如何将数据插入到字符串中 [英] How to insert data into string
问题描述
我有一个数据集 ds ,它被转换为字符串。
ds可能有20列,有时是30列。
现在我想插入一些文本字符串,如果它有更少的列(我不知道有多少列可能存在)
所以任何人都可以告诉我如何在最后将文本插入字符串。
**我不知道字符串的长度。
字符串例如:
I have a dataset ds which is converted to string.
ds may have 20 columns and sometimes 30 columns.
Now I want to insert some text string if it has less columns(which i don't know how many columns mayb exist)
So can anyone tell me how to insert text into string at the end.
**I don't know length the string.
string ex:
2034027436594110044004745638140274381004390000000 000000000000000015000 0000000gdfgdf00000000000 000000000000000000000 000000160510 dfgdfgdfgbdfghdfh8706216710500000000000000000 000000000 0000000000000 00000000000000000000000000000000000000000000 fghdfhdtgh 16051000000000000000000000000000000000000
<无线电通信/>
我不知道这有多长ytime。
但是我有什么方法可以在不知道长度的情况下在最后添加文字
我尝试了什么:
任何人都可以帮我解决这个问题。
我被困在这里。
I don't know how much length this will be everytime.
But is there any way I can add text at the end without knowing length
What I have tried:
Anyone please help me solve this.
I am stuck here.
推荐答案
**我不知道字符串的长度。
**I don't know length the string.
参见 String.Length Property(System) [ ^ ]
学习c#字符串函数!
将 ds
转换为strong并在之后添加列可能不是一个好主意,但为此,您需要告诉 DS
以及要添加的列。
See String.Length Property (System)[^]
Learn the c# strings functions!
Converting ds
to strong and adding columns after may not be a good idea, but for that, you need to tell what is DS
and what columns you want to add.
基本上,你不必知道字符串的大小来附加另一个字符串。
有很多方法可以做到这一点:
Basically, you do not have to know the size of a string to append it another string.
There are a bunch of ways to do that:
string a = "a";
string b = "b";
StringBuilder sb = new StringBuilder(a);
string concat1 = a + b;
string concat2 = string.Concat(a, b);
string concat3 = string.Format("{0}{1}", a, b);
string concat4 =
{A} {b}跨度>;
string concat5 = sb.Append(b).ToString();
"{a}{b}"; string concat5 = sb.Append(b).ToString();
全部 concat
变量将保持相同的值,这将是字符串变量a和b的串联。
All concat
variables will hold the same value, which will be the concatenation of both string variables a and b.
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