在php中执行mysql查询以获取列和行

Question

我需要根据单个表中选定的日期显示所有员工的出勤率。

这是我的MySQL查询：

 SELECT 
     GROUP_CONCAT(DISTINCT 
     CONCAT(
      'max(CASE WHEN ca.Date = ''', 
      date_format(Date, '%Y-%m-%d'), 
      ''' THEN coalesce(p.Out_time) END) AS `', 
      date_format(Date, '%Y-%m-%d'), '`' 
     ) 
    ) INTO @sql 
    FROM attendance 
    where Date>='2015-04-27' 
     and Date <= '2015-04-30'; 

    SET @sql 
     = CONCAT('SELECT ca.User_name, 
        ca.User_id, 
        ', @sql, ' 
       from 
       (
        select c.Date, a.User_name, a.User_id 
        from attendance c 
        cross join attendance a 
       ) ca 
       left join attendance p 
        on ca.User_id = p.User_id 
        and ca.Date = p.Date 
       where ca.Date>=''2015-04-27'' 
        and ca.Date <= ''2015-04-30'' 
       group by ca.User_name, ca.User_id 
       order by ca.User_id'); 

PREPARE stmt FROM @sql; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

它是在MySQL（它显示的结果）执行罚款，但同时在PHP它不显示任何内容执行，

$qqq=mysql_query($sss); 

while($rrr=mysql_fetch_array($qqq)) 
{ 
echo $rrr; 
} 

它扔这些警告：

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in D:\Xampp\htdocs\php\temp.php on line 118 

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\Xampp\htdocs\php\temp.php on line 119 

我的表结构是这样的：

CREATE TABLE `attendance` (
    `Id` int(11) NOT NULL AUTO_INCREMENT, 
    `User_id` varchar(50) NOT NULL, 
    `User_name` varchar(100) NOT NULL, 
    `Date` date DEFAULT NULL, 
    `In_time` time DEFAULT NULL, 
    `Out_time` time DEFAULT NULL, 
    `Time_diff` decimal(5,2) DEFAULT NULL, 
    PRIMARY KEY (`Id`), 
    UNIQUE KEY `Id_UNIQUE` (`Id`) 
) 

Answer 1

您可以创建过程并通过php执行它。

欲了解更多信息和执行参考链接：Prepare Statement Execution Using PHP

希望这会帮助你。