需要帮助建立一个mysql查询，将一个变量与两个表进行比较

Question

I have built 2 queries that compares the $project_id variable to the 2 tables MEETING & MEETING_AGENDA shown in the flow diagram below:

                                    +---------------------+
                +------------+      |    MEETING_AGENDA   |
                |  MEETING   |      +---------------------+
                +------------+      | meeting_agenda_id   |
                | meeting_id |----->| meeting_id          |
$project_id --->| project_id |      | meeting_agenda_name |
                +------------+      +---------------------+

The logic behind this flow diagram is, I have stored a variable called $project_id which acts as the input. If the $project_id = MEETING.project_id then I need to store a list of the MEETING.meeting_id 's. There are multiple duplicates of project_id 's in MEETING resulting in an array of meeting_id 's (just for clarification). If any of the MEETING.meeting_id 's = MEETING_AGENDA.meeting_id then print meeting_agenda_name .

My messy attempt (which works) looks like this:

$project_id = $_SESSION['PROJECT_ID'];
$query1 = 
"
SELECT meeting_id, meeting_project_id
FROM MEETING
WHERE project_id = $project_id
";

$result1 = mysqli_query($con, $query1) or die("Query error: " . mysqli_error($con));

while($row = mysqli_fetch_array($result1)){
  $meeting_ids = $row['meeting_id'];

  $query2 = 
  "
  SELECT *
  FROM MEEITNG_AGENDA
  WHERE meeting_id = $meeting_ids
  ";

  $result2 = mysqli_query($con, $query2) or die("Query error: " . mysqli_error($con));

  while($row2 = mysqli_fetch_array($result2)){
    echo $row2['meeting_agenda_name'] . "<br>"  
  }
}

I use 2 query's, I would like to clean this up into 1 query if possible. I have tried varies attempts at a single query but nothing has worked for me. Here is my latest attempt.

NEW QUERY:

 $query = 
 "
 SELECT MEETING.project_id, MEETING.meeting_id, MEETING_AGENDA.*
 FROM MEETING
 WHERE MEETING.project_id = $project_id
 INNER JOIN MEETING.project_id
 ON $project_id = MEETING.project_id
 ";

I apologize for my lack of mysql knowledge, but any help is appreciated.

Answer 1

You can get that information in a single query with the following:

SELECT m.project_id, m.meeting_id, ma.*
FROM MEETING m
INNER JOIN MEETING_AGENDA ma ON ma.meeting_id  = m.meeting_id 
WHERE m.project_id = $project_id