PHP MySQL - 仅当不为空时才进行内部联接

Question

我有一个使用 Inner Join 的 mysqli SELECT 查询，我注意到一个大问题：它不会选择条件列值为 null 的行（因为第二个表中不存在 NULL）。这是我的代码：

<?php

$sql = mysqli_connect(/* CONNECTION */);

$query =    "SELECT " .
            "e.EQUIPMENT_ID, " .
            "e.CUSTOMER_ID, " .
            "e.DESCRIPTION, " .
            "e.LOCATION, " .
            "e.JOB_SITE, " .
            "e.PROJECT_NAME, " .
            "jb.DESCRIPTION AS JOB_SITE_NAME " .
            "FROM equipments e " .
            "INNER JOIN jobsites jb ON jb.JOBSITE_ID = e.JOB_SITE " .
            "WHERE e.CUSTOMER_ID = 1 ".
            "ORDER BY e.EQUIPMENT_ID ASC";

$results = mysqli_query($sql, $query);

if(!isset($data)) $data = array(); $cc = 0;

while($info = mysqli_fetch_array($results, MYSQLI_ASSOC)){

    if(!isset($data[$cc])) $data[$cc] = array();

    ///// FROM TABLE equipments /////
    $data[$cc]['EQUIPMENT_ID'] = $info['EQUIPMENT_ID'];
    $data[$cc]['DESCRIPTION'] = $info['DESCRIPTION'];
    $data[$cc]['LOCATION'] = $info['LOCATION'];
    $data[$cc]['PROJECT_NAME'] = $info['PROJECT_NAME'];
    $data[$cc]['JOB_SITE_ID'] = $info['JOB_SITE'];

    ///// FROM TABLE jobsites /////
    $data[$cc]['JOB_SITE'] = $info['JOB_SITE_NAME'];

    $cc++;
}

print_r($data);

?>

所以，正如我所说，代码返回值，但前提是“设备”中的“JOB_SITE”列具有作业站点 ID（非空）。丑陋的解决方案是在表“jobsites”中创建一个名为“empty”的jobsite_id 的行，但如果我可以跳过这个，我会的。

只有当 e.JOB_SITE 不为空时，有没有办法加入？

Answer 1

您可以在 SQL 查询中使用LEFT JOIN。

$query =    "SELECT " .
            "e.EQUIPMENT_ID, " .
            "e.CUSTOMER_ID, " .
            "e.DESCRIPTION, " .
            "e.LOCATION, " .
            "e.JOB_SITE, " .
            "e.PROJECT_NAME, " .
            "jb.DESCRIPTION AS JOB_SITE_NAME " .
            "FROM equipments e " .
            "LEFT JOIN jobsites jb ON jb.JOBSITE_ID = e.JOB_SITE " .
            "WHERE e.CUSTOMER_ID = 1 ".
            "ORDER BY e.EQUIPMENT_ID ASC";

如果equipments 表中没有与jb.JOBSITE_ID 匹配的row，则此查询将为列JOB_SITE_NAME 返回NULL VALUE