数组中的问题.............. [英] problem in array..............

问题描述

我在C#中是新的，我有一个像这样的数组值

ID名称主题标记
1个jhone数学60
2史密斯科学50
3 clark math 90
1 jhone science 80
3克拉克科学56
1 jhone economics 75
2史密斯数学40
3克拉克经济学99

首先，我想对数组进行排序
ID名称主题标记
1个jhone数学60
1 jhone science 80
1 jhone economics 75

2史密斯科学50
2史密斯数学40

3 clark math 90
3克拉克科学56
3克拉克经济学99

array.sort命令正在运行..okk

现在我想连接这些值并像这样回答显示
1，jhone，数学60，科学80，经济学75
3，克拉克，数学90，科学56，经济学99

请记住，id 2没有显示连续性bcz id 2有2个主题..他的经济结果正在等待中..so只显示此结果

1，jhone，数学60，科学80，经济学75
3，克拉克，数学90，科学56，经济学99


请解决我的问题
在此先感谢

i m new in c# i have an array values like this

id name subject marks
1 jhone math 60
2 smith science 50
3 clark math 90
1 jhone science 80
3 clark science 56
1 jhone ecnomics 75
2 smith math 40
3 clark ecnomics 99

first i want to sort array like this
id name subject marks
1 jhone math 60
1 jhone science 80
1 jhone ecnomics 75

2 smith science 50
2 smith math 40

3 clark math 90
3 clark science 56
3 clark ecnomics 99

array.sort command working..okk

now i want to concatenate these values and answer show like this
1,jhone,math 60,science 80,ecnomics 75
3,clark,math 90,science 56,ecnomics 99

Remember id 2 not display contenate bcz id 2 have 2 subjects..his ecnomics result is pending..so only show this result

1,jhone,math 60,science 80,ecnomics 75
3,clark,math 90,science 56,ecnomics 99


please solve my problem
thanks in advance

推荐答案

您将必须编写自己的逻辑来做到这一点.

对结果进行排序后，您将需要一个循环来导航到数组中的每个值.声明一个字符串变量并连接所需的部分.仅当您找到3个具有相同ID的主题时，字符串中的值才会完整.

这不是很难.您只需要开始尝试.查看我的建议并开始实施.调试器将在执行此操作时派上用场.尝试过并遇到特定"问题时请回来.

祝你好运！

You will have to write your own logic to do it.

You will need a loop to navigate to each value in the array after you have sorted the results. Declare a string variable and concatenate the required parts. The value in the string will be complete only when you find 3 subjects for the same id.

It isn''t very hard. You just need to start trying. See what I have suggested and start implementing it. The debugger will come handy in doing this. Come back when you have tried and have ''specific'' issue.

Good luck!

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