代码会在每个循环中运行mysql_query($query)。

从以下位置删除：

mysql_query

执行：

while($rows = mysql_fetch_array(mysql_query($query)))
                          //HERE^

Answer 2

查看您的脚本，您仍可以进行一些改进。对于初学者，你的变量声明（$ connect_error）被创建 AFTER 。

虽然您应该采用mysqli PHP API代替您使用的MySQL函数，但已将其标记为已弃用。

# Where query results are stored
$database1 = array();

# Error Message
$connect_error = 'Sorry, there are some connection problems.';

# mySQL Loop No.1
while($rows = mysql_fetch_array(mysql_query("SELECT (name,db_host,db_user,db_pass) FROM servers"))) {
    foreach($rows as $key=>value) {
       $database1[$key] = $value;   
    }   
}

# Since this is a duplicate of existing row duplicate
$database1['table'] = $rows['db_user'];

# Connect to new db using details
$con1 = @mysql_connect($database1['host'], $database1['user'], $database1['password']) or die($connect_error);
if(!$con1) {
       exit;
}

# Select 2nd DB
mysql_select_db($database1['table']) or die($connect_error);

# mySQL Loop No.2
while($info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"))) {
    $total = $info['total'];
}

#Lastly with all data gotten you can show string
echo "No servidor ".$database1['name']." existem um total de ".$total. " contas registradas, alem disso";

请参阅进一步阅读：

http://www.php.net/manual/en/mysqli.quickstart.dual-interface.php

如何使用while连接两个mysql？

2 个答案: