Php 用另一个表中的数据填充表

对MySQL来说是新手，在这个问题上遇到了困难。 在我的php脚本中卡住了

这是emailQueue表（抱歉，现在还不能发布图片）：

这是学生桌：

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我想从Students表中获取所有CourseID，并将它们添加到各自StudentEmail行的第1、2、3列等。提前谢谢

您可以通过编程方式或使用类似于：

INSERT INTO db.table-destination (c1, c2, c3...)
SELECT a1, a2, a3... FROM db.table-source;

您也可以尝试选择进入

您可以尝试以下方法：

<?php

/* ESTABLISH CONNECTION */

$con=mysqli_connect("YourHost","Username","Password","NameOfYourDatabase");

if(mysqli_connect_errno()){

echo "Error".mysqli_connect_error();
}

$result=mysqli_query($con,"SELECT * FROM Students ORDER BY StudentEmail"); /* USE ORDER BY StudentEmail */
while($row=mysqli_fetch_array($result)){

$courseid=mysqli_real_escape_string($con,$row['CourseID']); /* STORE THE CURRENT COURSE NAME TO $courseid VARIABLE */

if(empty($studentemail)){
   $studentemail=mysqli_real_escape_string($con,$row['StudentEmail']);
   $id=1;
   /* $studentemail VARIABLE PURPOSE IS TO STORE StudentEmail, FOR LATER CONDITION */
}

else {

   if($row['StudentEmail']==$studentemail){ /* IF $studentemail IS THE SAME WITH THE CURRENT StudentEmail ROW, INCREMENT $id */
   $id=$id+1;
   }

   else { /* ELSE, WHERE $studentemail IS NOT THE SAME WITH THE CURRENT StudentEmail ROW, RESTART THE COUNT OF $id BACK TO 1 AND STORE THE NEW StudentEmail INTO $studentemail */
   $studentemail=mysqli_real_escape_string($con,$row['StudentEmail']);
   $id=1;
   }

}

mysqli_query($con,"UPDATE emailQueue SET '$id'='$courseid' WHERE StudentEmail='$studentemail'"); /* UPDATE QUERY */

} /* END OF WHILE LOOP */

?>

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请告诉我们到目前为止您尝试了什么，以及您遇到的任何错误。