asp.net中的messagebox面临异常 [英] messagebox in asp.net facing exception

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问题描述

System.Web.HttpContext.Current.Response.Write("<script>alert('Item Already Exists')</script>");





在asp.net c#我正在使用这个或showung消息框,但它给出了一个异常,即

对象引用没有设置为对象的实例



我还用过



ScriptManager.RegisterClientScriptBlock(this,this.GetType( ),警报,警报('你的消息');,真实);



但我认为它因为这个而需要控制事件.GetType()我没有使用控件这是一个简单的用户定义函数,我想在其中显示消息框,

请告诉我在哪里制作mistek。谢谢。



in asp.net c# i am using this or showung messagebox but its giving a exception that is
Object reference not set to an instance of an object.

I also used

ScriptManager.RegisterClientScriptBlock(this,this.GetType(), "alert", "alert('Your Message');", true);

but i think its requried control event because of this.GetType() and i am not using control for this it is a simple user defined function in which I want to show messagebox,
plz tell me where i did make mistek. thank you.

推荐答案

您没有显示带有对象引用未设置为对象实例消息的异常位置。



不用担心。这是检测和修复的最简单的案例之一。它只是意味着某些引用类型的某个成员/变量通过使用和它的实例(非静态)成员解除引用,这要求此成员/变量为非null,但实际上它似乎为null。只需在调试器下执行它,它将停止抛出异常的执行。在该行上设置一个断点,重新启动应用程序并再次到达这一点。评估下一行中涉及的所有引用,并查看哪一个为null,而不需要为null。解决这个问题之后,修复代码:要么确保将成员/变量正确初始化为非空引用,要么将其检查为null,如果为null,则执行其他操作。



另请参阅:想要在按钮点击时显示下一条记录。但是如果下一个记录功能的条件对象引用未设置为对象的实例则会出错。



有时候,你不能这样做在调试器下,由一个或另一个原因。一个非常讨厌的情况是,只有在调试信息不​​可用时构建软件时才会出现问题。在这种情况下,你必须使用更难的方式。首先,你需要确保你永远不会通过静默处理异常来阻止异常的传播(这是开发人员对自己的犯罪,但很常见)。您需要在每个线程的最顶层堆栈帧上捕获绝对所有异常。如果处理类型 System.Exception 的异常,则可以执行此操作。在处理程序中,您需要记录所有异常信息,尤其是 System.Exception.StackTrace

http://msdn.microsoft.com/en-us/library/system.exception.aspx

http://msdn.microsoft.com/en-us/library/ system.exception.stacktrace.aspx



堆栈跟踪只是一个字符串,显示从throw语句到处理程序的异常传播的完整路径。通过阅读,您总能找到目的。对于日志记录,使用类 System.Diagnostics.EventLog

http://msdn.microsoft.com/en-us/library/system.diagnostics.eventlog.aspx



祝你好运,

-SA
You did not show where the exception with the message "Object reference not set to an instance of an object" is thrown.

Not to worry. This is one of the very easiest cases to detect and fix. It simply means that some member/variable of some reference type is dereferenced by using and of its instance (non-static) members, which requires this member/variable to be non-null, but in fact it appears to be null. Simply execute it under debugger, it will stop the execution where the exception is thrown. Put a break point on that line, restart the application and come to this point again. Evaluate all references involved in next line and see which one is null while it needs to be not null. After you figure this out, fix the code: either make sure the member/variable is properly initialized to a non-null reference, or check it for null and, in case of null, do something else.

Please see also: want to display next record on button click. but got an error in if condition of next record function "object reference not set to an instance of an object".

Sometimes, you cannot do it under debugger, by one or another reason. One really nasty case is when the problem is only manifested if software is built when debug information is not available. In this case, you have to use the harder way. First, you need to make sure that you never block propagation of exceptions by handling them silently (this is a crime of developers against themselves, yet very usual). The you need to catch absolutely all exceptions on the very top stack frame of each thread. You can do it if you handle the exceptions of the type System.Exception. In the handler, you need to log all the exception information, especially the System.Exception.StackTrace:
http://msdn.microsoft.com/en-us/library/system.exception.aspx,
http://msdn.microsoft.com/en-us/library/system.exception.stacktrace.aspx.

The stack trace is just a string showing the full path of exception propagation from the throw statement to the handler. By reading it, you can always find ends. For logging, it's the best (in most cases) to use the class System.Diagnostics.EventLog:
http://msdn.microsoft.com/en-us/library/system.diagnostics.eventlog.aspx.

Good luck,
—SA


引用:

是在Web方法中

public static void QueryFunModbusRTU()

yes inside a Web Method
public static void QueryFunModbusRTU()

这就是它即将到来的原因。



做一件事。不要这样做以显示警告 WebMethods 旨在采取一些参数,并在某些操作后返回一些东西。所以,在完成所有操作之后,只需返回一个值或其他内容。然后在客户端 Ajax 成功,读取该值并显示 alert 。这是一个理念和完美的方式。

That is the reason why it is coming.

Do one thing. Don't do like this to show an alert. WebMethods are designed to take some paramter and return you something after some operation. So, after all your operation, just return a value or something. Then in client side Ajax success, read that value and show an alert. This is idea and perfect way.


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