我有点小麻烦。我的数据库里有两张桌子:
servico,与id,nome,profissao (水管工,电工,e),数据和本地。
canalizador,有id,nome,profissao (只有水管工),数据和本地。
我有这样的代码,这样我就可以到达"canalizador“表了:
mysql_connect($host,$user,$passwd) or die (mysql_error());
mysql_select_db($database);
$query = "SELECT * FROM canalizador WHERE profissao LIKE '$profissao'";
$result = mysql_query($query) or die(mysql_error());
$casaArray = array();但我需要一个代码,在那里我可以搜索所有的表格,如"canalizador“表,只需按下每个职业的按钮(Profissao)在菜单上。如果我这么做:
mysql_connect($host,$user,$passwd) or die (mysql_error());
mysql_select_db($database);
$query = "SELECT * FROM canalizador, carpinteiro WHERE profissao LIKE '$profissao'";
$result = mysql_query($query) or die(mysql_error());
$casaArray = array();我有一个错误:
where子句中的“profissao”列
发布于 2015-06-11 07:29:28
对两个表执行UNION ALL,在结果中使用LIKE (作为派生表):
select id, nome, profissao, data, local
from
(
select id, nome, profissao, data, local from servico
UNION ALL
select id, nome, profissao, data, local from canalizador
) dt
where dt.profissao LIKE '$profissao'或者,正如AsConfused所建议的那样,只是一个UNION ALL
select id, nome, profissao, data, local from servico
where profissao LIKE '$profissao'
UNION ALL
select id, nome, profissao, data, local from canalizador
where profissao LIKE '$profissao'发布于 2015-06-11 07:24:26
试试这个:
mysql_connect($host,$user,$passwd) or die (mysql_error());
mysql_select_db($database);
$query = "SELECT * FROM canalizador c1, carpinteiro c2 WHERE c1.profissao LIKE '$profissao'";
// ^ c1 or c2
$result = mysql_query($query) or die(mysql_error());
$casaArray = array();https://stackoverflow.com/questions/30774093
复制相似问题