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  • Should be simple: existing laptop with local user and outlook 2007 migrate on same computer to domain user with outlook 2007 emails intact

    - by bifpowell
    I have Dell Laptop with windows 7 64 bit and for the last year it's been just a machine with an account like: machine\john there are files in folders and stuff in c:\users\john and john uses outlook 2007 as a pop3 client and has identifiable local appdata pst files. Now I installed a server and want to have everything be domain-centric so I added this laptop to the domain with admin credentials and then logged in as a domain user as: domain\john.smith Now I want to duplicate machine\john (outlook emails mostly) to domain\john.smith. In the past I used the Files and Settings Xfer Wizard and done. I tried that here and it crunched away for a while, made the file, but the restore had no effect - it ran for a while, had a progress bar, but it's like nothing happened at all afterwards. I've rebooted the machine, logged in as domain administrator as the first user to log on after the restart and tried: c:\users\john xcopy c:\users\john c:\users\john.smith /V /C /F /H /K /Y /E ...and it copies some of it, but when it gets to c:\users\john.smith\appdata\local\application data it chokes "Access denied, unable to create directory" I also tried logging in as domain\john.smith and copying the entire directory that the PSTs are in from machine\john and a lot of the mail was there when I launched outlook after replacing the PSTs, but not all of them??? I got errors about files in use when doing this method, which I figure must be why not all the old emails are in the inbox?... There must be some extremely simple way to do what must be a very common requirement. Any guidance appreciated.

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  • Comparing columns in Excel

    - by Regan
    I needed to take columns A, B, and C and compare D, E and F. Here's an example: A B C D E F Jump Smith 5 Jump Smith 8 Run Naylor 2 Swim Fran 4 Swim Fran 7 Jog Dylan 1 Jump Fran 3 Jog Smith 4 So I want to match column A and B with D and E but still have both number related C for 2011 and F for 2012. Can anyone please help with that formula? My data is from A3-C4344 and D3 - D4470.

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  • Ways to have audio output without wires

    - by viraptor
    I'm trying to find a way of using my home speakers/amp without actually having to connect them. There are two laptops that use them normally (so I don't like changing the connection all the time) and I'd rather move the speakers to a place that's away from the couch. I'm not sure how to do this though... The options I can think of are: some kind of wireless jack-jack connection finally getting a media server Unfortunately I can't find any good product for the first solution. I've seen some headphones which have the receiver integrated and a separate transmitted, so in general the idea is already out there, just not the way I need ;) I've seen also http://www.miccus.com/products/blubridge-mini-jack, but I'd have to have a compatible receiver which I can't find on its own (maybe there's some application that the media server could use?). As far as media server goes... many of the plug servers look really interesting, but I'm not sure how to create an audio output and how to redirect the input really. None of the plug servers I've seen so far advertises the option of audio output jack port. I think this part could be fixed by getting one with an usb port and a separate cheap usb soundcard. I hope that input can be sorted out in some rather simple way. I've got Linux running on both laptops so I hope that would be possible to configure jack/pulse/whatever to use the remote endpoint, or even write a simple local-/dev/dsp:network:media-server-/dev/dsp forwarder. So the main question is... are there better ways? Are there any out of the box solutions? Or maybe this was already done by someone and described somewhere?

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  • Why no multiple instances of Firefox on Linux as on Windows?

    - by Jack
    On Windows If I run Firefox as user jack, and then try to start another instance of firefox I will be unable to, as one is already running. If I choose to run firefox as administrator, then I can have two instances of firefox, separate from each other side by side, because they are under different user accounts. This does not seem to be true on Linux. As user jack if I start firefox, like on windows I am unable to start a new instance. If I open a terminal and change to root, set XAUTHORITY to jacks .Xauthority and try to start firefox as root....I get the error that firefox is already running. Why is this? Please don't spare any technical details in your answers....thankyou.

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  • No sound through headset - only mic is working

    - by Kristis
    I noticed that no sound is being played to my headphones. The laptop has a conexant sound card together with the sounds apps provided. But the thing is that I also noticed that now instead of one playback device - 2 are presented: speakers and headphones. And while speakers do play sound nicely - even test sound are not played through headphone output. Also, headphone output does not have a jack assigned to it, while speakers have L R Rear panel Analog Jack(my laptop does not have a jack on the back - only on the right). Also - my headphones have a mic as well - and when I plug it in - the mic is working(using top panel digital jack),but the headphones themselves are not. And laptop is recognizing when audio device is plugged in. And I checked the headset on other devices - the headphones are working. I have tried updating drivers, rolling back drivers and completely uninstalling drivers and then restarting - nothing helped. I imagine that I somehow need to reconfigure the jack configuration just have no idea how and where. Any suggestions? Thanks

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  • Varnish: User specific pages

    - by jchong0707
    I'm new to Varnish and am interested in using it to speed up my web application I wanted to know if Varnish can handle caching and serving user specific content. For example if I have a page say for example /welcome which is dynamically generated in the backend and is user specific So if User John Smith shows up to /welcome it'll show in the page itself 'Welcome John Smith' and if Bob Smith shows up to /welcome it'll show 'Welcome Bob Smith' Ideally both of those /welcome pages will be cached for each unique User, is this something Varnish can do? (is this even a good Use Case of Varnish?) Thanks!

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  • ASUS N45SF - play subwoofer with audio connected

    - by Jaroslav Bucko
    I have notebook ASUS N45SF. It comes with dedicated subwoofer, which is connected to separate audio jack. When I connect any audio device to audio jack, internal speakers remain silent, but subwoofer too. I want to let subwoofer play even with audio device connected to jack. Are there any drivers or settings in OS, which would eneble this behaviour? I have Win7/Ubuntu dualboot so OS doesnt matter. Thanks

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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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  • No voice on front headphone port.

    - by asdacap
    I have a strange problem. It just happen recently, when I accidentally unplug my headphone. But I unplugged it before and nothing happen. Basically now, when I use my headphone through front jack, when playing videos, I can't hear voice. Only background music. Using kde sound setup, pressing front left and front right test button, result in a mono sound. No distinction between right and left. This only happen with front jack. Rear jack is working fine.

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  • Cannot start ubuntu-desktop

    - by Jack
    I am mainly a Windows technician and am trying to install ubuntu server. Everything worked fine and I can log in using the shell but when I installed ubuntu-desktop it just refuses to start? I did try startx and I get the message "server already running" I tried "start gdm" (what is this supposed to do?) and it comes back with "Job is already running: gdm" I know that the server version is not really for ubuntu-desktop but all our other servers are like that and I want it, is there any help out there? Ps. the server is running on a VM install that my IT department made for me and I connect to the machine shell using "Tera Ter Web 3.1" Thank you Jack

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  • Googlemail users can't email my email address

    - by Jack W-H
    Hi folks I have a GridServer account at MediaTemple. The address linked up to my MT account is [email protected]. My non-Google email address could email [email protected] just fine. But when my friend tried to email it from his gmail address, he got the following message: From: Mail Delivery Subsystem Date: Thu, Apr 15, 2010 at 12:02 PM Subject: Delivery Status Notification (Failure) To: [email protected] Delivery to the following recipient failed permanently: [email protected] Technical details of permanent failure: Google tried to deliver your message, but it was rejected by the recipient domain. We recommend contacting the other email provider for further information about the cause of this error. The error that the other server returned was: 550 550 relay not permitted (state 14). ----- Original message ----- MIME-Version: 1.0 Received: by 10.231.205.139 with HTTP; Thu, 15 Apr 2010 12:02:26 -0700 (PDT) In-Reply-To: <[email protected] References: <[email protected] Date: Thu, 15 Apr 2010 12:02:26 -0700 Received: by 10.231.169.144 with SMTP id z16mr211585iby.25.1271358147047; Thu, 15 Apr 2010 12:02:27 -0700 (PDT) Message-ID: Subject: Re: Hi Friend From: My Friend To: "[email protected]" Content-Type: multipart/alternative; boundary=0016e6d26c5abcb2a704844b22bf Does this work. Does this work. Does this work? On Thu, Apr 15, 2010 at 11:30 AM, [email protected] wrote: Hi Friend. Just testing the email address I set up for My Site. Could you please reply so I can check if it's working OK? Cheers Jack I thought it was just a fluke, but exactly the same thing happens when I use MY Gmail address that I also have. Can anyone shed some light on the problem? Jack

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  • Blue screen issue

    - by Jack
    I received several BSOD's that are recorded in the following logs: Problem signature: Problem Event Name: BlueScreen OS Version: 6.1.7601.2.1.0.256.48 Locale ID: 3081 Additional information about the problem: BCCode: 50 BCP1: FFFFF95FF8150C10 BCP2: 0000000000000008 BCP3: FFFFF95FF8150C10 BCP4: 0000000000000005 OS Version: 6_1_7601 Service Pack: 1_0 Product: 256_1 Files that help describe the problem: C:\Windows\Minidump\040412-20030-01.dmp C:\Users\Jack\AppData\Local\Temp\WER-33025-0.sysdata.xml ~~~~~ Problem signature: Problem Event Name: BlueScreen OS Version: 6.1.7601.2.1.0.256.48 Locale ID: 3081 Additional information about the problem: BCCode: 1e BCP1: 0000000000000000 BCP2: 0000000000000000 BCP3: 0000000000000000 BCP4: 0000000000000000 OS Version: 6_1_7601 Service Pack: 1_0 Product: 256_1 Files that help describe the problem: C:\Windows\Minidump\040412-32729-01.dmp C:\Users\Jack\AppData\Local\Temp\WER-64319-0.sysdata.xml It seems to occur at random. I have gone 2 months without a BSOD, then I have gone a week with 10+ without changing what I am doing. This is my system: Windows 7 Professional 64-bit Gigabyte GA-890GPA-UD3H AMD Phenom II x6 1090T Processor 3.2GHz 8GB Ram(4X 2GB) Radeon HD 7850 2TB HDD Thermaltake 500W PSU I'm not sure about what the BSOD says, it just counts to 100 by 5's then restarts the computer. It happens fast and I have tried to get a picture before but to no avail.

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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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  • LinkedIn type friends connection required in php

    - by Akash
    Hi, I am creating a custom social network for one of my clients. In this I am storing the friends of a user in the form of CSV as shown below in the user table uid user_name friends 1 John 2 2 Jack 3,1 3 Gary 2,4 4 Joey 3 In the above scenario if the logged in user is John and if he visits the profile page of Joey, the connection between them should appear as John-Jack-Gary-Joey I am able to establish the connection at level 1 i.e If Jack visits Joey's profile I am able to establish the following : Jack-Gary-Joey But for the 2nd level I need to get into the same routine of for loops which I know is not the right solution + I am not able to implement that as well. So, can someone please help me with this? Thanks in Advance, Akash P:S I am not in a position to change the db architecture :(

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  • Read only file system

    - by Jack Moon
    I'm running Ubuntu 12.10, Upon opening any shell I get the following error: /home/jack/.rbenv/libexec/rbenv-init: line 87: cannot create temp file for here-document: Read-only file system I realised this wasn't simply a rbenv issue, as any file I try to write to returns an error saying the system is Read-only. I don't know how else to describe my problem, each time I boot up the system goes through a disk check, where it supposedly fixes several errors in my disk. Here is my /etc/fstab # <file system> <mount point> <type> <options> <dump> <pass> proc /proc proc nodev,noexec,nosuid 0 0 # / was on /dev/sda1 during installation UUID=1cc4b2ab-a984-4516-ac25-6d64f5050244 / ext4 errors=remount-ro 0 1 # swap was on /dev/sda5 during installation UUID=4e0dfeae-701a-43ce-b5c6-65f15ab3d8e3 none swap sw 0 0 The entire file system is read-only. I've tried the following sudo fsck.ext4 -f /dev/sda1 which gave the following (shortened) output /dev/sda1: ***** FILE SYSTEM WAS MODIFIED ***** /dev/sda1: ***** REBOOT LINUX ***** /dev/sda1: 1257080/45268992 files (1.0% non-contiguous), 50696803/181051904 blocks

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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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  • Activity Indicator not displaying based on whether the UIWebView is loading or not...

    - by Jack W-H
    Hi folks Sorry if this is an easy one. Basically, here is my code: MainViewController.h: // // MainViewController.h // Site // // Created by Jack Webb-Heller on 19/03/2010. // Copyright __MyCompanyName__ 2010. All rights reserved. // #import "FlipsideViewController.h" @interface MainViewController : UIViewController <UIWebViewDelegate, FlipsideViewControllerDelegate> { IBOutlet UIWebView *webView; IBOutlet UIActivityIndicatorView *spinner; } - (IBAction)showInfo; @property(nonatomic,retain) UIWebView *webView; @property(nonatomic,retain) UIActivityIndicatorView *spinner; @end MainViewController.m: // // MainViewController.m // Site // // Created by Jack Webb-Heller on 19/03/2010. // Copyright __MyCompanyName__ 2010. All rights reserved. // #import "MainViewController.h" #import "MainView.h" @implementation MainViewController @synthesize webView; @synthesize spinner; - (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil { if (self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil]) { // Custom initialization } return self; } // Implement viewDidLoad to do additional setup after loading the view, typically from a nib. - (void)viewDidLoad { NSURL *siteURL; NSString *siteURLString; siteURLString=[[NSString alloc] initWithString:@"http://www.site.com"]; siteURL=[[NSURL alloc] initWithString:siteURLString]; [webView loadRequest:[NSURLRequest requestWithURL:siteURL]]; [siteURL release]; [siteURLString release]; [super viewDidLoad]; } - (void)flipsideViewControllerDidFinish:(FlipsideViewController *)controller { [self dismissModalViewControllerAnimated:YES]; } - (void)webViewDidFinishLoad:(UIWebView *)webView { [spinner stopAnimating]; spinner.hidden=FALSE; NSLog(@"viewDidFinishLoad went through nicely"); } - (void)webViewDidStartLoad:(UIWebView *)webView { [spinner startAnimating]; spinner.hidden=FALSE; NSLog(@"viewDidStartLoad seems to be working"); } - (IBAction)showInfo { FlipsideViewController *controller = [[FlipsideViewController alloc] initWithNibName:@"FlipsideView" bundle:nil]; controller.delegate = self; controller.modalTransitionStyle = UIModalTransitionStyleFlipHorizontal; [self presentModalViewController:controller animated:YES]; [controller release]; } - (void)didReceiveMemoryWarning { // Releases the view if it doesn't have a superview. [super didReceiveMemoryWarning]; // Release any cached data, images, etc that aren't in use. } - (void)viewDidUnload { // Release any retained subviews of the main view. // e.g. self.myOutlet = nil; } - (void)dealloc { [spinner release]; [webView release]; [super dealloc]; } @end Unfortunately nothing is ever written to my log, and for some reason the Activity Indicator never seems to appear. What's going wrong here? Thanks folks Jack

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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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