好吧,所以我天生就害怕一切,MySQL+----+-----------------+------++----+---23 |+----+-----------------+------+$raw = mysql_query("SELECT * FROM example") or die(mysql_error());
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我下载了MySQL5.5.34的源代码,并按照说明进行了编译。然后,在/usr/local/ mysql /bin文件夹中,我执行了./mysqld_-user=,mysql服务器立即启动并关闭。/usr/local/mysql/data/p2020rdb.err读取
mysqld_safe Starting mysqld daemon with databases from /usr/local
我有以下索引:年龄是一个可以为空的字段。first=x and last=y and age=zSELECT * FROM mytable WHERE first=x and last=y and age=z and gender=wSELECT * FROM mytable WHERE first=x and last=
我有php列表的项目。它工作正常,但最近的记录在增加,突然它没有在列表中显示记录。我尝试过内部连接,它的记录较少,工作正常。 SELECT farmer.*, state.state_name, videos.video_url, employee.name AS nm LEFT JOIN state ON state.id = farmer.state