Tag Archives: net

Solution to msxml3.dll error ‘800c0008’ the system cannot locate the resource specified

msxml3.dll ?펳 ‘800 c0005’

The system always locate The resource specified.

/plugins. Asp??А 165

this is usually used when collecting information or statically updating a web page with XMLHTTP, and I’ve used similar errors before

set objXmlHttp = server.createobject (” microsoft.xmlhttp “) ‘is generally used in versions below msxml2.6

‘ set objXmlHttp = server.createobject (” msxml2.xmlhttp “) ‘supported by msxml3.dll+

is used instead of
, but now every error is reported.

has searched all the technical forums and search engines I know on the Internet, and there may be a variety of situations that could cause such a problem. Here is a summary:

(a) that is the most basic, your server does not support XMLHTTP or component version low, you need to download the corresponding component. This is not required by the average server, and XMLHTTP is supported in most cases, so this is a rare case.

(2) msxml3.dll file may be corrupted or the file is not authorized enough (this case is less likely), if the file is corrupted, you can download the corresponding system environment’s msxml3.dll file to replace (2003 system seems to be under c:// Windows /system32), need to close the corresponding service such as iis and so on to overwrite.

(three) fireproof or anti-virus software has disabled the corresponding port, the specific how to set it depends on different software, you can try to close the firewall or anti-virus software and then test to determine whether it is the problem.

(4) try re-registering the component. Start & gt; > Run: RegsvR32 MSxml3.DLL

(5) DNS resolution error, check your network Settings or correct DNS access

(6) check your IP filter Settings (this is not very clear)

(7) resource failure of the other party. For example, when collecting some information, the other party does not have the error that leads to program termination, which may also be the case. This error occurs when the XMLHTTP component’s Open method is called, followed by the Send method. An 8000005 error occurs when the URL parameter of the open method is not accessible. And if such an error occurs, the application terminates and cannot continue. Most programs are written like this:

Function functionName(pararm …)

Dim Http

= Set Http Server. CreateObject (” MSXML2. XMLHTTP. 4.0 “)

With Http

.open “GET”,HttpUrl,False

.Send

End With

If Http.Readystate< > 4 then

Set Http=Nothing

……

Exit function

End if

End Function

most programs use the XMLHTTP Readystate property to determine the return status from the server. This is not always the case, and many times using the ReadyState property does not actually detect errors in the program flow. When an error is encountered, the program will still be terminated. In fact, modify the above code, it is possible to achieve the process of skipping the execution of the program encountered errors, so that the program continues to run. The modified code is as follows:

Function functionName(pararm …)

Dim Http

= Set Http Server. CreateObject (” MSXML2. XMLHTTP. 4.0 “)

With Http

.open “GET”,HttpUrl,False

.Send

End With

On Error Resume Next

If Http.Status< > 200 then

Set Http=Nothing

……

Exit function

End if

End Function
When the
Send method produces an error, the ReadyState value may be 4, but the Status value must not be 200. Well, I’ve tracked ReadyState and Status’s worth to previous results many times. There may be mistakes. I haven’t found them so far.

hope the above program solution can help you!! If you have a better solution, please let me know.

, which I illustrate with msxml2.xmlhttp.4.0, is also suitable for other versions of the XMLHTTP component. To check which versions of the XMLHTTP component you have installed on your system, look under HKEY_CLASSES_ROOT in the registry.

(8) go to the server and set your IE security options (try setting the level to medium or low). Just add the domain name you want in the trusted website.

(9) I have tried all of the above methods after looking at them, but none of them worked out. So I went to Microsoft and downloaded MSXML4.0 Service Pack 2 (Microsoft XML Core Services) and Hotfix for MSXML4.0 Service Pack 2 – KB832414 – Simplified Chinese (KB832414_MSxml4.0_x86.exe) and installed it. The problem was finally solved!!

(10) if your case, I can solve the above methods, it is back, you continue to look for other possible to online, if you don’t want to continue to find you reinstall your iis or simply reshipment system (do this before you can try to restart the system to see, may probably as if lucky can use again

Oci-22053: overflow error problem

When connecting Oracle in.NET and using dataAdapter fill, the exception of OCI-22053: overflow error occurs, because the data type precision of.NET is smaller than that of Oracle, such as
Select sysdate- ISSUetime as Timediff from sometable returns an excessive precision
Select TRUNc (sysdate-issuetime,2) as Timediff from sometable.
There are other ways to do this online:
http://excel.cnblogs.com/archive/2005/11/14/276202.html using the new ODP.NET da. SafeMapping. Add (field name 1, typeof (System. String)); Method converts all fields to String. But I went to the Oracle website and found that the 9207 version of ODAC is 80M… I’m too lazy to download.

 
The Oracle numeric data type can store up to 38 bytes of precision.
the Oracle value
may become too large when converting the Oracle value to the common language runtime data type. This causes an Oracle OCI-22053 overflow error.
The solution is to use the round function.
How to use the Oracle Round function (rounded)
Description: Returns a value that is rounded to the specified number of decimal digits.
SELECT ROUND( number, [ decimal_places ] ) FROM DUAL
Parameters:
Number: The number you want to process
Decimal_places: Round up and take a few decimal places (default is 0)
Sample :
Select round (123.456) from dual; The back 123
Select round(123.456, 0) from Dual; The back 123
Select round(123.456, 1) from Dual; The back 123.5
Select round(123.456, 2) from Dual; The back 123.46
Select round(123.456, 3) from Dual; The back 123.456
Select round(-123.456, 2) from dual; The back to 123.46
 
From:
http://blog.csdn.net/xeonol/article/details/726133