在php中匹配未转义的字符(match unescaped characters in php)

 我认为你的第二个触发器需要使用实际表中的值，而不是插入/删除的表来填充日志表 - 插入/删除将始终具有未更改的原始值，而更改的值将显示在表中。 使第二个触发器成为“After”触发器，因此您不必使用sp_settriggerorder。 像这样，例如： 
 
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TRIGGER [dbo].[trg_Trig1] 
   ON  [dbo].[TestTable] 
   FOR INSERT
AS 
BEGIN
    -- SET NOCOUNT ON added to prevent extra result sets from
    -- interfering with SELECT statements.
    SET NOCOUNT ON;

    update TestTable
    set [value] = 10
    where [value] = 25

END

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TRIGGER [dbo].[trg_Trig2]
   ON  [dbo].[TestTable] 
   AFTER INSERT
AS 
BEGIN
    -- SET NOCOUNT ON added to prevent extra result sets from
    -- interfering with SELECT statements.
    SET NOCOUNT ON;

    -- Insert statements for trigger here
    insert into log_TestTable
    (id, description, [value])
    select tt.id, tt.description, tt.[value]
    from inserted i
        LEFT JOIN TestTable tt
            ON tt.id = i.id

END

I think your second trigger needs to use the values from the actual table, not the inserted/deleted tables to populate the log table - inserted/deleted will always have the unaltered, original values, while your altered values will appear in the table. Make the second trigger an "After" trigger, so you will not have to use the sp_settriggerorder. Like this, for example:
 
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TRIGGER [dbo].[trg_Trig1] 
   ON  [dbo].[TestTable] 
   FOR INSERT
AS 
BEGIN
    -- SET NOCOUNT ON added to prevent extra result sets from
    -- interfering with SELECT statements.
    SET NOCOUNT ON;

    update TestTable
    set [value] = 10
    where [value] = 25

END

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TRIGGER [dbo].[trg_Trig2]
   ON  [dbo].[TestTable] 
   AFTER INSERT
AS 
BEGIN
    -- SET NOCOUNT ON added to prevent extra result sets from
    -- interfering with SELECT statements.
    SET NOCOUNT ON;

    -- Insert statements for trigger here
    insert into log_TestTable
    (id, description, [value])
    select tt.id, tt.description, tt.[value]
    from inserted i
        LEFT JOIN TestTable tt
            ON tt.id = i.id

END