Php sql查询中几乎不需要修改

下面是我的sql查询，在该查询中，facility_id i wat分配来自$fid的值。请让我知道以下代码有什么问题，以便我可以相应地删除它们

$fid = 101;
$q = 'select r_present,r_reminder 
    from z_events 
    where resident_id = ? 
        and r_added > date_sub(curdate(), interval 37 day) 
        and fascility_id='."$fid". '
    group by r_added 
    order by r_added 
    desc limit 4';

---

小的打字错误在fid和组之间添加空格，如下所示：

$q = 'select r_present,r_reminder 
from z_events 
where resident_id = ? 
    and r_added > date_sub(curdate(), interval 37 day) 
    and fascility_id='.$fid. ' 
group by r_added 
order by r_added 
desc limit 4';

下一个建议是：无需在$fid中加上双引号

“

这里我假设您的field facility\u id是整数类型。如果是字符串，则在其上添加双引号，如：

$q = 'select r_present,r_reminder 
from z_events 
where resident_id = ? 
    and r_added > date_sub(curdate(), interval 37 day) 
    and fascility_id="'.$fid. '" 
group by r_added 
order by r_added 
desc limit 4';

---

小的打字错误在fid和组之间添加空格，如下所示：

$q = 'select r_present,r_reminder 
from z_events 
where resident_id = ? 
    and r_added > date_sub(curdate(), interval 37 day) 
    and fascility_id='.$fid. ' 
group by r_added 
order by r_added 
desc limit 4';

下一个建议是：无需在$fid中加上双引号

“

这里我假设您的现场设施id是整数类型。如果是字符串，则向其添加双引号，如：

$q = 'select r_present,r_reminder 
from z_events 
where resident_id = ? 
    and r_added > date_sub(curdate(), interval 37 day) 
    and fascility_id="'.$fid. '" 
group by r_added 
order by r_added 
desc limit 4';

请尝试一下：

$fid = 101;
    $q = "select r_present,r_reminder from z_events where resident_id = ? and r_added > date_sub(curdate(), interval 37 day) and fascility_id=$fid group by r_added order by r_added desc limit 4";

请尝试一下：

$fid = 101;
    $q = "select r_present,r_reminder from z_events where resident_id = ? and r_added > date_sub(curdate(), interval 37 day) and fascility_id=$fid group by r_added order by r_added desc limit 4";

---

当不使用字符串连接时，最好使用

{$fid}

。）当不使用字符串连接时，最好使用

{$fid}

。）