我从我的Linux shell连接到mysql。我时不时地运行一个太大的SELECT查询。打印啊打印啊，我已经知道这不是我的意思。我想停止查询。

按Ctrl+C(几次)完全杀死mysql，带我回到shell，所以我必须重新连接。

是否有可能在不杀死mysql本身的情况下停止查询?

我正在使用以下代码发送电子邮件。代码在我的本地机器上正常工作。但是在生产服务器上，我得到了错误消息

var fromAddress = new MailAddress("mymailid@gmail.com");
var fromPassword = "xxxxxx";
var toAddress = new MailAddress("yourmailid@yourdoamain.com");

string subject = "subject";
string body = "body";

System.Net.Mail.SmtpClient smtp = new System.Net.Mail.SmtpClient
{
    Host = "smtp.gmail.com",
    Port = 587,
    EnableSsl = true,
    DeliveryMethod = System.Net.Mail.SmtpDeliveryMethod.Network,
    UseDefaultCredentials = false,
    Credentials = new NetworkCredential(fromAddress.Address, fromPassword)       
};

using (var message = new MailMessage(fromAddress, toAddress)
{
    Subject = subject,
    Body = body
})

smtp.Send(message);

在我的Gmail A/c上，我从生产服务器运行代码后收到了以下电子邮件

Hi , Someone recently used your password to try to sign in to your Google Account mymailid@gmail.com. This person was using an application such as an email, client or mobile device. We prevented the sign-in attempt in case this was a hijacker trying to access your account. Please review the details of the sign-in attempt: Friday, 3 January 2014 13:56:08 o'clock UTC IP Address: xxx.xx.xx.xxx (abcd.net.) Location: Philadelphia PA, Philadelphia, PA, USA If you do not recognise this sign-in attempt, someone else might be trying to access your account. You should sign in to your account and reset your password immediately. Reset password If this was you and you are having trouble accessing your account, complete the troubleshooting steps listed at http://support.google.com/mail?p=client_login Yours sincerely, The Google Accounts team

我正在尝试为Ruby安装PostgreSQL的pg gem。

我发出了以下命令:

gem install pg

我使用RVM安装Ruby 1.9.2。

上面的命令显示了以下错误。

错误是:

Building native extensions.  This could take a while...

ERROR:  Error installing pg:

ERROR: Failed to build gem native extension.

/home/User/.rvm/rubies/ruby-1.9.2-preview3/bin/ruby extconf.rb

checking for pg_config... yes
checking for libpq-fe.h... yes
checking for libpq/libpq-fs.h... yes
checking for PQconnectdb() in -lpq... no
checking for PQconnectdb() in -llibpq... no
checking for PQconnectdb() in -lms/libpq... no
Can't find the PostgreSQL client library (libpq)

*** extconf.rb failed ***

Could not create Makefile due to some reason, probably lack of
necessary libraries and/or headers.  Check the mkmf.log file for more
details.  You may need configuration options.

Provided configuration options:
 --with-opt-dir
 --without-opt-dir
 --with-opt-include
 --without-opt-include=${opt-dir}/include
 --with-opt-lib
 --without-opt-lib=${opt-dir}/lib
 --with-make-prog
 --without-make-prog
 --srcdir=.
 --curdir
 --ruby=/home/User/.rvm/rubies/ruby-1.9.2-preview3/bin/ruby
 --with-pg
 --without-pg
 --with-pg-config
 --without-pg-config
 --with-pg-dir
 --without-pg-dir
 --with-pg-include
 --without-pg-include=${pg-dir}/include
 --with-pg-lib
 --without-pg-lib=${pg-dir}/lib
 --enable-static-build
 --disable-static-build
 --with-pqlib
 --without-pqlib
 --with-libpqlib
 --without-libpqlib
 --with-ms/libpqlib
 --without-ms/libpqlib

Gem files will remain installed in /home/user/.rvm/gems/ruby-1.9.2-preview3/gems/pg-0.9.0 for inspection.

Results logged to /home/user/.rvm/gems/ruby-1.9.2-preview3/gems/pg-0.9.0/ext/gem_make.out

我不知道是什么错误…

我正在调整基地在一次git拉之后，调整基地。我有一些文件合并冲突。我如何接受特定文件的“他们的”更改或“我的”更改?

$ git status
# Not currently on any branch.
# You are currently rebasing.
#   (fix conflicts and then run "git rebase --continue")
#   (use "git rebase --skip" to skip this patch)
#   (use "git rebase --abort" to check out the original branch)
#
# Changes to be committed:
#   (use "git reset HEAD <file>..." to unstage)
#
#       modified:  CorrectlyMergedFile
#
# Unmerged paths:
#   (use "git reset HEAD <file>..." to unstage)
#   (use "git add <file>..." to mark resolution)
#
#       both modified: FileWhereIWantToAcceptTheirChanges
#       both modified: FileWhereIWantToAcceptMyChanges

通常我只是打开文件或合并工具，手动接受所有“他们的”或“我的”更改。但是，我怀疑我错过了一个方便的git命令。

另外，请注意，只有当我看到哪些文件发生冲突或冲突可能是什么时，我才能为每个文件选择合并策略。

我以前很轻松地使用过工会;今天当我读到这篇文章并知道这个代码时，我很震惊

union ARGB
{
    uint32_t colour;

    struct componentsTag
    {
        uint8_t b;
        uint8_t g;
        uint8_t r;
        uint8_t a;
    } components;

} pixel;

pixel.colour = 0xff040201;  // ARGB::colour is the active member from now on

// somewhere down the line, without any edit to pixel

if(pixel.components.a)      // accessing the non-active member ARGB::components

实际上是未定义的行为，即从工会成员中读取除最近写的人以外的内容会导致未定义的行为。如果这不是联合的预期用途，那么什么才是?谁能详细解释一下吗?

更新:

我想事后澄清一些事情。

The answer to the question isn't the same for C and C++; my ignorant younger self tagged it as both C and C++. After scouring through C++11's standard I couldn't conclusively say that it calls out accessing/inspecting a non-active union member is undefined/unspecified/implementation-defined. All I could find was §9.5/1: If a standard-layout union contains several standard-layout structs that share a common initial sequence, and if an object of this standard-layout union type contains one of the standard-layout structs, it is permitted to inspect the common initial sequence of any of standard-layout struct members. §9.2/19: Two standard-layout structs share a common initial sequence if corresponding members have layout-compatible types and either neither member is a bit-field or both are bit-fields with the same width for a sequence of one or more initial members. While in C, (C99 TC3 - DR 283 onwards) it's legal to do so (thanks to Pascal Cuoq for bringing this up). However, attempting to do it can still lead to undefined behavior, if the value read happens to be invalid (so called "trap representation") for the type it is read through. Otherwise, the value read is implementation defined. C89/90 called this out under unspecified behavior (Annex J) and K&R's book says it's implementation defined. Quote from K&R: This is the purpose of a union - a single variable that can legitimately hold any of one of several types. [...] so long as the usage is consistent: the type retrieved must be the type most recently stored. It is the programmer's responsibility to keep track of which type is currently stored in a union; the results are implementation-dependent if something is stored as one type and extracted as another. Extract from Stroustrup's TC++PL (emphasis mine) Use of unions can be essential for compatness of data [...] sometimes misused for "type conversion".

最重要的是，这个问题(它的标题从我的提问开始就没有改变)是为了理解联合的目的而提出的，而不是关于标准允许什么。例如，使用继承来实现代码重用当然是c++标准允许的，但这并不是将继承引入c++语言特性的目的或初衷。这就是为什么安德烈的回答仍然被人们所接受的原因。

我试图使用python发送电子邮件(Gmail)，但我得到以下错误。

Traceback (most recent call last):  
File "emailSend.py", line 14, in <module>  
server.login(username,password)  
File "/usr/lib/python2.5/smtplib.py", line 554, in login  
raise SMTPException("SMTP AUTH extension not supported by server.")  
smtplib.SMTPException: SMTP AUTH extension not supported by server.

Python脚本如下所示。

import smtplib

fromaddr = 'user_me@gmail.com'
toaddrs  = 'user_you@gmail.com'
msg = 'Why,Oh why!'
username = 'user_me@gmail.com'
password = 'pwd'
server = smtplib.SMTP('smtp.gmail.com:587')
server.starttls()
server.login(username,password)
server.sendmail(fromaddr, toaddrs, msg)
server.quit()

假设我想杀死所有包含amarok这个词的进程。我可以打印出我想要执行的命令。但是我如何让shell执行它们呢。ie。

ps aux | grep -ie amarok | awk '{print "kill -9 " $2}'
Output:
kill -9 3052
kill -9 3071
kill -9 3076
kill -9 3077
kill -9 3079
kill -9 3080
kill -9 3082
kill -9 3083
kill -9 3084
kill -9 3085
kill -9 3086
kill -9 3087
kill -9 3088
kill -9 3089
kill -9 4031

我做了一件蠢事，结果引起了冲突。我解决了冲突，现在一切都好了(我也使用了mergetool)。

当我用git commit file.php -m "message"提交解析文件时，我得到了错误:

fatal: cannot do a partial commit during a merge.

我以前遇到过同样的问题，在commit中使用-a工作得很好。我认为这不是完美的方式，因为我不想提交所有的更改。我想用单独的注释单独提交文件。我该怎么做呢?为什么git不允许用户在合并后单独提交文件?对于这个问题，我找不到满意的答案。

在网上搜索了一段时间后，我找不到任何像这样的地址:

宾夕法尼亚大街东南1200号 华盛顿，哥伦比亚特区，20003年

并将其转换为可点击的链接:

http://maps.google.com/maps?f=q&source=s_q&hl=en&q=1200+Pennsylvania+Ave+SE,+Washington,+District+of+Columbia,+20003&sll=37.0625,-95.677068&sspn=44.118686,114.169922&ie=UTF8&cd=1&geocode=FT5MUQIdIDlp-w&split=0&ll=38.882147,-76.99017&spn=0.01064,0.027874&z=16&iwloc=A

jQuery或PHP是首选或者任何有用的信息。

我正在使用Automapper，我有以下场景: 类OrderModel有一个名为“ProductName”的属性，该属性不在数据库中。 所以当我尝试用:

Mapper.CreateMap<OrderModel, Orders>(); 

它会生成一个异常:

" Project.ViewModels.OrderModel上的以下1个属性没有被映射:

我已经在AutoMapper的Wiki上阅读了相反的情况(额外的属性是在目的地，而不是在实际上是我的情况下的源)

我如何避免自动程序使这个属性的映射?