小编ris*_*p89的帖子

我有一个python字典,其定义不适合单行.谁能告诉我python词典的指南.我目前有这个对我的眼睛不好看.

initialstate =  {
                    'state':grid,
                    'f':find_manhattan_distance(grid,goal),
                    'g':0,
                    'h':find_manhattan_distance(grid,goal),
                    'ancestor': None
                }

我正在使用scrapy来抓取域下的所有网页.

我见过这个问题.但是没有解决方案.我的问题似乎与此类似.我的crawl命令输出如下所示:

scrapy crawl sjsu2012-02-22 19:41:35-0800 [scrapy] INFO: Scrapy 0.14.1 started (bot: sjsucrawler)
2012-02-22 19:41:35-0800 [scrapy] DEBUG: Enabled extensions: LogStats, TelnetConsole, CloseSpider, WebService, CoreStats, MemoryUsage, SpiderState
2012-02-22 19:41:35-0800 [scrapy] DEBUG: Enabled downloader middlewares: HttpAuthMiddleware, DownloadTimeoutMiddleware, UserAgentMiddleware, RetryMiddleware, DefaultHeadersMiddleware, RedirectMiddleware, CookiesMiddleware, HttpCompressionMiddleware, ChunkedTransferMiddleware, DownloaderStats
2012-02-22 19:41:35-0800 [scrapy] DEBUG: Enabled spider middlewares: HttpErrorMiddleware, OffsiteMiddleware, RefererMiddleware, UrlLengthMiddleware, DepthMiddleware
2012-02-22 19:41:35-0800 [scrapy] DEBUG: Enabled item pipelines: 
2012-02-22 19:41:35-0800 [sjsu] INFO: Spider opened
2012-02-22 19:41:35-0800 [sjsu] INFO: Crawled 0 pages (at 0 pages/min), …

这是我的代码.我的parse_item方法没有被调用.

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector

class SjsuSpider(CrawlSpider):

    name = 'sjsu'
    allowed_domains = ['sjsu.edu']
    start_urls = ['http://cs.sjsu.edu/']
    # allow=() is used to match all links
    rules = [Rule(SgmlLinkExtractor(allow=()), follow=True),
             Rule(SgmlLinkExtractor(allow=()), callback='parse_item')]

    def parse_item(self, response):
        print "some message"
        open("sjsupages", 'a').write(response.body)

如何使Scala对象具有线程安全性.

class Stack {
    case class Node(value: Int, var next: Node)

    private var head: Node = null
    private var sz = 0

    def push(newValue: Int) {
        head = Node(newValue, head)
        sz += 1
    }

    def pop() = {
        val oldNode = head
        head = oldNode.next
        oldNode.next = null
        sz -= 1
    }

    def size = sz    //I am accessing sz from two threads
}

这个类显然不是线程安全的.我想让它成为线程安全的.

提前致谢,

生命值

我第一次使用http://typesafe.com/stack/,我创建了简单的akka​​项目.我的scala版本是2.9.2我收到以下错误.

[info] Done updating.
[info] Compiling 1 Scala source to /Users/hrishikeshparanjape/git-public/web-service/target/scala-2.9.2/classes...
[info] 'compiler-interface' not yet compiled for Scala 2.9.2. Compiling...
sbt appears to be exiting abnormally.
  The log file for this session is at /var/folders/26/hqgjyf0j7192hmjdsz17f3v80000gn/T/sbt2587622650679130928.log
java.lang.OutOfMemoryError: PermGen space
    at java.util.concurrent.FutureTask$Sync.innerGet(FutureTask.java:222)
    at java.util.concurrent.FutureTask.get(FutureTask.java:83)
    at sbt.CompletionService$$anon$1.take(CompletionService.scala:29)
    at sbt.Execute.next$1(Execute.scala:74)
    at sbt.Execute.processAll(Execute.scala:77)
    at sbt.Execute.runKeep(Execute.scala:57)
    at sbt.EvaluateTask$.run$1(EvaluateTask.scala:109)
    at sbt.EvaluateTask$.runTask(EvaluateTask.scala:124)
    at sbt.Aggregation$$anonfun$7.apply(Aggregation.scala:87)
    at sbt.Aggregation$$anonfun$7.apply(Aggregation.scala:85)
    at sbt.EvaluateTask$.withStreams(EvaluateTask.scala:87)
    at sbt.Aggregation$.runTasks(Aggregation.scala:85)
    at sbt.Aggregation$$anonfun$applyDynamicTasks$1.apply(Aggregation.scala:141)
    at sbt.Aggregation$$anonfun$applyDynamicTasks$1.apply(Aggregation.scala:136)
    at sbt.Command$$anonfun$applyEffect$2$$anonfun$apply$3.apply(Command.scala:64)
    at sbt.Command$$anonfun$applyEffect$2$$anonfun$apply$3.apply(Command.scala:64)
    at sbt.Command$.process(Command.scala:92)
    at sbt.MainLoop$$anonfun$next$1$$anonfun$apply$1.apply(Main.scala:121)
    at sbt.MainLoop$$anonfun$next$1$$anonfun$apply$1.apply(Main.scala:121)
    at sbt.State$$anon$1.process(State.scala:154) …

我有以下实体:

@RooEntity
Class X {
    @NotNull
    private Locale locale;
}

是否可以在数据库中存储Locale对象的toString()表示,并且在检索时我仍然可以获取Locale对象？

我可以在这里使用@Basic注释吗？

有没有办法在方案/球拍中使用网络服务来托管列表

我有一个数组(树)像:

['Humidity\n', ['High', ['Outlook\n', ['Sunny', ['Temp\n', ['Hot', ['Wind\n', ['Weak', ['No']], ['Strong', ['No']]]], ['Mild', ['No']]]], ['Overcast', ['Temp\n', ['Hot', ['Yes']], ['Mild', ['Yes']]]], ['Rain', ['Temp\n', ['Mild', ['Wind\n', ['Weak', ['Yes']], ['Strong', ['No']]]]]]]], ['Normal', ['Wind\n', ['Weak', ['Outlook\n', ['Rain', ['Yes']], ['Overcast', ['Temp\n', ['Cool', ['Yes']], ['Hot', ['Yes']]]], ['Sunny', ['Yes']]]], ['Strong', ['Temp\n', ['Cool', ['No']], ['Mild', ['Outlook\n', ['Rain', ['Yes']], ['Sunny', ['Yes']]]]]]]]]

我想把它打印成树.这是我的尝试

def printTree(treeToPrint, tab):
    print tab + treeToPrint[0]
    newTab = tab + "--"
    nextTree = treeToPrint[1:]
    for tree in nextTree:
        print newTab + tree[0]
        printTree(tree[1], newTab + "--") …

我的EmailHandler类如下：

Class EmailHandler {
    ...
    @Autowired
        protected Configuration config;
    }
    ...
    protected Template getTemplate(String templateName, Locale locale) throws IOException, TemplateException {
        return config.getTemplate(templateName, locale);
    }
    ...
}

在我的applicationcontext.xml中，我有

<bean id="freemarkerConfiguration" class="org.springframework.ui.freemarker.FreeMarkerConfigurationFactoryBean">
    <property name="templateLoaderPath" value="classpath:/templates/email"/>
</bean>

我的模板目录结构如下：

--src
----main
------resources
--------templates
----------email
------------template1.ftl
------------template2.ftl
------------template3.ftl
------------template4.ftl
------------multilanguage
--------------fr
----------------template1.ftl

目前，getTemplate始终使用（string，Locale.US）进行调用。但是，将来，我希望能够使用（string，Locale.FR）调用getTemplate。

以下是我的问题：1.如何更改目录结构以指定法语模板。2. config.getTemplate（templateName，locale）是什么？到底是做什么的？该方法如何在模板目录中找到Locale.US的模板？3.我想从email / multilanguage / fr目录加载我的法语模板。我怎么做？

谢谢，

里士

我可以如下指定log4j格式化程序来打印spring mvc应用程序中的当前线程id

log4j.appender.FILE.layout.ConversionPattern=%d{ISO8601} %p %t %c - %m%n

如何在我的spring应用程序中以编程方式获取当前线程ID.更确切地说,我想在拦截控制器方法的方面获取当前线程id.我的方面如下:

@Configurable
@Aspect
public class TimingAspect {

    @Autowired
    private HttpServletRequest httpServletRequest;

    //Generic performance logger for any mothod
    private Object logPerfomanceInfo(ProceedingJoinPoint joinPoint) {
       // do something with thread id
       // do something with httprequest
        ...
    }

    // Performance info for API calls
    @Around("execution(* package.controller.*.*(..))")
    public Object logAroundApis(ProceedingJoinPoint joinPoint) throws Throwable {
        return logPerfomanceInfo(joinPoint);
    }
}