在选择上获得重复记录

Question

我有4张桌子：

clients, pricelist, result and result_details

我想要生成一个transaction_id, name, test_name and result出现在其上的表单。我为name做了一个选择，但我找不到test_name和result的选择。

我想要这样的东西[pdf] [5]

$result=mysqli_query($con,"SELECT r.transaction_Id,r.client_id,r.result_id,r.test_id,t.test_id,t.result,p.test_id,p.Name,c.Id,c.surName,c.firstName,c.dob,c.Gender FROM result r 
                                JOIN result_details t 
                                on r.test_id = t.test_id
                                left JOIN pricelist p ON r.test_id = p.test_id 
                                left JOIN clients c ON r.client_id = c.Id")or die(mysqli_error());


$r=mysqli_query($con,"SELECT DISTINCT id,firstName,surName,Gender,dob FROM clients join result WHERE result.client_id=clients.Id")or die(mysqli_error());

while ($row = mysqli_fetch_array ($r)) {

    $id=$row['id'];
    $clientfirstname=$row['firstName'];
    $surname=$row['surName'];
    $gender=$row['Gender'];
    $dob=$row['dob'];
    $t=mysqli_query($con,"SELECT transaction_Id FROM transaction WHERE clientId=$id") or die(mysqli_error());

    while ($test = mysqli_fetch_array ($t)) {
        $transaction=$test['transaction_Id'];
        $m=mysqli_query($con,"SELECT test_id,result_id from result where client_id=$id")or die(mysqli_error());
        while ($q = mysqli_fetch_array ($m)) {    
            $test_id=$q['test_id'];
            $result_id=$q['result_id'];
            $z=mysqli_query($con,"SELECT Name from pricelist where test_id=$test_id")or die(mysqli_error());
            while ($x = mysqli_fetch_array ($z)) {    
                $name=$x['Name'];
                $query=mysqli_query($con,"SELECT result from result_details where result_id=$result_id")or die(mysqli_error());
                while ($query1 = mysqli_fetch_array ($query)) {    
                    $result=$query1['result'];
                }
            }
        }
    }
}                       

Answer 1

只需添加：

HTTP(s)

GROUP BY将具有相同field1，field2，...和Charge的所有行分组在一起，然后HAVING COUNT（1）= 1仅显示只有1个祖先的行。我希望它能起作用