Question

我首先搜索所有问题信息。来自＆＃34;问题＆＃34;表格包括标题，内容，用户等守则：

$sql  = "select * FROM question where id>0 ORDER BY id ASC";  
    $result1 = mysql_query($sql);
    $res=Array();

然后我想从＆＃34;用户＆＃34;中搜索用户的观点。表。所以我必须从result1搜索每一行中每个用户的点守则：

while($rows=mysql_fetch_assoc($result1))
    {
        $res[]=$rows;
        $user = $rows['user'];
        $sql2 = "select point from user where name='$user'";
        $result2 = mysql_query($sql2);  
    }

我的问题是如何结合所有用户＆＃39;将问题信息（result1）与point（result2）放在一起，这样我就可以为每一行返回一个json。

Answer 1

最好在这里加入我给你的查询。我希望它可以帮到你

select * from question q,user u where q.id>0 ORDER BY id ASC

Answer 2

尝试这样的事情：使用左连接

select question.*,user.point FROM question left join user on user.name= question.name where id>0 ORDER BY id ASC

Answer 3

使用左连接，因为我理解这项工作为你

$sql  = "SELECT q.*, u.point AS point FROM question AS q LEFT JOIN user AS u ON q.user = u.name WHERE q.id > 0 ORDER BY q.id ASC";  
$result = mysql_query($sql);

php mysql查询组合

3 个答案: