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我目前正在尝试通过 PHP 执行一个存储过程（两个 INSERTS 和一个 SELECT）。存储过程有效（不会引发错误）。这些值甚至被插入到给定的表中。我们将结果存储在 JSON 对象中，但它返回一个空数组。当我们执行只包含一个 SELECT 的存储过程时，它会返回正确的值。我们表中的所有名称都已更改（可能遗漏了一些但我们的代码运行）。谁能解释为什么我们的阵列没有填充？

这是我们的存储过程

ALTER procedure 
As
BEGIN

INSERT INTO TABLE1
(Loan_ID, Doc_ID, Borrower_Name, Docs_Drawn, Funder, First_Payment, Interest_Rate, Loan_Amount)
Select Loan_ID, 
        (SELECT (COUNT(*)) FROM TABLE WHERE A.Loan_ID = Loan_ID) AS Doc_ID,
        Borrower ,[Date_Docs_Drawn]
      ,[Funder]
      ,[Payment_Date]
      ,[Interest_Rate]
      ,[Loan_Amount] from VIEW A

 INSERT INTO TABLE2
(Loan_ID, Submitted_By, Event_Class, Event_Type, Event_Date, Doc_ID)
SELECT A.[Loan_ID], 
    'Program' AS Submitted_By
    ,'Collateral' AS Event_Class
      , 'Outstanding' AS Event_Type
     , CURRENT_TIMESTAMP AS Event_Date 
     , Doc_ID                                       
  FROM TABLE1 AS A
  WHERE NOT (A.Loan_ID = ANY (SELECT Loan_ID FROM TABLE2))

    SELECT [Loan_ID],[Doc_ID],[Borrower_Name]
      ,[Funder]
      ,[Docs_Drawn]
      ,[First_Payment]
      ,[Loan_Amount]
      ,[Interest_Rate]
    FROM TABLE1
    WHERE Loan_ID <> ALL
    (SELECT Loan_ID FROM TABLE2
    WHERE Event_Type <> 'Outstanding')
    ORDER BY Funder

END

这是我们PHP的粗略草图

$query = "exec storedProcedure";
$result=sqlsrv_query($conn, $query);
$table = array();
while($row = sqlsrv_fetch_array($result)) {
    $table[] = $row;
}
echo json_encode($table);
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1 回答 1

2

尝试SET NOCOUNT ON在您的存储过程中使用。您可能受到从 SQL Server PHP 驱动程序返回给 PHP 的结果的限制，因此您没有从 SELECT 语句中获取数据。

于 2013-07-11T16:39:12.660 回答