我有 4 张 table :
clients, pricelist, result and result_details
我想生成一个表单，其中显示 transaction_id、name、test_name 和 result。我已经选择了 name，但无法找到 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'];
}
}
}
}
}
最佳答案
只需添加:
GROUP BY AcctNo, OrderDate, Charge
HAVING COUNT(1) = 1
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。我希望它会起作用
关于php - 选择时获取重复记录，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41891309/