prepare()に失敗しました。prepare()が失敗しました。 MySQLでSQL文を準備しようとするとエラーが発生する
私は、このSQL文を準備しようとすると、このエラーが来ている:$sqlQuery
の
$stmt = $statement_handler->prepare($sqlQuery);
エコーは次のとおりです。
SELECT
distinct(reservation.reservationid),reservation.name,reservation.age
as p_age,reservation.gender as
p_gender,reservation.lessonregistrationno,reservation.mobile,reservation.email,reservation.arrivaldate,reservation.arrivaltime,reservation.departuredate,reservation.departuretime,centers.name
as
centername,reservation.purposeofvisitid,reservation.otherpurpose,reservation.additionalinformation,reservation.status,reservation.membertype
as
type,reservation.iskriyaban,reservation.sharedacco,reservation.citizenship,reservation.total_guest,reservation.created,reservationmember.name
as
member_name,reservationmember.age,reservationmember.lessonregistrationnumber,reservationmember.gender,reservationmember.membertype,relationships.relationship,countries.name
as country,purposeofvisits.purpose FROM reservation LEFT JOIN centers
ON reservation.centerid = centers.centerid LEFT JOIN purposeofvisits
on purposeofvisits.purposeofvisitid = reservation.purposeofvisitid
LEFT JOIN countries ON countries.id = reservation.countryid LEFT JOIN
reservationmember on reservationmember.reservationid =
reservation.reservationid LEFT JOIN relationships on
reservationmember.relationshipid = relationships.relationshipid WHERE
1 AND ((reservation.arrivaldate >= '2016-10-01')) AND (
(reservation.arrivaldate <= '2016-10-10')) ORDER BY
reservation.status desc, reservation.arrivaldate asc,
reservation.reservationid asc LIMIT 0, 25
バックエンドは、MySQLで、フロントエンドはPHPです。
エラーメッセージは何ですか? –
どのようなエラー??? 。 – devpro
「WHERE 1」は何が必要ですか? – devpro