mysql - 基于选择查询更新查询

Question

我确实想将“UPDATE”和“SELECT”写入一个查询中。

我需要检查设置字段。对于此操作，我在 TABLE1 上使用 SELECT 查询，然后如果没有结果，则更新 TABLE2 的另一个字段。例如:

$res = mysqli_query($con , "select sID FROM schedule where (dayID = '{$this->dID}' AND patientID = '')");
$rep = mysqli_fetch_array($res);
if(count($rep) == 0)
mysqli_query($con,"update days set schFilled = 1  where dID = '{$this->dID}'");
else   
mysqli_query($con,"update days set schFilled = 0 where dID = '{$this->dID}'");

我想用一个查询来运行它们，事实上我想要这样的东西:(whit CASE 也可以写第二个更新)

update days set schFilled = 0 where( (select sID FROM schedule where (dayID = '{$this->dID}' AND patientID = '') IS NULL) AND (dID = '{$this->dID}'))

Answer 1

像这样使用 mysqli 对象:

 $res = mysqli_query($con , "your select");
 if($res->num_rows === 0) {
      //no res
 }
 else {
      //else
 }