method

Page not found (404) Request Method: GET Request URL: http://127.0.0.1:8000/

匿名 (未验证) 提交于 2019-12-03 02:38:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I was following the django documentation and making a simple poll app. I have come across the following error : Using the URLconf defined in mysite.urls, Django tried these URL patterns, in this order: ^polls/ ^admin/ The current URL, , didn't match any of these." settings.py ROOT_URLCONF = 'mysite.urls' mysite/mysite/urls.py from django.conf.urls import include,url from django.contrib import admin urlpatterns = [ url(r'^polls/',include('polls.urls')), url(r'^admin/', admin.site.urls),] mysite/polls/urls.py from django.conf.urls import url

Smail代码调试LogUtil

旧时模样 提交于 2019-12-03 02:37:27
.class public Lutil/SLog; .super Ljava/lang/Object; .source "SLog.java" # static fields .field private static final DEFAULT_TAG:Ljava/lang/String; = "cylee" .field private static mTag:Ljava/lang/String; # direct methods .method static constructor <clinit>()V .locals 1 .prologue .line 10 const-string v0, "cylee" sput-object v0, Lutil/SLog;->mTag:Ljava/lang/String; return-void .end method .method public constructor <init>()V .locals 0 .prologue .line 8 invoke-direct {p0}, Ljava/lang/Object;-><init>()V return-void .end method .method public static log(B)V .locals 2 .param p0, "i" # B .prologue

Change one character in a string?

匿名 (未验证) 提交于 2019-12-03 02:28:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: What is the easiest way in Python to replace a character in a string like: text = "abcdefg" ; text [ 1 ] = "Z" ; ^ 回答1: Don't modify strings. Work with them as lists; turn them into strings only when needed. >>> s = list ( "Hello zorld" ) >>> s [ 'H' , 'e' , 'l' , 'l' , 'o' , ' ' , 'z' , 'o' , 'r' , 'l' , 'd' ] >>> s [ 6 ] = 'W' >>> s [ 'H' , 'e' , 'l' , 'l' , 'o' , ' ' , 'W' , 'o' , 'r' , 'l' , 'd' ] >>> "" . join ( s ) 'Hello World' Python strings are immutable (i.e. they can't be modified). There are a lot of reasons for this.

Java反射机制调用私有方法

ε祈祈猫儿з 提交于 2019-12-03 02:13:43
1、获取目标类: 每个类都有一个class属性,通过实体类的class属性获取: Class clazz = Person.class 通过对象获取。 Person p1 = new Person("sqc") Class clazz = p1.getClass() Class clazz = Class.forName("org.mooctest.Person") 2、获取目标方法并执行 Method method = clazz.getDeclaredMethod("sortByValue", Room []. class ); // 函数名,参数类型 method.setAccessible( true ); method.invoke(clazz.newInstance(), (Object)rooms); //参数二:方法传入的具体值 来源: https://www.cnblogs.com/sqchao/p/11770870.html

static initialization in interface

匿名 (未验证) 提交于 2019-12-03 02:05:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: When I tried to write something like this: public interface MyInterface { static { System.out.println("Hello!"); } } the compiler could not compile it. But when I wrote something like this: interface MyInterface { Integer iconst = Integer.valueOf(1); } and decompiled it, I saw static initialization: public interface MyInterface{ public static final java.lang.Integer i; static {}; Code: 0: iconst_1 1: invokestatic #1; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer; 4: putstatic #2; //Field i:Ljava/lang/Integer; 7: return } Could you

Use of ESAPI getValidInput method

匿名 (未验证) 提交于 2019-12-03 01:37:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I was not able to use the of method present under ESAPI class' java.lang.String getValidInput(java.lang.String context, java.lang.String input, java.lang.String type, int maxLength, boolean allowNull) throws ValidationException, IntrusionException Parameters: type - The regular expression name that maps to the actual regular expression from "ESAPI.properties". How to pass parameter type from ESAPI.properties file? Any example to use properties file value from which I can refer? 回答1: Here's an example call where I'm validating the "to"

How do I get the shipping method the user has chosen during checkout?

匿名 (未验证) 提交于 2019-12-03 01:29:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I want to get the name of the shipping method the user has chosen during checkout. Does anyone know how to retrieve that info? This will get it to some extent but it is cached: Mage::getSingleton('checkout/session')->getQuote()->getShippingAddress()->getShippingDescription(); When I am on the onestep checkout and I go back to the shipping tab and change the shipping, it is still holding the old shipping method. I need to figure out how to get the current one. 回答1: Foreword Constructed from Magento app/code/core/Mage/Checkout/Block/Onepage

Expected type: java.lang.Integer, actual value: org.hibernate.id.IdentifierGeneratorHelper$2

匿名 (未验证) 提交于 2019-12-03 01:00:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: i have this error in my code when i m trying to call save method this method is woking fine while i m on JBoss server , now i move on tomcat server 7 ERROR: HHH000123: IllegalArgumentException in class: com.sb.core.beans.CompanyStudies,setter method of property: studyId Aug 5, 2013 9:24:52 AM org.hibernate.property.BasicPropertyAccessor$BasicSetter set ERROR: HHH000091: Expected type: java.lang.Integer, actual value: org.hibernate.id.IdentifierGeneratorHelper$2 org.hibernate.PropertyAccessException: IllegalArgumentException occurred while

checkstyle Method is not designed for extension - needs to be abstract, final or empty [duplicate]

匿名 (未验证) 提交于 2019-12-03 00:59:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: This question already has an answer here: Checkstyle “Method Not Designed For Extension” error being incorrectly issued? 3 answers I have an interface public interface LdapConnectionFactory { LdapConnectionWrapper getConnectionWrapper () throws LDAPException ; } and the implementation class LdapConnectionFactoryImpl which implements the getConnectionWrapper() method. public class LdapConnectionFactoryImpl implements LdapConnectionFactory { ... public LdapConnectionWrapper getConnectionWrapper () throws LDAPException { ... } } When

Why show error IllegalStateException when setting MediaRecorder?

匿名 (未验证) 提交于 2019-12-03 00:56:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: My code setting MediaRecorder, it show error at row set Quality mMediaRecorder = new MediaRecorder(); // Step 1: Unlock and set camera to MediaRecorder mCamera.stopPreview(); mCamera.unlock(); mMediaRecorder.setCamera(mCamera); mMediaRecorder.setAudioSource(MediaRecorder.AudioSource.CAMCORDER); mMediaRecorder.setVideoSource(MediaRecorder.VideoSource.DEFAULT); mMediaRecorder.setOutputFormat(MediaRecorder.OutputFormat.MPEG_4); mMediaRecorder.setProfile(CamcorderProfile .get(CamcorderProfile.QUALITY_HIGH)); mMediaRecorder.setAudioEncoder