PHP 在一个查询中显示来自四个表的数据（即：LEFT JOIN）

Question

我有三个 table：

我想显示 'Event Details'，它显示参加员工的详细信息（来自 'attending_employees table > 相应员工详细信息表格的员工 ID 列表 'employee' table），他们属于哪个团队（来自 club_teams table）以及来自 'club_events' table 的活动详情）。

目前我正在使用多个 mysqli 查询来显示此信息，但是无法在一个查询中从数据库中提取数据（即：LEFT JOIN）。非常感谢您的帮助！

以下是我目前正在使用的查询：

$query = msqli_query($con, "SELECT * FROM attending_employees")or die(mysqli_error($con));
if(mysqli_num_rows($query) > 0){
    while($attending = mysqli_fetch_array($query)){
        foreach($attending['club_event']){
            $eventid = $attending['club_event'];
            $query = msqli_query($con, "SELECT * FROM club_events WHERE club_event_id = '$eventid'")or die(mysqli_error($con));
            while($event_details = mysqli_fetch_array($query)){
                // Echo event details
            }
        }foreach($attending['employee']){
            $empid = $attending['employee'];
            $query = msqli_query($con, "SELECT * FROM employees WHERE employee_id = '$empid'")or die(mysqli_error($con));
            while($event_employees = mysqli_fetch_array($query)){
                // Echo employee details
            }
        }foreach($attending['team']){
            $teamid = $attending['team'];
            $query = msqli_query($con, "SELECT * FROM club_teams WHERE clb_team_id = '$teamid'")or die(mysqli_error($con));
            while($event_team = mysqli_fetch_array($query)){
                // Echo team details
            }
        }
    }
}

这种方法非常低效且浪费，因为它会检索重复数据（即：'attending_employees' table 中所有重复的“club_event_id”。）

Answer 1

试试这个：

$query = msqli_query(
    $con, 
    "SELECT"
    . " attending_employees.*"
    . ", club_events.*"
    . ", employees.*"
    . ", club_teams.*"
    . " FROM"
    . " attending_employees"
    . " LEFT JOIN club_events ON club_events.club_event_id = attending_employees.club_event"
    . " LEFT JOIN employees ON employees.employee_id = attending_employees.employee"
    . " LEFT JOIN club_teams ON club_teams.clb_team_id = attending_employees.team"
    ) or die(mysqli_error($con));

PHP 在一个查询中显示来自四个表的数据（即：LEFT JOIN）

PHP displaying data from four tables in one query (ie: LEFT JOIN)

php

mysql

sql

mysqli

left-join