如何为学生创建自动生成ID？ [英] How to creat auto generate id for student?

问题描述

我想要我的学生ID示例-140200001

听到14是02年是修补编码它是学校ID并生成ID来自00001到2,3，.....那样但是'0000'也是固定数字数字

note（在这个ID-changeble数字是yy它在一年中的变化和123 ...作为序列号明智而来。剪切'020000'是修复代码不改变这个id）

plz帮我怎么生成....？



i need - ist student id-140200001

然后下一个学生ID生成-140200002

然后下一个id generet-140200003

然后140200004 ......就像那个

解决方案

你不能简单地按照你的算法吗？您只需要永久存储（例如通过序列化）最后生成的后缀。

检查方法只需传递参数year，bla bla并获取STUDENTID

 string GetObjectID（int sequence，string yr =14，string d =02）
 {
 string getZero =; 
 for（int i = 0; i< 5  -  sequence.ToString（）。Length; i ++）
 {
 getZero + =0; 
} 
返回yr + d + getZero + sequence; 
} 
 

这将为

返回1：140200001

10：140200010

适用于99：140200099

.....

for 999：140200999

i want to my student id example-140200001
Hear 14 is year 02 is fixcode it is a school id and generate id comming 00001 to 2,3,.....like that but '0000' is also fix digit number
note( In This ID-changeble number is yy its change in year wise and 123... coming as serial number wise.hear '020000' is fix code its not change in this id )
plz help me how to generat....?

i need - ist student id-140200001
then next student id generate-140200002
then next id generet-140200003
then 140200004...... like that

解决方案

Cannot you simply follow your algorithm? You just need to permanently store (for instance via serialization) the last generated suffix.

Check a method just pass the parameter year, bla bla and get STUDENTID

string GetObjectID(int sequence, string yr = "14", string d = "02")
  {
      string getZero = "";
      for (int i = 0; i < 5 - sequence.ToString().Length; i++)
      {
          getZero += "0";
      }
      return yr + d + getZero + sequence;
  }

This will return like
for 1 : 140200001
For 10: 140200010
For 99: 140200099
.....
for 999: 140200999

这篇关于如何为学生创建自动生成ID？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！