If you are interested in using java tutorials to educate yourself from home, you are probably interested in learning how to tell the difference between valuable resource materials, and those which are outdated or incorrect.  Learning to evaluate the quality of available tutorials is both an art and a science, and is best accomplished by paying attention to some of the individual components which make up a quality tutorial.  We will take a look at four of the most important:

·         Good organization – The tutorial needs to have a well-developed structure, which comprehensively details the content it will deliver, and is very easy for users to navigate.  A good organizational structure is indicative of a polished educational thought process, and is more important than you may think in the development of a good tutorial.

·         Valuable content – For good java tutorials, the content should be structured around accomplishing individual tasks.  It should do so by providing clear instruction to the reader, and be concise and to the point as well.  The delivery of quality content is the primary purpose of any tutorial.

·         Attractive appearance – Attention needs to be paid to the manner in which the tutorial is presented.  They should always strive to be visually appealing and not overly busy, so as to distract from communicating the message.  A clean and simple presentation also helps to emphasize the content.

As the cloud buzz is getting louder with every passing day you are tempted to take the big leap into the cloud but may have restrained yourself paranoid by ad infinitum cloud security discussions floating on the web. No one can deny the fact that your data is the lifeblood your business. So, undoubtedly its security is of paramount importance for survival of your business.  As cloud computing is a paradigm shift from the traditional ways of using computing resources, you must understand its practical security aspects.

Is Cloud Computing Safe?

There can’t be a binary answer (Yes or No) to this question. But with my experience and as an authority on the subject I can tell you that technologies enabling Cloud services are not in any way less secure than the traditional or on-premise hosting model.  Also, with the evolution of technology, the cloud providers are getting matured and almost all the providers are offering built-in security, privacy, data backups and risk management as a part of their core service.If you are not a big IT company then you must ask yourself:

·         Can an on-premise solution or a traditional hosting provider match the same level of standard security and privacy requirement as provided by the specialist cloud provider whose core competency lies in providing state of the art security and privacy?

The short answer is, yes and no. It depends upon who you are. The purpose of this entry is to help you determine, yes or no.

Full disclosure. This entry is created on a Mac mini. And doing so on Windows 8 (Release Preview). If you are a developer, in my humble opinion you need to test on all platforms you expect your app to run or you are not much of a developer.

To be successful you need to leave politics in geographical territory known as Washington DC. My definition of that is: 14 mi.² of real estate surrounded by reality.

Only in politics can we afford to take sides. Those of us in IT, especially developers need to do our best to be all things to all people. Certainly this is a technical impossibility. However in our game we can get some points for at least being serviceable if not outstanding.

Unlike Java, Python does not have a string contains method.  Instead, use the in operator or the find method.  The in operator finds treats the string as a word list whereas the find method looks for substrings.  In the example shown below, 'is' is a substring of this but not a word by itself.  Therefore, find recoginizes 'is' in this while the in operator does not.

s = "This be a string"
if s.find("is") == -1:
    print "No 'is' here!"
else:
    print "Found 'is' in the string."
    
if "is" in s:
    print "No 'is' here!"
else:
    print "Found 'is' in the string."

#prints out the following:
Found 'is' in the string
No 'is' here!