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  • Venture Capital SEO (Search Engine Optimization)

    Back in the day, venture capitalists relied on tips, networking and brute force to find companies to invest in. Typically, more of their time was spent actively searching for leads than marketing their services. The internet changed all of that. A quick resource for start-up companies looking for companies to invest in them, today's venture capitalists have been forced to change the way they do business - now they must spend more of their time advertising.

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  • Search Engine Optimization For Beginners

    Learning about the SEO techniques can be scary and a bit overwhelming to say the least. If you do not know anything about SEO now is the time to learn, before trying to do it with just a little bit of knowledge, you will fail if you do it that way.

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  • 7 Steps to Search Engine Optimization

    Internet Marketing is a new media and latest trend to advertising which moving away from conventional advertising like print media, electronic media and etc. Internet marketing is a holistic and effective approach on internet to reach wider and more defined target marketing.

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  • Search Engine Optimization

    The best way a company can inform its public about itself is through its website. Most corporations and firms today have a website to their name. The internet has proved itself to be an open market for all kinds of products; the only problem being the trouble of attracting internet users.

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  • Search Engine Optimization Terms

    By the time you complete this multiple lesson tutorial, you'll know just what it takes to score top search engine positions for your Web sites. You'll understand how search engines crawl the Web, how they rank Web sites, and how they find previously undiscovered sites. You'll master the important HTML tags that are your key to getting your sites on a search engine's radar, and you'll see why it's important to amass as many potential keywords as possible.

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  • xslt apply-templates in second level

    - by m00sila
    I cannot wrap <panel> tags to second level individual items as shown in Expected result bellow. But with the xslt i have written, all 1.x get into single node. Please help me. Source xml <root> <step id="1"> <content> <text> 1.0 Sample first level step text </text> </content> <content/> <step> <content> <text> 1.1 Sample second level step text </text> </content> </step> <step> <content> <text> 1.2 Sample second level step text </text> </content> </step> <step> <content> <text> 1.3 Sample second level step text </text> </content> </step> </step> </root> Expected output <panel> <panel> 1.0 Sample first level step text </panel> <panel> 1.1 Sample second level step text </panel> <panel> 1.2 Sample second level step text </panel> <panel> 1.3 Sample second level step text </panel> </panel> My XSLT <xsl:template match="/"> <panel> <xsl:apply-templates/> </panel> </xsl:template> <xsl:template match="root/step" > <panel> <panel> <xsl:apply-templates select ="content/text/node()"></xsl:apply-templates> </panel> <panel> <xsl:apply-templates select ="step/content/text/node()"></xsl:apply-templates> </panel> </panel> </xsl:template>

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  • Configurable ruby logger setup: Logger.new().level = variable

    - by Daniel
    Hi, I want to change the logging level of an application (ruby). require 'logger' config = { :level => 'Logger::WARN' } log = Logger.new STDOUT log.level = Kernel.const_get config[:level] Well, the irb wasn't happy with that and threw "NameError: wrong constant name Logger::WARN" in my face. Ugh! I was insulted. I could do this in a case/when to solve this, or do log.level = 1, but there must be a more elegant way! Does anyone have any ideas? -daniel

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  • How to use second level cache for lazy loaded collections in Hibernate?

    - by Chandru
    Let's say I have two entities, Employee and Skill. Every employee has a set of skills. Now when I load the skills lazily through the Employee instances the cache is not used for skills in different instances of Employee. Let's Consider the following data set. Employee - 1 : Java, PHP Employee - 2 : Java, PHP When I load Employee - 2 after Employee - 1, I do not want hibernate to hit the database to get the skills and instead use the Skill instances already available in cache. Is this possible? If so how? Hibernate Configuration <session-factory> <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property> <property name="hibernate.connection.password">pass</property> <property name="hibernate.connection.url">jdbc:mysql://localhost/cache</property> <property name="hibernate.connection.username">root</property> <property name="hibernate.dialect">org.hibernate.dialect.MySQLInnoDBDialect</property> <property name="hibernate.cache.use_second_level_cache">true</property> <property name="hibernate.cache.use_query_cache">true</property> <property name="hibernate.cache.provider_class">net.sf.ehcache.hibernate.EhCacheProvider</property> <property name="hibernate.hbm2ddl.auto">update</property> <property name="hibernate.show_sql">true</property> <mapping class="org.cache.models.Employee" /> <mapping class="org.cache.models.Skill" /> </session-factory> The Entities with imports, getters and setters Removed @Entity @Table(name = "employee") @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) public class Employee { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private String name; public Employee() { } @ManyToMany @JoinTable(name = "employee_skills", joinColumns = @JoinColumn(name = "employee_id"), inverseJoinColumns = @JoinColumn(name = "skill_id")) @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) private List<Skill> skills; } @Entity @Table(name = "skill") @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) public class Skill { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private String name; } SQL for Loading the Second Employee and his Skills Hibernate: select employee0_.id as id0_0_, employee0_.name as name0_0_ from employee employee0_ where employee0_.id=? Hibernate: select skills0_.employee_id as employee1_1_, skills0_.skill_id as skill2_1_, skill1_.id as id1_0_, skill1_.name as name1_0_ from employee_skills skills0_ left outer join skill skill1_ on skills0_.skill_id=skill1_.id where skills0_.employee_id=? In that I specifically want to avoid the second query as the first one is unavoidable anyway.

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  • Why is WPO(whole-program optimization) not doing any improvements in my program size? (FPC 2.4.0)

    - by Gregory Smith
    I use FPC 2.4.0 for WinXP(binary from the official page), also tryed with same version but compiled from source on my comp. I put something like this: I:\pascal\fpc-2.4.0.source\fpc-2.4.0\compiler\ppc386 -FWserver-1.wpo -OWsymbolliveness -CX -XX -Xs- -al -Os -oServer1.o Server I:\pascal\fpc-2.4.0.source\fpc-2.4.0\compiler\ppc386 -FWserver-2.wpo -OWsymbolliveness -Fwserver-1.wpo -Owsymbolliveness -CX -XX -Xs- -al -Os -oServer2.o Server ..(up to 100 times) but always same .wpo files, and same .o sizes(.s, assembly files change intermittently) I also not(through compiler messages), that not used variables are still alive. Also tryed -OWall -owall What am i doing wrong?

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  • How do you get average of sums in SQL (multi-level aggregation)?

    - by paxdiablo
    I have a simplified table xx as follows: rdate date rtime time rid integer rsub integer rval integer primary key on (rdate,rtime,rid,rsub) and I want to get the average (across all times) of the sums (across all ids) of the values. By way of a sample table, I have (with consecutive identical values blanked out for readability): rdate rtime rid rsub rval ------------------------------------- 2010-01-01 00.00.00 1 1 10 2 20 2 1 30 2 40 01.00.00 1 1 50 2 60 2 1 70 2 80 02.00.00 1 1 90 2 100 2010-01-02 00.00.00 1 1 999 I can get the sums I want with: select rdate,rtime,rid, sum(rval) as rsum from xx where rdate = '2010-06-01' group by rdate,rtime,rid which gives me: rdate rtime rid rsum ------------------------------- 2010-01-01 00.00.00 1 30 (10+20) 2 70 (30+40) 01.00.00 1 110 (50+60) 2 150 (70+80) 02.00.00 1 190 (90+100) as expected. Now what I want is the query that will also average those values across the time dimension, giving me: rdate rtime ravgsum ---------------------------- 2010-01-01 00.00.00 50 ((30+70)/2) 01.00.00 130 ((110+150)/2) 02.00.00 190 ((190)/1) I'm using DB2 for z/OS but I'd prefer standard SQL if possible.

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  • Pure Java open-source libraries for portfolio selection (= constrained, non-linear optimization)?

    - by __roland__
    Does anyone know or has experience with a pure Java library to select portfolios or do some similar kinds of quadratic programming with constraints? There seems to be a lot of tools, as already discussed elsewhere - but what I would like to use is a pure Java implementation. Since I want to call the library from within another open-source software with a BSD-ish license I would prefer LGPL over GPL. Any help is appreciated. If you don't know such libraries, which is the most simple algorithm you would suggest to implement? It has to cope with an inequality constraint (all x_i = 0) and an equality constraint (sum of all x_i = 1).

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  • MDX: How To Aggregate Hierarchy Level Members With Same Name

    - by Dave Frautnick
    Greetings, I am new to MDX, and am having trouble understanding how to perform an aggregation on a hierarchy level with members that have the same names. This query is particular to Microsoft Analysis Services 2000 cubes. I have a given hierarchy dimension with levels defined as follows: [Segment].[Flow].[Segment Week] Within the [Segment Week] level, I have the following members: [Week- 1] [Week- 2] [Week- 3] ... [Week- 1] [Week- 2] [Week- 3] The members have the same names, but are aligned with a different [Flow] in the parent level. So, the first occurrence of the [Week- 1] member aligns with [Flow].[A] while the second occurrence of [Week- 1] aligns with [Flow].[B]. What I am trying to do is aggregate all the members within the [Segment Week] level that have the same name. In SQL terms, I want to GROUP BY the member names within the [Segment Week] level. I am unsure how to do this. Thank you. Dave

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  • Mysql optimization question - How to apply AND logic in search and limit on results in one query?

    - by sandeepan-nath
    This is a little long but I have provided all the database structures and queries so that you can run it immediately and help me. Run the following queries:- CREATE TABLE IF NOT EXISTS `Tutor_Details` ( `id_tutor` int(10) NOT NULL auto_increment, `firstname` varchar(100) NOT NULL default '', `surname` varchar(155) NOT NULL default '', PRIMARY KEY (`id_tutor`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=41 ; INSERT INTO `Tutor_Details` (`id_tutor`,`firstname`, `surname`) VALUES (1, 'Sandeepan', 'Nath'), (2, 'Bob', 'Cratchit'); CREATE TABLE IF NOT EXISTS `Classes` ( `id_class` int(10) unsigned NOT NULL auto_increment, `id_tutor` int(10) unsigned NOT NULL default '0', `class_name` varchar(255) default NULL, PRIMARY KEY (`id_class`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=229 ; INSERT INTO `Classes` (`id_class`,`class_name`, `id_tutor`) VALUES (1, 'My Class', 1), (2, 'Sandeepan Class', 2); CREATE TABLE IF NOT EXISTS `Tags` ( `id_tag` int(10) unsigned NOT NULL auto_increment, `tag` varchar(255) default NULL, PRIMARY KEY (`id_tag`), UNIQUE KEY `tag` (`tag`), KEY `id_tag` (`id_tag`), KEY `tag_2` (`tag`), KEY `tag_3` (`tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=18 ; INSERT INTO `Tags` (`id_tag`, `tag`) VALUES (1, 'Bob'), (6, 'Class'), (2, 'Cratchit'), (4, 'Nath'), (3, 'Sandeepan'), (5, 'My'); CREATE TABLE IF NOT EXISTS `Tutors_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_tutor` int(10) default NULL, KEY `Tutors_Tag_Relations` (`id_tag`), KEY `id_tutor` (`id_tutor`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Tutors_Tag_Relations` (`id_tag`, `id_tutor`) VALUES (3, 1), (4, 1), (1, 2), (2, 2); CREATE TABLE IF NOT EXISTS `Class_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_class` int(10) default NULL, `id_tutor` int(10) NOT NULL, KEY `Class_Tag_Relations` (`id_tag`), KEY `id_class` (`id_class`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Class_Tag_Relations` (`id_tag`, `id_class`, `id_tutor`) VALUES (5, 1, 1), (6, 1, 1), (3, 2, 2), (6, 2, 2); Following is about the tables:- There are tutors who create classes. Tutor_Details - Stores tutors Classes - Stores classes created by tutors And for searching we are using a tags based approach. All the keywords are stored in tags table (while classes/tutors are created) and tag relations are entered in Tutor_Tag_Relations and Class_Tag_Relations tables (for tutors and classes respectively)like this:- Tags - id_tag tag (this is a a unique field) Tutors_Tag_Relations - Stores tag relations while the tutors are created. Class_Tag_Relations - Stores tag relations while any tutor creates a class In the present data in database, tutor "Sandeepan Nath" has has created class "My Class" and "Bob Cratchit" has created "Sandeepan Class". 3.Requirement The requirement is to return tutor records from Tutor_Details table such that all the search terms (AND logic) are present in the union of these two sets - 1. Tutor_Details table 2. classes created by a tutor in Classes table) Example search and expected results:- Search Term Result "Sandeepan Class" Tutor Sandeepan Nath's record from Tutor Details table "Class" Both the tutors from ... Most importantly, there should be only one mysql query and a LIMIT applicable on the number of results. Following is a working query which I have so far written (It just applies OR logic of search key words instead of the desired AND logic). SELECT td . * FROM Tutor_Details AS td LEFT JOIN Tutors_Tag_Relations AS ttagrels ON td.id_tutor = ttagrels.id_tutor LEFT JOIN Classes AS wc ON td.id_tutor = wc.id_tutor INNER JOIN Class_Tag_Relations AS wtagrels ON td.id_tutor = wtagrels.id_tutor LEFT JOIN Tags AS t ON t.id_tag = ttagrels.id_tag OR t.id_tag = wtagrels.id_tag WHERE t.tag LIKE '%Sandeepan%' OR t.tag LIKE '%Nath%' GROUP BY td.id_tutor LIMIT 20 Please help me with anything you can. Thanks

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