Question

php mysql 查询一个表 相同字段两次的一个问题

tb :
tb_id, integer, not null
tb_content, varchar(50),not null

$string = "select tb1.tb_content, tb2.tb_content from tb as tb1, tb as tb2 where tb1.tb_id =1 and tb2.tb_id=4;";
$query = mysql_query($string, $db_connection);
while($array=mysql_fetch_array($query))
{
print("<br>".$array[tb1.tb_content]."<br>");
}
不输出
请问怎样解决？

Answer 1

首先你要确认数据库里面存在tb_id为1和4的记录。

其次你输出$array[tb1.tb_content]是不对的，两个字段都会叫做$array[tb_content]，你可以用$array[0]、$array[1]来输出，或者语句里面定义别名：
$string = "select tb1.tb_content content1, tb2.tb_content content2 from tb as tb1, tb as tb2 where tb1.tb_id =1 and tb2.tb_id=4";
这样就可以输出$array[content1]、$array[content2]

另外最好程序中随时检测并输出错误信息，我修改语句如下：
if ($query = mysql_query($string, $db_connection)){
...while ...
}else echo "执行 $string 错误：".mysql_error();

Answer 2

$string
=
"select
tb1.tb_content,
tb2.tb_content
from
tb
as
tb1,
tb
as
tb2
where
tb1.tb_id
=1
and
tb2.tb_id=4;";
print("<br>".$array[tb1.tb_content]."<br>");
修改成:$string
=
"select
tb1.tb_content
as
content1,
tb2.tb_content
as
content2
from
tb
as
tb1,
tb
as
tb2
where
tb1.tb_id
=1
and
tb2.tb_id=4";
print("<br>".$array['content1']."<br>");
print("<br>".$array[tb1.tb_content]."<br>");
这样打印是不对的，至少应该是print("<br>".$array['tb1.tb_content']."<br>");