我正在用php和後端sql server 2005創建一個報告。我在sql服務器上寫了這個運行正常的blew query,同時在php中執行時給出錯誤。php中的查詢執行錯誤
$query = "select distinct t1.VisitDate,t1.ReceiptNo,t1.VisitorNo,p.PatientName,STUFF((
SELECT f.Particular + ', ' from CollectionPointTable t2,FeesTable f
where t1.ReceiptNo = t2.ReceiptNo and f.ID = t2.TestID
FOR XML PATH(''), TYPE).value('.', 'VARCHAR(MAX)') ,1,0,'') Tests , sum(t1.payment) as Amount
from CollectionPointTable t1,patientinformationtable p
where p.visitorno = t1.visitorno and convert(varchar(10),t1.VisitDate,103) = '$date'
group by t1.Receiptno,p.patientname,t1.VisitorNo,t1.VisitDate
order by ReceiptNo";
$stid = sqlsrv_query($conn, $query);
$Today_patient_list = sqlsrv_fetch_array($stid,SQLSRV_FETCH_ASSOC);
即其通過錯誤給出
警告該錯誤:sqlsrv_fetch_array()預計參數1是資源,在布爾C中給出:\ wampnew \ WWW \日誌\ viewTodaysCollection.php上線65
請問您首先檢查您的sqlsrv_connect?你的連接是否完成? – KinjalMistry
'sqlsrv_query'可能返回了一個錯誤。在'fetch_array'中使用'$ stid'之前檢查它的返回值。 (也許在你的'$ query'中有錯誤) – Pierre
用'var_dump'打印出來檢查'$ stid'的值,你會發現它是布爾類型 –