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Search found 2008 results on 81 pages for 'greg smith'.

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  • Performance tuning of tabular data models in Analysis Services

    - by Greg Low
    More and more practical information around working with tabular data models is starting to appear as more and more sites get deployed.At SQL Down Under, we've already helped quite a few customers move to tabular data models in Analysis Services and have started to collect quite a bit of information on what works well (and what doesn't) in terms of performance of these models. We've also been running a lot of training on tabular data models.It was great to see a whitepaper on the performance of these models released today.Performance Tuning of Tabular Models in SQL Server 2012 Analysis Services was written by John Sirmon, Greg Galloway, Cindy Gross and Karan Gulati. You'll find it here: http://msdn.microsoft.com/en-us/library/dn393915.aspx

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  • Copying content on webpages in safari. To HTML

    - by Carl Smith
    Hi, is there an easier way to copy and paste website content in html? Want to copy and look like this. Product Information: Length: S / M / L Material: Polyester and Elasthane Brand: Roxana Exclusive Style: Basque But when i paste it into my content box it looks like this- Product Information Length: S / M / L Material: Polyester and Elasthane Brand: Roxana Exclusive Style: Basque Then i need to edit it in the html editor to rearrange it. Is the some sort of app or plugin that i can get so i can turn the text of the page into html so it looks right straight away when i copy it into my content box? If that makes any sense? Thanks Carl Smith :-)

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  • Gateway IP Returns to Zero

    - by Robert Smith
    When you set a static IP under Ubuntu 12.04.1, you must supply the desired machine IP and the gateway IP, all using the Network Manager. When I first entered them and rebooted, everything worked great. On the second boot, however, Firefox could find no Web page. Upon checking, I discovered that the gateway IP had returned to zero. Now, no matter how often I resupply it, it returns to zero immediately after NM "saves" it: that is, appears as zero when redisplayed. The only way I can get to the Internet is to restore DHCP operation. I need to use static IP for access to my home network. Would appreciate any suggestion. --Robert Smith

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  • Ways to ensure unique instances of a class?

    - by Peanut
    I'm looking for different ways to ensure that each instance of a given class is a uniquely identifiable instance. For example, I have a Name class with the field name. Once I have a Name object with name initialised to John Smith I don't want to be able to instantiate a different Name object also with the name as John Smith, or if instantiation does take place I want a reference to the orginal object to be passed back rather than a new object. I'm aware that one way of doing this is to have a static factory that holds a Map of all the current Name objects and the factory checks that an object with John Smith as the name doesn't already exist before passing back a reference to a Name object. Another way I could think of off the top of my head is having a static Map in the Name class and when the constructor is called throwing an exception if the value passed in for name is already in use in another object, however I'm aware throwing exceptions in a constructor is generally a bad idea. Are there other ways of achieving this?

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  • Oracle VM Deep Dives

    - by rickramsey
    "With IT staff now tasked to deliver on-demand services, datacenter virtualization requirements have gone beyond simple consolidation and cost reduction. Simply provisioning and delivering an operating environment falls short. IT organizations must rapidly deliver services, such as infrastructure-as-a-service (IaaS), platform-as-a-service (PaaS), and software-as-a-service (SaaS). Virtualization solutions need to be application-driven and enable:" "Easier deployment and management of business critical applications" "Rapid and automated provisioning of the entire application stack inside the virtual machine" "Integrated management of the complete stack including the VM and the applications running inside the VM." Application Driven Virtualization, an Oracle white paper That was published in August of 2011. The new release of Oracle VM Server delivers significant virtual networking performance improvements, among other things. If you're not sure how virtual networks work or how to use them, these two articles by Greg King and friends might help. Looking Under the Hood at Virtual Networking by Greg King Oracle VM Server for x86 lets you create logical networks out of physical Ethernet ports, bonded ports, VLAN segments, virtual MAC addresses (VNICs), and network channels. You can then assign channels (or "roles") to each logical network so that it handles the type of traffic you want it to. Greg King explains how you go about doing this, and how Oracle VM Server for x86 implements the network infrastructure you configured. He also describes how the VM interacts with paravirtualized guest operating systems, hardware virtualized operating systems, and VLANs. Finally, he provides an example that shows you how it all looks from the VM Manager view, the logical view, and the command line view of Oracle VM Server for x86. Fundamental Concepts of VLAN Networks by Greg King and Don Smerker Oracle VM Server for x86 supports a wide range of options in network design, varying in complexity from a single network to configurations that include network bonds, VLANS, bridges, and multiple networks connecting the Oracle VM servers and guests. You can create separate networks to isolate traffic, or you can configure a single network for multiple roles. Network design depends on many factors, including the number and type of network interfaces, reliability and performance goals, the number of Oracle VM servers and guests, and the anticipated workload. The Oracle VM Manager GUI presents four different ways to create an Oracle VM network: Bonds and ports VLANs Both bond/ports and VLANS A local network This article focuses the second option, designing a complex Oracle VM network infrastructure using only VLANs, and it steps through the concepts needed to create a robust network infrastructure for your Oracle VM servers and guests. More Resources Virtual Networking for Dummies Download Oracle VM Server for x86 Find technical resources for Oracle VM Server for x86 -Rick Follow me on: Blog | Facebook | Twitter | Personal Twitter | YouTube | The Great Peruvian Novel

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  • Making Use of Plan Explorer in my own Environment

    - by Jonathan Kehayias
    Back in October 2010, I briefly blogged about the SQL Sentry Plan Explorer in my blog post wrap up for SQL Bits 7 and how impressed I was with what I saw from a Alpha demo standpoint from Greg Gonzalez ( Blog | Twitter ) while I was at SQLBits 7 in York.  To be 100% honest and transparent, Greg gave me early access to this tool after discussing it at SQLBits 7, and I had the opportunity to test a number of pre-Beta releases where I was able to offer significant feedback and submit bugs in the...(read more)

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  • Making Use of Plan Explorer in my own Environment

    - by Jonathan Kehayias
    Back in October 2010, I briefly blogged about the SQL Sentry Plan Explorer in my blog post wrap up for SQL Bits 7 and how impressed I was with what I saw from a Alpha demo standpoint from Greg Gonzalez ( Blog | Twitter ) while I was at SQLBits 7 in York.  To be 100% honest and transparent, Greg gave me early access to this tool after discussing it at SQLBits 7, and I had the opportunity to test a number of pre-Beta releases where I was able to offer significant feedback and submit bugs in the...(read more)

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  • #SQLMug - Like a collectors set of 5 x geeky SQL Mugs?

    - by Greg Low
    Hi Folks,For a while, I've been wanting to get some great SQL mugs printed for SQL Down Under but I need further inspiration so here's your chance to get a collectors set of 5 SQL mugs:Send me (greg @ sqldownunder . com) a great line to go onto the mugs, along with your country and a delivery address. I'll pick the best 5 and get mugs printed with those sayings. If you're one of the 5, I'll send you a collectors set with one of each of the 5. Simple enough?Here are some ideas I've already received to get you started:Chuck Norris gets NULL. Nothing compares to him either.ALTER MUG  SET SINGLE_USER  WITH ROLLBACK IMMEDIATE;DENY CONTROL  ON OBJECT::MUG  TO public;knock knock who's there? sp_ sp_who? spid 1, spid 2, spid 3, spid 4... ALTER DATABASE CriticalDB SET ChuckNorrisMode = ON WITH NOWAIT;I'll probably cut off new entries around the end of April.

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  • What is the effect of this order_by clause?

    - by bread
    I don't understand what this order_by clause is doing and whether I need it or not: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date order by i.order_date desc; This produces this data: 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10101 John Gray 30-Jun-1999 Raft 58.00 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10101 John Gray 02-Jan-2000 Lantern 16.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 While if I remove the order_by clause completely, as in this query: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date; I get these results: 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10101 John Gray 02-Jan-2000 Lantern 16.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10101 John Gray 30-Jun-1999 Raft 58.00 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 I'm not sure what the order_by is doing here and if it's having the intended effects.

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  • Tricky SQL query involving consecutive values

    - by Gabriel
    I need to perform a relatively easy to explain but (given my somewhat limited skills) hard to write SQL query. Assume we have a table similar to this one: exam_no | name | surname | result | date ---------+------+---------+--------+------------ 1 | John | Doe | PASS | 2012-01-01 1 | Ryan | Smith | FAIL | 2012-01-02 <-- 1 | Ann | Evans | PASS | 2012-01-03 1 | Mary | Lee | FAIL | 2012-01-04 ... | ... | ... | ... | ... 2 | John | Doe | FAIL | 2012-02-01 <-- 2 | Ryan | Smith | FAIL | 2012-02-02 2 | Ann | Evans | FAIL | 2012-02-03 2 | Mary | Lee | PASS | 2012-02-04 ... | ... | ... | ... | ... 3 | John | Doe | FAIL | 2012-03-01 3 | Ryan | Smith | FAIL | 2012-03-02 3 | Ann | Evans | PASS | 2012-03-03 3 | Mary | Lee | FAIL | 2012-03-04 <-- Note that exam_no and date aren't necessarily related as one might expect from the kind of example I chose. Now, the query that I need to do is as follows: From the latest exam (exam_no = 3) find all the students that have failed (John Doe, Ryan Smith and Mary Lee). For each of these students find the date of the first of the batch of consecutively failing exams. Another way to put it would be: for each of these students find the date of the first failing exam that comes after their last passing exam. (Look at the arrows in the table). The resulting table should be something like this: name | surname | date_since_failing ------+---------+-------------------- John | Doe | 2012-02-01 Ryan | Smith | 2012-01-02 Mary | Lee | 2012-01-04 Ann | Evans | 2012-02-03 How can I perform such a query? Thank you for your time.

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  • Odd SQL Results

    - by Ryan Burnham
    So i have the following query Select id, [First], [Last] , [Business] as contactbusiness, (Case When ([Business] != '' or [Business] is not null) Then [Business] Else 'No Phone Number' END) from contacts The results look like id First Last contactbusiness (No column name) 2 John Smith 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number I'd expect record 2 to also show No Phone Number If i change the "[Business] is not null" to [Business] != null then i get the correct results id First Last contactbusiness (No column name) 2 John Smith No Phone Number 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number Normally you need to use is not null rather than != null. whats going on here?

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  • Creating an excel macro to sum lines with duplicate values

    - by john
    I need a macro to look at the list of data below, provide a number of instances it appears and sum the value of each of them. I know a pivot table or series of forumlas could work but i'm doing this for a coworker and it has to be a 'one click here' kinda deal. The data is as follows. A B Smith 200.00 Dean 100.00 Smith 100.00 Smith 50.00 Wilson 25.00 Dean 25.00 Barry 100.00 The end result would look like this Smith 3 350.00 Dean 2 125.00 Wilson 1 25.00 Barry 1 100.00 Thanks in advance for any help you can offer!

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  • Simple Select Statement on MySQL Database Hanging

    - by AlishahNovin
    I have a very simple sql select statement on a very large table, that is non-normalized. (Not my design at all, I'm just trying to optimize while simultaneously trying to convince the owners of a redesign) Basically, the statement is like this: SELECT FirstName, LastName, FullName, State FROM Activity Where (FirstName=@name OR LastName=@name OR FullName=@name) AND State=@state; Now, FirstName, LastName, FullName and State are all indexed as BTrees, but without prefix - the whole column is indexed. State column is a 2 letter state code. What I'm finding is this: When @name = 'John Smith', and @state = '%' the search is really fast and yields results immediately. When @name = 'John Smith', and @state = 'FL' the search takes 5 minutes (and usually this means the web service times out...) When I remove the FirstName and LastName comparisons, and only use the FullName and State, both cases above work very quickly. When I replace FirstName, LastName, FullName, and State searches, but use LIKE for each search, it works fast for @name='John Smith%' and @state='%', but slow for @name='John Smith%' and @state='FL' When I search against 'John Sm%' and @state='FL' the search finds results immediately When I search against 'John Smi%' and @state='FL' the search takes 5 minutes. Now, just to reiterate - the table is not normalized. The John Smith appears many many times, as do many other users, because there is no reference to some form of users/people table. I'm not sure how many times a single user may appear, but the table itself has 90 Million records. Again, not my design... What I'm wondering is - though there are many many problems with this design, what is causing this specific problem. My guess is that the index trees are just too large that it just takes a very long time traversing the them. (FirstName, LastName, FullName) Anyway, I appreciate anyone's help with this. Like I said, I'm working on convincing them of a redesign, but in the meantime, if I someone could help me figure out what the exact problem is, that'd be fantastic.

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  • Is it possible to use ContainsTable to get results for more than one column?

    - by LockeCJ
    Consider the following table: People FirstName nvarchar(50) LastName nvarchar(50) Let's assume for the moment that this table has a full-text index on it for both columns. Let's suppose that I wanted to find all of the people named "John Smith" in this table. The following query seems like a perfectly rational way to accomplish this: SELECT * from People p INNER JOIN CONTAINSTABLE(People,*,'"John*" AND "Smith*"') Unfortunately, this will return no results, assuming that there is no record in the People table that contains both "John" and "Smith" in either the FirstName or LastName columns. It will not match a record with "John" in the FirstName column, and "Smith" in the LastName column, or vice-versa. My question is this: How does one accomplish what I'm trying to do above? Please consider that the example above is simplified. The real table I'm working with has ten columns and the input I'm receiving is a single string which is split up based on standard word breakers (space, dash, etc.)

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  • Parsing complex string using regex

    - by wojtek_z
    My regex skills are not very good and recently a new data element has thrown my parser into a loop Take the following string "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write" Previously I had the following for my regex : [+\\-/] Which would turn the result into USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read But now I have values with dashes in them which is causing bad output New string looks like "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write/FUNCTION=Read-Write" Which gives me the following result , and breaks the key = value structure. USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read Write Can someone help me formulate a valid regex for handling this or point me to some key / value examples. Basically I need to be able to handle + - / signs in order to get combinations.

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  • Are there any well known algorithms to detect the presence of names?

    - by Rhubarb
    For example, given a string: "Bob went fishing with his friend Jim Smith." Bob and Jim Smith are both names, but bob and smith are both words. Weren't for them being uppercase, there would be less indication of this outside of our knowledge of the sentence. Without doing grammar analysis, are there any well known algorithms for detecting the presence of names, at least Western names?

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  • Need help joining tables...

    - by yuudachi
    I am a MySQL newbie, so sorry if this is a dumb question.. These are my tables. student table: SID (primary) student_name advisor (foreign key to faculty.facultyID) requested_advisor (foreign key to faculty.facultyID) faculty table: facultyID (primary key) advisor_name I want to query a table that shows everything in the student table, but I want advisor and requested_advisor to show up as names, not the ID numbers. so like it displays like this on the webpage: Student Name: Jane Smith SID: 860123456 Current Advisor: John Smith Requested advisor: James Smith not like this Student Name: Jane Smith SID: 860123456 Current Advisor: 1 Requested advisor: 2 SELECT student.student_name, SID, student_email, faculty.advisor_name FROM student INNER JOIN faculty ON student.advisor = faculty.facultyID; this comes out close, but I don't know how to get the requested_advisor to show up as a name.

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  • Setting variables in shell script by running commands

    - by rajya vardhan
    >cat /tmp/list1 john jack >cat /tmp/list2 smith taylor It is guaranteed that list1 and list2 will have equal number of lines. f(){ i=1 while read line do var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` echo $i,$var1,$var2 i=`expr $i+1` echo $i,$var1,$var2 done < $INFILE } So output of f() should be: 1,john,smith 2,jack,taylor But getting 1,p,p 1+1,p,p If i replace following: var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` with this: var1=`head -$i /tmp/vip_list|tail -1` var2=`head -$i /tmp/lb_list|tail -1` Then output: 1,john,smith 1,john,smith Not an expert of shell, so please excuse if sounds childish :)

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  • Creating one row of information in excel using a unique value

    - by user1426513
    This is my first post. I am currently working on a project at work which requires that I work with several different worksheets in order to create one mail master worksheet, as it were, in order to do a mail merge. The worksheet contains information regarding different purchases, and each purchaser is identified with their own ID number. Below is an example of what my spreadsheet looks like now (however I do have more columns): ID Salutation Address ID Name Donation ID Name Tickets 9 Mr. John Doe 123 12 Ms. Jane Smith 100.00 12 Ms.Jane Smith 300.00 12 Ms. Jane Smith 456 22 Mr. Mike Man 500.00 84 Ms. Jo Smith 300.00 What I would like to do is somehow sort my data so that everythign with the same unique identifier (ID) lines up on the same row. For example ID 12 Jane Smith - all the information for her will show up under her name matched by her ID number, and ID 22 will match up with 22 etc... When I merged all of my spreadsheets together, I sorted them all by ID number, however my problem is, not everyone who made a donation bought a ticket or some people just bought tickets and nothing us, so sorting doesn't work. Hopefully this makes sense. Thanks in advance.

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  • XEROX Phaser 3160N installation on UBUNTU 12.04 LTE machine

    - by Greg Verrall
    I have recently had windows XP die on one of my machines, and have installed Linux UBUNTU. The OS works great, except for installing the Xerox Phaser 3160N printer. The OS can find and install the network printer, but when I print a test page, it tells me “Internal Error – Please use the correct driver”. I have the correct drivers, as your support team have sent me the link, (http://www.support.xerox.com/support/phaser-3160/file-download/enau.html?operatingSystem=linux&fileLanguage=en_GB&contentId=105724&from=downloads&viewArchived=false) but I cannot install these drivers to run the printer. These are the instructions from the online guide for installing on a Linux machine: 1. Make sure that the machine is connected to your network and powered on. Also, your machine’s IP address should have been set. 2. Insert the supplied software CD into your CD-ROM drive. 3. Double-click CD-ROM icon that appears on your Linux desktop. 4. Double-click the Linux folder. 5. Double-click the install.sh icon. 6. The Xerox Installer window opens. Click Continue. 7. The Add printer wizard window opens. Click Next. 8. Select Network printer and click Search button. 9. The Printer’s IP address and model name appears on list field. 10. Select your machine and click Next. I get as far as step 5, and step 6 never happens, if it did, it would be very easy from there. There are options to add additional software to UBUNTU, however it does not recognise the installation CD as valid when I try to add it as a source. Any ideas on who can help me? regards, Greg Verrall

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  • Run script after switching user account "to the same account"

    - by Peter Sivák
    In Ubuntu, when I click on Switch User Account... and then choose the same account to log in (for example if my name is John Smith, I click on switch user account and then log into the John Smith account again), how can I run a script after that? (I know, that I can run a script after "first" login by putting it in /etc/profile file, but this script is not executed again when I choose switch user account and then immediately log in back to the same account.)

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  • GDL Presents: Women Techmakers with JESS3

    GDL Presents: Women Techmakers with JESS3 Join Leslie, COO and Co-founder of JESS3, in conversation with Megan Smith and Betsy Masiello, as they discuss Leslie's experience growing a design business from two employees to a transnational operation. Hosts: Megan Smith - Vice President, Google [x] | Betsy Masiello - Policy Manager Guest: Leslie Bradshaw - President, COO and Co-founder, JESS3 From: GoogleDevelopers Views: 0 3 ratings Time: 01:00:00 More in Science & Technology

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  • Greasemonkey @require jQuery not working "Component not available"

    - by Greg K
    I've seen the other question on here about loading jQuery in a Greasemonkey. Having tried that method, with this require statement inside my ==UserScript== tags: // @require http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js I still get the following error message in Firefox's error console: Error: Component is not available Source File: file:///Users/greg/Library/Application%20Support/ Firefox/Profiles/xo9xhovo.default/gm_scripts/myscript/jquerymin.js Line: 36 This stops my greasemonkey code from running. I've made sure I included the @require for jQuery and saved my js file before installing it, as required files are only loaded on installation. Code: // ==UserScript== // @name My Script // @namespace http://www.google.com // @description My test script // @include http://www.google.com // @require http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js // ==/UserScript== GM_log("Hello"); I have Greasemonkey 0.8.20091209.4 installed on Firefox 3.5.7 on my Macbook Pro, Leopard (10.5.8). I've cleared my cache (except cookies) and have disabled all other plugins except Flashblock 1.5.11.2, Web Developer 1.1.8 and Adblock Plus 1.1.3. My config.xml with my Greasemonkey script installed: <UserScriptConfig> <Script filename="myscript.user.js" name="My Script" namespace="http://www.google.com" description="My test script" enabled="true" basedir="myscript"> <Include>http://www.google.com</Include> <Require filename="jquerymin.js"/> </Script> I can see jquerymin.js sat in the gm_scripts/myscript/ directory. Additionally, is it common for this error to occur in the console when installing a Greasemonkey script? Error: not well-formed Source File: file:///Users/Greg/Documents/myscript.user.js Line: 1, Column: 1 Source Code: // ==UserScript==

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  • PHPUnit installed but class PHPUnit_TestCase not found

    - by Greg K
    Talk about falling at the first hurdle. My test script: <?php require_once('PHPUnit/Framework.php'); class TransferResponseTest extends PHPUnit_TestCase { ... } Running my test case: $ phpunit TransferResponseTest Fatal error: Class 'PHPUnit_TestCase' not found in /Volumes/Data/greg/code/syndicate/tests/TransferResponseTest.php on line 5 $ php -i | grep include_path include_path => .:/usr/lib/php => .:/usr/lib/php $ ls -l /usr/lib/php/PHPUnit/ total 8 drwxr-xr-x 16 root wheel 544 27 Mar 19:03 Extensions drwxr-xr-x 28 root wheel 952 27 Mar 19:03 Framework -rw-r--r-- 1 root wheel 3193 27 Mar 19:03 Framework.php drwxr-xr-x 8 root wheel 272 27 Mar 19:03 Runner drwxr-xr-x 5 root wheel 170 27 Mar 19:03 TextUI drwxr-xr-x 32 root wheel 1088 27 Mar 19:03 Util I copied /etc/php.ini-default to /etc/php.ini and explicitly specified the include path as /usr/lib/php/ with an end / but still no success. $ php -i | grep include_path include_path => .:/usr/lib/php/ => .:/usr/lib/php/ $ phpunit TransferResponseTest.php PHP Fatal error: Class 'PHPUnit_TestCase' not found in /Volumes/Data/greg/code/syndicate/tests/TransferResponseTest.php on line 5 $ phpunit --version PHPUnit 3.4.11 by Sebastian Bergmann. Any ideas?

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  • ArchBeat Link-o-Rama for December 11, 2012

    - by Bob Rhubart
    Good To Know - Conflicting View Objects and Shared Entity | Andrejus Baranovskis Oracle ACE Director Andrejus Baranovskis shares his thoughts—and a sample application—dealing with an "interesting ADF behavior" encountered over the weekend. Patching Oracle Exalogic - Updating Linux on the Compute Nodes - Part 1 | Jos Nijhoff Jos Nijhoff launches a series of posts the deal with "patching the operating system on the modified Sun Fire X4170 M2 servers...dubbed compute nodes in Exalogic terminology." Expanding on requestaudit - Tracing who is doing what...and for how long | Kyle Hatlestad "One of the most helpful tracing sections in WebCenter Content (and one that is on by default) is the requestaudit tracing," says Oracle Fusion Middleware A-Team architect Kyle Hatlestad. Get up close and technical in his post. Oracle Data Integrator Presentation from NYOUG Webinar | Gurcan Orhan Oracle ACE Director and award-winning data warehouse architect Gurcan Orhan shares his presentation from the recent NYOUG LI SIG. SOA 11g Technology Adapters – ECID Propagation | Greg Mally "Many SOA Suite 11g deployments include the use of the technology adapters for various activities including integration with FTP, database, and files to name a few," says Oracle Fusion Middleware A-Team member Greg Mally. "Although the integrations with these adapters are easy and feature rich, there can be some challenges from the operations perspective." Greg's post focuses on technical tips for dealing with one of these challenges. Missing Duties for RUP3 upgrade in Fusion Applications Richard from the Oracle Fusion Middleware A-Team explains how to safely apply policy store changes in thirteen easy steps. Thought for the Day "Well over half of the time you spend working on a project (on the order of 70 percent) is spent thinking, and no tool, no matter how advanced, can think for you." — Frederick P. Brooks Source: SoftwareQuotes.com

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