相关疑难解决方法(0)

我安装了XAMPP 1.7.4(用PHP 5.3.5),问题是PHP没有显示任何错误信息.例如,如果我连接到mysql_connect()没有参数的MYSQL ,PHP将不会抱怨所需的字段.

为什么是这样？

如何配置PHP以显示错误？

我写了一个触发器.

USE [TEST]
GO
/****** Object:  Trigger [dbo].[TR_POSTGRESQL_UPDATE_YC]    Script Date: 05/26/2010 08:54:03 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

ALTER TRIGGER [dbo].[TR_POSTGRESQL_UPDATE_YC] ON [dbo].[BCT_CNTR_EVENTS]

FOR INSERT
AS 
BEGIN


DECLARE @MOVE_TIME varchar(14);
DECLARE @MOVE_TIME_FORMATED varchar(20);
DECLARE @RELEASE_NOTE varchar(32);
DECLARE @CMR_NUMBER varchar(15);
DECLARE @MOVE_TYPE varchar(2);

SELECT @MOVE_TIME = inserted.move_time
      ,@MOVE_TYPE = inserted.move_type
      ,@RELEASE_NOTE = inserted.release_note
      ,@CMR_NUMBER = inserted.cmr_number FROM inserted


IF(@MOVE_TYPE = 'YC')
    BEGIN

    SET @MOVE_TIME_FORMATED = SUBSTRING(@MOVE_TIME,1,4) + '-' + SUBSTRING(@MOVE_TIME,5,2) + '-' + SUBSTRING(@MOVE_TIME,7,2) + ' 00:00:00'

    --UPDATE …