php - 让我的代码在没有新表的情况下变得更好“某天表”

Question

我的代码运行良好，但我不想将其插入新表中我想在同一个表中查询Examdata并获得相同的结果

$g="INSERT into someday select * from Examdata join (SELECT Student_ID, count(*), exam_dates FROM Examdata group by Student_ID, exam_dates having count(*) > 1) Examdata1 on Examdata.`Student_ID`=Examdata1.`Student_ID` and Examdata1.`exam_dates`=Examdata.`exam_dates` ORDER BY Examdata.exam_dates ASC";
mysqli_query($connection,$g);
$students = $connection->query("SELECT distinct Student_ID from someday")->fetch_all(MYSQLI_ASSOC);

foreach ($students as $student) {

  $sql = $connection->query("SELECT Student_ID,Subject_name,exam_days,exam_dates FROM someday where Student_ID=".$student["Student_ID"]." ;")->fetch_all(MYSQLI_ASSOC);
  echo "<br> ".$student["Student_ID"].":<br> ";
foreach ($sql as $sq) {
            echo $sq["exam_dates"]."  - <br> ";
            echo $sq["exam_days"]."  - <br> ";
            echo $sq["Subject_name"]."  - <br> ";
        }
}

标签： phpmysql

Answer 1

试试这个 -

$sql = $connection->query("SELECT Student_ID,Subject_name,exam_days,exam_dates FROM examdata order by student_ID,exam_dates ASC;")->fetch_all(MYSQLI_ASSOC);
$student = "";
foreach ($sql as $sq) {
    if ($student != $sq["Student_ID"]) {
        $student = $sq["Student_ID"];
        echo "<br> ".$sq["Student_ID"].":<br> ";
    }
            echo $sq["exam_dates"]."  - <br> ";
            echo $sq["exam_days"]."  - <br> ";
            echo $sq["Subject_name"]."  - <br> ";
}