同时声明并分配多个字符串变量

Question

我正在声明一些空字符串，因此以后不会抛出错误。

我读过这是正确的方法：

string Camnr = Klantnr = Ordernr = Bonnr = Volgnr = Omschrijving = Startdatum = Bonprioriteit = Matsoort = Dikte = Draaibaarheid = Draaiomschrijving = Orderleverdatum = Regeltaakkode = Gebruiksvoorkeur = Regelcamprog = Regeltijd = Orderrelease = "";

但这不起作用。我收到此错误：Klantnr does not exist in the current context。

我做错了什么？

Answer 1

你可以这样做：

string Camnr, Klantnr, Ordernr, Bonnr, Volgnr;// and so on.
Camnr = Klantnr = Ordernr = Bonnr = Volgnr = string.Empty;

首先，您必须定义变量，然后才能使用它们。

Answer 2

你可以这样做：

string Camnr = "", Klantnr = "", ... // or String.Empty

或者您可以先将它们全部声明，然后在下一行中使用您的方式。

Answer 3

我称之为连接声明的一个例子：

string Camnr = "",
        Klantnr = "",
        Ordernr = "",
        Bonnr = "",
        Volgnr = "",
        Omschrijving = "",
        Startdatum = "",
        Bonprioriteit = "",
        Matsoort = "",
        Dikte = "",
        Draaibaarheid = "",
        Draaiomschrijving = "",
        Orderleverdatum = "",
        Regeltaakkode = "",
        Gebruiksvoorkeur = "",
        Regelcamprog = "",
        Regeltijd = "",
        Orderrelease = "";

只需2美分，希望它可以帮到某个人。

Answer 4

尝试：

 string Camnr, Klantnr, Ordernr, Bonnr, Volgnr, Omschrijving;
 Camnr = Klantnr = Ordernr = Bonnr = Volgnr = Omschrijving = string.Empty;

Answer 5

尝试

string     Camnr , Klantnr , Ordernr , Bonnr , Volgnr , Omschrijving , Startdatum ,    Bonprioriteit , Matsoort , Dikte , Draaibaarheid , Draaiomschrijving , Orderleverdatum , Regeltaakkode , Gebruiksvoorkeur , Regelcamprog , Regeltijd , Orderrelease ;

然后

Camnr = Klantnr = Ordernr = Bonnr = Volgnr = Omschrijving = Startdatum = Bonprioriteit = Matsoort = Dikte = Draaibaarheid = Draaiomschrijving = Orderleverdatum = Regeltaakkode = Gebruiksvoorkeur = Regelcamprog = Regeltijd = Orderrelease = "";

Answer 6

string Camnr , Klantnr , Ordernr , Bonnr , Volgnr , Omschrijving , Startdatum , Bonprioriteit , Matsoort , Dikte , Draaibaarheid , Draaiomschrijving , Orderleverdatum , Regeltaakkode , Gebruiksvoorkeur , Regelcamprog , Regeltijd , Orderrelease;
Camnr = Klantnr = Ordernr = Bonnr = Volgnr = Omschrijving = Startdatum = Bonprioriteit = Matsoort = Dikte = Draaibaarheid = Draaiomschrijving = Orderleverdatum = Regeltaakkode = Gebruiksvoorkeur = Regelcamprog = Regeltijd = Orderrelease = string.Empty;

Answer 7

现在可以通过以下方式声明和初始化多个变量：

var (anInt, aFloat, aBoolean, aChar, aString, anArray, aRecordType, anObjectType) = 
    (1, 2.14, true, 'a', "C# is awesome!", new[] { "Asia", "Europe" } , new Country { Name = "India"}, new City { Name = "Kolkata"} ); 
Console.WriteLine(anInt);
Console.WriteLine(aFloat);
Console.WriteLine(aBoolean);
Console.WriteLine(aChar);
Console.WriteLine(aString);
Array.ForEach(anArray, Console.WriteLine);
Console.WriteLine(aRecordType.Name);
Console.WriteLine(anObjectType.Name);

以下是所需自定义类型的定义：

internal record Country { internal string Name {get; set;}}
internal class City { internal string Name {get; set;}}

这已经在 .NET 5/C# 9 上测试过。

Answer 8

string a = "", b = a , c = a, d = a, e = a, f =a;

Answer 9

^{所有信息都在现有答案中，但是我个人希望提供一个简洁的摘要，因此请尝试一下；这些命令使用int变量来简化，但是它们类似地适用于任何类型，包括string。}

要声明多个变量和 ：

• 其中之一：初始化它们每个 ：

int i = 0, j = 1; // declare and initialize each; `var` is NOT supported as of C# 8.0

• 或：将它们全部初始化为相同的值 ：

int i, j;    // *declare* first (`var` is NOT supported)
i = j = 42;  // then *initialize* 

// Single-statement alternative that is perhaps visually less obvious:
// Initialize the first variable with the desired value, then use 
// the first variable to initialize the remaining ones.
int i = 42, j = i, k = i;

---

什么是不起作用：

• 您不能在上述语句中使用var ，因为var仅适用于（a）具有初始化值的声明（类型来自该声明）可以推断出），并且（b）从C＃8.0开始，如果该声明是语句中仅 个声明（否则您将获得编译错误error CS0819: Implicitly-typed variables cannot have multiple declarators）。

• 仅在多声明语句中的 last 变量初始化仅 last变量 之后放置初始化值：

  int i, j = 1; // initializes *only* j

Answer 10

相当陈旧的问题，但是有人回去 这不像上面的其他答案一样紧凑，但使用Visual Studio多行选择快捷键[Alt + Shift +↑]（或其他方向）可以更容易阅读和输入

string Camnr = string.Empty;
string Klantnr = string.Empty;

在新行上输入所有变量名称。在它们前面多选一个类型＆＃34;字符串＆＃34;。在它们后面进行多选并键入＆＃34; = string.Empty;＆＃34;。

Answer 11

只是提醒：不允许在多个声明中使用隐式类型var。可能存在以下编译错误。

var Foo = 0, Bar = 0;

  

类型简单的变量不能有多个声明符

类似地，

var Foo, Bar;

  

类型明确的变量必须初始化