Question

我需要将数据插入到表中没有退出的facilities_details表中，下面的代码无法正常运行。

$facility = $_POST["facilities_cb"];
foreach ($facility as $feature)
{
    mysqli_query($conn,"INSERT INTO facilities_details(facilitiestype_id,hotel_id) values ('$feature','$hid') WHERE NOT EXISTS (SELECT facilitiestype_id FROM facilities_details)");
}

Answer 1

没有此类语法WHERE无法与INSERT一起使用。如果您想在数据存在时忽略INSERT，则可以使用INSERT IGNORE

INSERT IGNORE INTO facilities_details(facilitiestype_id,hotel_id)
VALUES ('$feature','$hid')

Answer 2

试试这样。

IF NOT EXISTS
  (SELECT facilitiestype_id
   FROM facilities_details)
INSERT INTO facilities_details(facilitiestype_id,hotel_id)
VALUES ('$feature',
        '$hid');

Answer 3

试试这个

insert into facilities_details (facilitiestype_id, hotel_id) 
select $feature, $hid where not exists (
    select 1 from facilities_details where facilitiestype_id = $feature and hotel_id = $hid
)

或者

insert into facilities_details (facilitiestype_id, hotel_id) 
select $feature, $hid where not exists (
    select 1 from facilities_details where facilitiestype_id = $feature
)

使用NOT EXISTS插入值

3 个答案: