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WebService Error: Unable to load endpoint configuration portion of contract for XXX because multiple endpoint configurations for the contract were found

WebService Error: Unable to load endpoint configuration portion of contract for XXX because multiple endpoint configurations for the contract were foundWebService Error: Unable to load endpoint configuration portion of contract for XXX because multiple endpoint configurations for the contract were foundThis issue occurs after you update the WebService server.This error occurs at run time when the service re

Undefined symbols for Architecture xxx, summary of errors

Simply put, Undefined symbols is basically equivalent to Java classnotfoundexception, the most common reasons for this are:1, build when no Add the frameworkBuild, compile stage no problem, but link error undefined symbols for architecture xxx (here XXX may be armv7s,armv7 or arm64, depending on the configuration, will say later). The workaround is to add xxx.framework to the build phases link Binary with l

Can load a file or assembly "XXX" or one of its dependencies, the system cannot find the specified file

Can load a file or assembly "XXX" or one of its dependencies, the system cannot find the specified filehttp://blog.csdn.net/pplcheer/article/details/7796211There are always problems in doing projects, and this is the problem today. I used the EXT control in my project and when I added a refresh button, I reported a "failure to load the file or assembly" Coolite.Ext.Web "or one of its dependencies", and then I re-found my Coolite installation file and

About playmaker play animation appears warning the animationclip ' xxx ' used by the Animati ...

Reprinted from the network: This prompt appears: the animationclip ' xxx ' used by the Animation component ' xxx ' must is marked as Legacy. Baidu, said to be in the rig panel, set Animation type for legacy, but my animation is not the import model of animation, no rig related parameters can be set later found in fact is the problem of animation type, click your animation section, Properties are as followsA

No module named "XXX" solution

When executing the. py file with Jenkins Today, always prompt modulenotfounderror:no module named ' XXX ', baffled by its solution. But in the Pycharm is able to execute successfully, think and then run the. py file in the terminal, still prompt modulenotfounderror:no module named ' XXX '. It was later found that the package was not found when executing the. py file on the command line because we did not sa

Table 'xxx' is marked as crashed and shoshould be retried red

An error "Table 'xxx' is marked as crashed and shocould be retried red" is found. Quickly search the Internet. It was easy to modify this serious error:1. Go to phpMyAdmin to manage MySQL2. Select your database on the left.3. Check the 'xxx' table in the error message on the right.4. Scroll down the screen to the bottom. There is a drop-down menu (with selected :) and select "repair table" Table '

Xxx is not in the sudoers file. This incident will be reported. solution, sudoersfile. this

Xxx is not in the sudoers file. This incident will be reported. solution, sudoersfile. this 1. Switch to the root user The su command switches between common users and root users, and the password set during installation must be entered;2. Add the write permission for the sudo file. The command is:Chmod u + w/etc/sudoers3. Edit the sudoers fileVi/etc/sudoersFind this line of root ALL = (ALL) ALL, add xxx AL

Mysql on linux reports Table & #39; xxx & #39; doesn & #39; t exist error solution, mysqlexist

Mysql on linux reports Table 'xxx' doesn' t exist error solution, mysqlexistMysql on linux reports Table 'xxx' doesn' t exist error SolutionProblem: The program reports the error of Table 'xxx' doesn' t exist, but the database shows that the Table already exists and the letters are not spelled correctly.Cause: mysql in linux is case sensitive by default.Solution:

CentOS MySQL error: Can't connect to MySQL server on 'xxx' (13)

CentOS MySQL error: Can't connect to MySQL server on 'xxx' (13)Environment: System Version: CentOS release 6.5 (Final) Database Version: 5.1.73 Problem description: Use the client to remotely log on to Mysql on the CentOS 6.5 server. The error "Can't connect to MySQL server on 'xxx' (13)" is returned) Solution: Generally, you can perform the following settings:1. log on to the MySQL console.# Mysql-u root-p

Cs0433: The type "XXXXX" exists in both XXX. dll and yyy. dll.

Cs0433: The type "XXXXX" exists in both XXX. dll and yyy. dll. In use. NET programming, we generally do not care too much about the inherits = "XXX" in @ page. It is generated by vs itself and can be used directly, this poses a major risk when a website is published ,. net Framework will use this method to create the page. If you use the same inherits value in multiple different pages, multiple methods w

The ole db access interface "sqlncli" of "XXX" on the linked server returns a message "Logon timeout expired"

The ole db access interface "sqlncli" of the link server "XXX" returns the message "Logon timeout expired ". The ole db access interface "sqlncli" of the linked server "XXX" returns a message "unable to complete the logon process due to delay in the pre-Logon response ". Message 258, level 16, state 1, 0th rows Shared Memory provision Program: Time-out error [258]. The

ubuntu--"XXX is not in the sudoers file. This incident'll be reported "error resolution

Ubuntu under the ordinary user with sudo execution command times title The error, the solution is to add permissions to the user in the/etc/sudoers file. As follows:1. Switch to the root userThe 2./etc/sudoers file is read-only by default and is also for root, so you need to add the Write permission to the sudoers file first, the command is:chmod u+w/etc/sudoers3. Edit the Sudoers fileVi/etc/sudoersFind this line root all= (all) all, add the following to him

"Failed to find the type name or namespace name xxx ..." error while resolving Visual Studio writing C #

You have a wonderful problem, when Visual Studio writes C # with a hint: "Could not find the type name or namespace name XXX (whether a using directive or assembly reference is missing)" Check it out,1, neither I do not have this class,2, nor did I write "Using XXX;" To reference namespaces,3, more than I referenced the other. dll files, forget to add references,4. There is no spelling error in the end. It'

<ios> <xcode> Solve the problem of can ' t map file, errno=22 file ' xxx ' for Architecture arm64 (ARMV7)! __ios </xcode> </ios>

Today, when introducing a static framework library, the compiler encountered the "Can" T map file, errno=22 file ' xxx ' for Architecture arm64 (ARMV7) Problem! Cause I was in the main project, build setting-----"Other linker flags-force_load the framework Library There are several ways to do this online: 1, find the reference project's build setting----"Build active architecture only----" set to No I was no, so it didn't work out. 2 and finally

Error 21002:[SQL-DMO] user "xxx" already exists

Label:Original: Error 21002:[SQL-DMO] user "xxx" already existsReprinted from: http://topic.csdn.net/u/20120829/09/faf381b6-d4e9-4f97-a21f-d1adfd43fbe4.html?seed=1190913404r= 79566866#r_79566866MSSQL2000 when you migrate a database or restore a database, create a new user name or specify a database logon user error message: "Error 21002:[SQL-DMO" user "xxx" already exists "Cause:This is what we often call "

MySQL Copy database table mode Backup, after restore prompt table xxx ' doesn ' t exist

MySQL is very powerful, support direct copy database file Quick Backup, where is the database file?Open the MySQL configuration file My.iniand locate the datadir node, asDatadir= "D:/program files/mysql/mysql Server 5.1/data"Enter the above folder, you can see the new database folder in MySQL, each folder is named after the database name, you want to back up which database, the folder copy away. To restore the database, copy it to the data directory, it is so simple!However, after a copy of a da

Android Build path contains duplicate entry: ' src ' for Project XXX

When the project was imported, it encountered this problem: Android Build path contains duplicate entry: ' src ' for Project XXX,Workaround: Locate the project's Project.Properties file, Modify the TARGET=ANDROID-17, modify the following number, and change it to the corresponding number of the SDK we installedThe reason: The version of the SDK I installed does not match the version of the SDK for the project you are importing.Android Build path contai

Springmvc problem xxx is not mapped [from XX] or unknown Entity:com.spring.main.pojo.Person

Springmvc problem xxx is not mapped [from XX] or unknown Entity:com.spring.main.pojo.PersonIt took a day till the wee hours to find a place not configuredAt the beginning of the HQL statement is wrong, but found no, online search to search is said Pojo class name written in the table name, so that the low-level error may have, examined n times, found not, but why will still report the same error? This error was found when I used the Get method to get

swift:missing argument label ' xxx ' in the call

Http://www.cnblogs.com/breezemist/p/4326644.htmlToday in the use of Swift found that the write Func always required to write out the second parameter of the external variable name, do not understand, feel and the book said the function is not the same, checked, and finally found the reason: written in the class inside the function called method, is a special functoin, the system will automatically fill in the external variable name, see the following connection Http://stackoverflow.com/questions

Yii2-the layout is determined by views/page/xxx. php.

The idea is as follows: Sometimes I think the parameter in views is used to determine which layout should be used. How can this problem be solved? Frontendviewspagexxx. php $ this- amp; gt; layout #039; main #039; error amp; lt ;? Phpinclude ( quot; .. layoutsmain. php quot ;);? Amp; gt; what should I do when an error is reported ?... The idea is as follows: Sometimes I think the parameter in views is used to determine which layout to use. How can we do this? Frontend/views/page/

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