部分加入Crystal Report [英] Partial Join In Crystal Report

问题描述

希望你们都做得很好，



我在水晶报告中遇到问题



Roll＃ =从Record_Roll_Number表中查看Std_Roll_Number字段的前10个字符



即Std_Roll_Number = 1212121212-00000-00000



学生姓名 =将上述Roll＃字段的信息链接到Record_Student表，拉出Student_FullName字段。



即在我的Record_Student表中，它有一列Short_Roll_number，其中包含此格式的值1212121212



现在我想在水晶报告中加入这两张表，以获得与学生全名相关的理想结果。



我没有找到如何加入这两个领域的方法，或者如何获得学生的名字。







这是我数据表的结构









Record_Roll_Number包含以下栏目

Std_Roll_Number ----------- Previous_Degree ------ Group_Name ---- --- Joining_Date ------- Old_Roll_No

1212121212-00000-00000 ----入学-------- IT ----------- --- 1-1-2012 ----------- 13141552





和表Record_Student包含以下栏目



Short_Roll_number -------- Student_FullName ------- Std_FatherName ---------城市--- ---- Phone_Number

1212121212 ------------ ABCDEFGH ---------------- GHIJKLMN ------ ------ XYZ ----------- 1010101010







这是表结构，测试值以及

Hi,
Hope you all are doing well,

I am facing a problem in crystal report

"Roll # = from the Record_Roll_Number table look at the first 10 characters of the Std_Roll_Number field

i.e Std_Roll_Number = 1212121212-00000-00000

Student Name = link the information found for the Roll # field above to the Record_Student table and pull the Student_FullName field."

i.e in my "Record_Student" table it has a column "Short_Roll_number" which contain the value in this format "1212121212"

now i want to join these two table in crystal report, to get my desired result related to student full name.

I am not finding a way how to join these two fields, or how to get student name.



This is the structure of my data table




Record_Roll_Number contain the following column
Std_Roll_Number-----------Previous_Degree------Group_Name-------Joining_Date-------Old_Roll_No
1212121212-00000-00000----Matriculation--------IT--------------1-1-2012-----------13141552


and table "Record_Student" contain the following column

Short_Roll_number--------Student_FullName-------Std_FatherName---------City-------Phone_Number
1212121212------------ABCDEFGH----------------GHIJKLMN------------XYZ-----------1010101010



this is the table structure, and the test value as well

推荐答案

考虑这种最佳方式

创建一列 Std_Roll_Number Record_Student

或

创建一列表格中的Short_Roll_number Record_Roll_Number

存储a在表格中存储数据时的适当值（在附加列上）。



完成上述任何操作后，您可以加入查询和表格中的表格。生成所需的输出。



Crystal报表要求列具有匹配的列数据类型（当您在报表设计器中建立关系时）。



除此之外，你还可以通过这种方式获得额外的好处。



编辑

-------- -------

Consider this best way
creating a column Std_Roll_Number in the table Record_Student
OR
creating a column Short_Roll_number in the table Record_Roll_Number
Store appropriate values(on additional column) when you store data in tables.

After done any of the above, you could join the tables in query & generate desired output.

Crystal reports requires the tables with matching datatypes of columns(When you do relationship in reports designer).

Besides you'll get additional benefits from this way.

EDIT
---------------

引用：

这是一个简单的方法，我不喜欢实现这个方法，我想简单地应用水晶报告中的连接，这是我客户的要求，所以请帮我用水晶报告来执行此操作。

This is a simple way, i don't like to implement this method, i want to apply simply a join from crystal report, it is the requirement from my client, so help me to perform this using crystal reports.

你不喜欢简单的方法？无论如何，我会给你另一个选择。



您是否在报告中使用数据库连接？如果是，那么就这样走吧。

来自SQL查询的C＃Crystal Reports [ ^ ]

通过这种方式，您需要对查询进行更改，如下所示。使用此查询绑定一个数据表&另一个带有其他查询的数据表。

You don't like the simple way? Anyway I'll give you an another option.

Are you using Database connection for your reports? IF yes then go with this way.
C# Crystal Reports from SQL Query[^]
In this way you need to do changes in query like below. Bind one datatable using this query & another datatable with other query.

SELECT LEFT(Std_Roll_Number,10) Std_Roll_Number, other fields FROM Record_Student

如果您的报告没有使用数据库，那么就这样使用。

没有数据库的C＃Crystal Reports [ ^ ]

同样的方式（在sql查询方式之上），你需要用提取的<$ c $绑定一个数据表c> Std_Roll_Number 。



就是这样。

If your reports not using database then go with this way.
C# Crystal Reports without database[^]
Same way(above sql query way), you need to bind one datatable with extracted Std_Roll_Number.

That's it.

我解决了我的问题自己，



只需应用Link-Sub报告的概念。



Std_Roll_Number和Short_Roll_编号并获取学生姓名

I solved my problem Own self,

Simply apply the concept of Link-Sub Report.

the "Std_Roll_Number" and "Short_Roll_number" and fetch the Student Name

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