我的代码:
$song = mysql_query("SELECT COUNT(*) FROM `".$this->prefix."playlist`
WHERE `time` > DATE_SUB(CURRENT_TIMESTAMP, INTERVAL 20 MICROSECOND)
AND `long` < DATE_SUB(CURRENT_TIMESTAMP, INTERVAL 20 MICROSECOND)");
if(mysql_result($song, 0, 0) == 0)
{
return "כלום";
} else {
$query = mysql_query("SELECT * FROM `".$this->prefix."playlist`
WHERE `time` > DATE_SUB(CURRENT_TIMESTAMP, INTERVAL 20 MICROSECOND)
AND `long` < DATE_SUB(CURRENT_TIMESTAMP, INTERVAL 20 MICROSECOND)");
$song = mysql_fetch_assoc($query);
return $song["song"];
}
它应该返回现在播放的歌曲。
MYDB:
1 till the worlds ends 2011-06-04 11:20:10 2011-06-04 11:21:10 zRafael
时间:
11:20
它返回:“כלום”而不是歌曲名称
这是我的表:
CREATE TABLE IF NOT EXISTS `playlist` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`song` varchar(150) NOT NULL,
`time` varchar(150) NOT NULL,
`long` varchar(150) NOT NULL,
`by` varchar(150) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;
--
-- Dumping data for table `playlist`
--
INSERT INTO `playlist` (`id`, `song`, `time`, `long`, `by`) VALUES
(1, 'till the worlds ends', '2011-06-04 11:40:10', '10', 'zRafael');
答案 0 :(得分:0)
我会重写这个以避免第二个查询,这应该可行
再次更新 - 您的日期字段为VARCHAR,而不是DATETIME,因此我之前的查询无效。
$rs = mysql_query("SELECT song FROM `".$this->prefix."playlist`
WHERE CAST(`time` AS DATETIME) >= NOW()
AND CAST(`long` AS DATETIME) < NOW()");
if (mysql_num_rows($rs) == 0)
return "כלום";
else
return mysql_result($rs, 0);
实际上你应该修复你的表定义。
ALTER TABLE `playlist`
MODIFY COLUMN `time` DATETIME NOT NULL
, MODIFY COLUMN `long` DATETIME NOT NULL;
然后你不需要那些演员表,你的数据库查找会更快。