我有两个表,我需要在这里显示第二个表数据来比较名称。这在我用名称获取的两个表中很常见,但它不起作用。
表1字段为id、name、email表2字段为id、name、gender
我需要在一列中显示数据,如姓名、电子邮件、性别
这是代码
$query = "(select name,email from table1 LEFT JOIN table2 ON table1.name=table2.name)";
$fetch = mysql_query($query);
while($list = mysql_fetch_assoc($fetch))
{
$name = $list['name'];
$email = $list['email'];
$gender = $list['gender'];
echo "Name:" . $name . "Email:" . $email . "Gender:" . $gender;
}
您正在选择名称和电子邮件,因此定义它,以便从维奇表中获得以下值:
像这样:
table1.name,table1.email,table2.gender
使用此更改查询:
$query = "(select table1.name,table1.email,table2.gender from table1 LEFT JOIN table2 ON table1.name=table2.name)";
试试这个,
$query = "(SELECT t1.name,t1.email, t2.gender FROM from table1 t1, table2 t2 where t1.name = t2.name)"