我构建了2个查询，将$project_id变量与2个表MEETING＆amp; MEETING_AGENDA显示在下面的流程图中：
+---------------------+
+------------+ | MEETING_AGENDA |
| MEETING | +---------------------+
+------------+ | meeting_agenda_id |
| meeting_id |----->| meeting_id |
$project_id --->| project_id | | meeting_agenda_name |
+------------+ +---------------------+
这个流程图背后的逻辑是，我存储了一个名为$project_id的变量作为输入。如果是$project_id = MEETING.project_id，那么我需要存储MEETING.meeting_id的列表。 project_id中有多个MEETING重复项，导致meeting_id数组（仅供澄清）。如果MEETING.meeting_id = MEETING_AGENDA.meeting_id中的任何一个meeting_agenda_name，则打印$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>"
}
}
。
我的混乱尝试（有效）看起来像这样：
$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
";
我使用2个查询，如果可能，我想将其清理为1个查询。我在单个查询中尝试了不同的尝试，但没有任何对我有用。这是我最近的尝试。
NEW QUERY：
{{1}}
我为我缺乏mysql知识而道歉，但感谢任何帮助。
答案 0 :(得分：1)
您可以使用以下内容在单个查询中获取该信息：
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