Search Results

Search found 6141 results on 246 pages for 'joe the person'.

Page 5/246 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • What is the best answer for: "my Internet is not working"?

    - by Maciek Sawicki
    Hi, I look for work in IT Support. One of interview questions is: what would you first say if user call You and tell my Internet is not working? I think about it a lot and still don't know what is correct answerer nor what answer my future employer expects. My choice would be something like: What part of Internet? (but more polite). For example I could ask for opening web page that works on my PC. Please give only serious answers. If You want BOFH or "The website is down" style answers I can create separate question for that.

    Read the article

  • Capitalization of Person names in programming

    - by Albert
    Hey all, Is anyone aware of some code/rules on how to capitalize the names of people correctly? John Smith Johan van Rensburg Derrick von Gogh Ruby de La Fuente Peter Maclaurin Garry McDonald (these may not be correct, just some sample names and how the capitalization could be/work) This seems like a losing battle... If anyone has some code or rules on when and how to capitalize names, let me know :) Cheers, Albert

    Read the article

  • Need an algorithm to group several parameters of a person under the persons name

    - by QuickMist
    Hi. I have a bunch of names in alphabetical order with multiple instances of the same name all in alphabetical order so that the names are all grouped together. Beside each name, after a coma, I have a role that has been assigned to them, one name-role pair per line, something like whats shown below name1,role1 name1,role2 name1,role3 name1,role8 name2,role8 name2,role2 name2,role4 name3,role1 name4,role5 name4,role1 ... .. . I am looking for an algorithm to take the above .csv file as input create an output .csv file in the following format name1,role1,role2,role3,role8 name2,role8,role2,role4 name3,role1 name4,role5,role1 ... .. . So basically I want each name to appear only once and then the roles to be printed in csv format next to the names for all names and roles in the input file. The algorithm should be language independent. I would appreciate it if it does NOT use OOP principles :-) I am a newbie.

    Read the article

  • Presence icon only showing for first person

    - by James123
    I am trying to show my colleagues in my custom webpart. So I adding presence Icon to each of colleague. It is showing fine when colleague is 1 only. If We have colleague more than 1 Presence Icon showing for 1st colleague you can dropdow that Icon also but other colleagues it is show simple Presense Icon (grayout) (not drop down is comming). code is like this. private static Panel GetUserInfo(UserProfile profile,Panel html, int cnt) { LiteralControl imnrc = new LiteralControl(); imnrc.Text = "<span style=\"padding: 0 5px 0 5px;\"><img border=\"0\" valign=\"middle\" height=\"12\" width=\"12\" src=\"/_layouts/images/imnhdr.gif\" onload=\"IMNRC('" + profile[PropertyConstants.WorkEmail].Value.ToString() + "')\" ShowOfflinePawn=1 id=\"IMID[GUID]\" ></span>"; html.Controls.Add(imnrc); html.Controls.Add(GetNameControl(profile)); //html.Controls.Add(new LiteralControl("<br>")); return html; } private static Control GetNameControl(UserProfile profile) { //bool hasMySite = profile[PropertyConstants.PublicSiteRedirect].Value == null ? false : true; bool hasMySite =string.IsNullOrEmpty(profile.PublicUrl.ToString()) ? false : true; string name = profile[PropertyConstants.PreferredName].Value.ToString(); if (hasMySite) { HyperLink control = new HyperLink(); control.NavigateUrl = String.IsNullOrEmpty(profile.PublicUrl.ToString()) ? null : profile.PublicUrl.ToString(); control.Style.Add("text-decoration","none"); control.Text = name; return control; } else { LiteralControl control = new LiteralControl(); control.Text = name; return control; } } http://i700.photobucket.com/albums/ww5/vsrikanth/presence-1.jpg

    Read the article

  • Splitting a person's name into forename and surname

    - by Nick
    ok so basically I am asking the question of their name I want this to be one input rather than Forename and Surname. Now is there any way of splitting this name? and taking just the last word from the "Sentence" e.g. name = "Thomas Winter" print name.split() and what would be output is just "Winter"

    Read the article

  • Is being a programmer a younger person's job?

    - by Saobi
    After you get old, say past 30 or 40. Can you still keep up with the young coders from your company, those fresh out of school, who can code for 15+ hours on 10 cans of redbulls (most people in Google, Facebook, etc) ? And given the lightning speed with which today's programming frameworks and architectures evolve, can you keep up with the most up to date stuff and be as proficient at them as the next college grad? I know for jobs like unix/c/embedded programming, it might be that the older the better. But for programming jobs in say web development, social media, search engine technology, etc. Do you become less and less competitive career-wise versus youngsters? For example, most coders in Google and Facebook, I believe are under 25 years old. In other words, once you reach a certain age, would it be unwise to continue to be a coder, and is it better to try becoming a project manager or architect?

    Read the article

  • Cluster analysis on two columns that contain name of person in R

    - by Alka Shah
    I am a beginner in R. I have to do cluster analysis in data that contains two columns with name of persons. I converted it in data frame but it is character type. To use dist() function the data frame must be numeric. example of my data: Interviewed.Type interviewed.Relation.Type 1. An1 Xuan 2. An2 The 3. An3 Ngoc 4. Bui Thi 5. ANT feed 7. Bach Thi 8. Gian1 Thi 9. Lan5 Thi . . . 1100. Xung Van I will be grateful for your help.

    Read the article

  • SharePoint 2007 Approval Workflow - Any other person can approve requirement

    - by dsibley
    How can I use SharePoint's Content Approval to enforce the rule that any two people in a group can make a change? We have a policy library that any two individuals can update (a modifier and an approver). If I do a group parallel approval, the modifier can approve their own work (I believe). Unfortunately, I don't have access to Visual Studio or even SharePoint Designer (as these have been disabled by corporate).

    Read the article

  • Show facebook status stream of a dedicated person on a website

    - by Pascal
    Hi all, I've been stomping at this all day :( I want to display a status feed of both twitter and facebook on my website. For twitter, this is not a problem, as my account is public, I can easily get the feed. Facebook however, is a whole different story! I can't seem to find an easy way to just get my last status updates and display it on my website. The code I currently have, needs authentication of the visitor, and I don't want that! This is my current code: $facebook = new Facebook($api_key, $secret); $stream = $facebook->api_client->fql_query("SELECT message,source,time FROM status WHERE uid = $user_id LIMIT 6"); I've seen several possible solutions, including the RSS feed, but as Facebook keeps changing the way their site works, none of the previous methods I've seen (including those from as late as januari) currently work! Is there anybody who can provide me with a currently working solution to this (simple?) problem?

    Read the article

  • jQuery question from a person who can't javascript

    - by Evilalan
    So I'm trying to adapt this Dropdown menu on Joomla the styles work great as expected so I'll post the javascript includes on the head of my website: <script type='text/javascript' src='js/jquery.js'></script> <script type='text/javascript' src='js/dropdown.js'></script> <script type='text/javascript'> $(function() { $('.menu').droppy(); }); </script> <script type='text/javascript'> $(function() { $('.menu').droppy({speed: 100}); }); </script> ok I don't know why its is not working I'll post the dropdown.js should I post the jQuery too? it's really big! $.fn.droppy = function(options) { options = $.extend({speed: 250}, options || {}); this.each(function() { var root = this, zIndex = 1000; function getSubnav(ele) { if (ele.nodeName.toLowerCase() == 'li') { var subnav = $('> ul', ele); return subnav.length ? subnav[0] : null; } else { return ele; } } function getActuator(ele) { if (ele.nodeName.toLowerCase() == 'ul') { return $(ele).parents('li')[0]; } else { return ele; } } function hide() { var subnav = getSubnav(this); if (!subnav) return; $.data(subnav, 'cancelHide', false); setTimeout(function() { if (!$.data(subnav, 'cancelHide')) { $(subnav).slideUp(options.speed); } }, 500); } function show() { var subnav = getSubnav(this); if (!subnav) return; $.data(subnav, 'cancelHide', true); $(subnav).css({zIndex: zIndex++}).slideDown(options.speed); if (this.nodeName.toLowerCase() == 'ul') { var li = getActuator(this); $(li).addClass('hover'); $('> a', li).addClass('hover'); } } $('ul, li', this).hover(show, hide); $('li', this).hover( function() { $(this).addClass('hover'); $('> a', this).addClass('hover'); }, function() { $(this).removeClass('hover'); $('> a', this).removeClass('hover'); } ); }); }; My question here is: Why is it not working! I know that this is really complex (I don't anything about JavaScript) but if you help me I'll post a tutorial and edited files that will help a lot of people! By the way I've download jQuery from the original site so I don't think that this can be the problem! Thanks in advance!

    Read the article

  • How to calculate the size of a project in the days-person unit of measurement?

    - by Will Marcouiller
    Once in a while I have read here and there the size of a project expressed in a matter of days-person or person-day. I may understand what this means, but I don't know on what do people base themselves to calculate it. What are the variables considered into this calculation? How these variables are used in the calculation formula? Otherwise, how to estimate it grossly, when something is missing from the formula's variables? Thanks! =)

    Read the article

  • xslt cookbook example not working

    - by Liza dawson
    Hi I am working on this from xslt cookbook type my.xml <?xml version="1.0" encoding="UTF-8"?> <people> <person name="Al Zehtooney" age="33" sex="m" smoker="no"/> <person name="Brad York" age="38" sex="m" smoker="yes"/> <person name="Charles Xavier" age="32" sex="m" smoker="no"/> <person name="David Williams" age="33" sex="m" smoker="no"/> <person name="Edward Ulster" age="33" sex="m" smoker="yes"/> <person name="Frank Townsend" age="35" sex="m" smoker="no"/> <person name="Greg Sutter" age="40" sex="m" smoker="no"/> <person name="Harry Rogers" age="37" sex="m" smoker="no"/> <person name="John Quincy" age="43" sex="m" smoker="yes"/> <person name="Kent Peterson" age="31" sex="m" smoker="no"/> <person name="Larry Newell" age="23" sex="m" smoker="no"/> <person name="Max Milton" age="22" sex="m" smoker="no"/> <person name="Norman Lamagna" age="30" sex="m" smoker="no"/> <person name="Ollie Kensington" age="44" sex="m" smoker="no"/> <person name="John Frank" age="24" sex="m" smoker="no"/> <person name="Mary Williams" age="33" sex="f" smoker="no"/> <person name="Jane Frank" age="38" sex="f" smoker="yes"/> <person name="Jo Peterson" age="32" sex="f" smoker="no"/> <person name="Angie Frost" age="33" sex="f" smoker="no"/> <person name="Betty Bates" age="33" sex="f" smoker="no"/> <person name="Connie Date" age="35" sex="f" smoker="no"/> <person name="Donna Finster" age="20" sex="f" smoker="no"/> <person name="Esther Gates" age="37" sex="f" smoker="no"/> <person name="Fanny Hill" age="33" sex="f" smoker="yes"/> <person name="Geta Iota" age="27" sex="f" smoker="no"/> <person name="Hillary Johnson" age="22" sex="f" smoker="no"/> <person name="Ingrid Kent" age="21" sex="f" smoker="no"/> <person name="Jill Larson" age="20" sex="f" smoker="no"/> <person name="Kim Mulrooney" age="41" sex="f" smoker="no"/> <person name="Lisa Nevins" age="21" sex="f" smoker="no"/> </people> type generic-attr-to-csv.xslt <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:csv="http://www.ora.com/XSLTCookbook/namespaces/csv"> <xsl:param name="delimiter" select=" ',' "/> <xsl:output method="text" /> <xsl:strip-space elements="*"/> <xsl:template match="/"> <xsl:for-each select="$columns"> <xsl:value-of select="@name"/> <xsl:if test="position( ) != last( )"> <xsl:value-of select="$delimiter/> </xsl:if> </xsl:for-each> <xsl:text>&#xa;</xsl:text> <xsl:apply-templates/> </xsl:template> <xsl:template match="/*/*"> <xsl:variable name="row" select="."/> <xsl:for-each select="$columns"> <xsl:apply-templates select="$row/@*[local-name(.)=current( )/@attr]" mode="csv:map-value"/> <xsl:if test="position( ) != last( )"> <xsl:value-of select="$delimiter"/> </xsl:if> </xsl:for-each> <xsl:text>&#xa;</xsl:text> </xsl:template> <xsl:template match="@*" mode="map-value"> <xsl:value-of select="."/> </xsl:template> </xsl:stylesheet> type my.xsl <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:csv="http://www.ora.com/XSLTCookbook/namespaces/csv"> <xsl:import href="generic-attr-to-csv.xslt"/> <!--Defines the mapping from attributes to columns --> <xsl:variable name="columns" select="document('')/*/csv:column"/> <csv:column name="Name" attr="name"/> <csv:column name="Age" attr="age"/> <csv:column name="Gender" attr="sex"/> <csv:column name="Smoker" attr="smoker"/> <!-- Handle custom attribute mappings --> <xsl:template match="@sex" mode="csv:map-value"> <xsl:choose> <xsl:when test=".='m'">male</xsl:when> <xsl:when test=".='f'">female</xsl:when> <xsl:otherwise>error</xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet> using the apache xalan parser D:\Test>java org.apache.xalan.xslt.Process -in my.xml -xsl my.xsl -out my.csv [Fatal Error] generic-attr-to-csv.xslt:15:6: The value of attribute "select" associated with an element type "xsl:v alue-of" must not contain the '<' character. file:///D:/Test/generic-attr-to-csv.xslt; Line #15; Column #6; org.xml.sax.SAXParseException: The value of attribut e "select" associated with an element type "xsl:value-of" must not contain the '<' character. java.lang.NullPointerException at org.apache.xalan.transformer.TransformerImpl.createSerializationHandler(TransformerImpl.java:1171) at org.apache.xalan.transformer.TransformerImpl.createSerializationHandler(TransformerImpl.java:1060) at org.apache.xalan.transformer.TransformerImpl.transform(TransformerImpl.java:1268) at org.apache.xalan.transformer.TransformerImpl.transform(TransformerImpl.java:1251) at org.apache.xalan.xslt.Process.main(Process.java:1048) Exception in thread "main" java.lang.RuntimeException at org.apache.xalan.xslt.Process.doExit(Process.java:1155) at org.apache.xalan.xslt.Process.main(Process.java:1128) Any ideas what am i doing wrong

    Read the article

  • Creating classed in JavaScript

    - by Renso
    Goal:Creating class instances in JavaScript is not available since you define "classes" in js with object literals. In order to create classical classes like you would in c#, ruby, java, etc, with inheritance and instances.Rather than typical class definitions using object literals, js has a constructor function and the NEW operator that will allow you to new-up a class and optionally provide initial properties to initialize the new object with.The new operator changes the function's context and behavior of the return statement.var Person = function(name) {   this.name = name;};   //Init the personvar dude= new Person("renso");//Validate the instanceassert(dude instanceof Person);When a constructor function is called with the new keyword, the context changes from global window to a new and empty context specific to the instance; "this" will refer in this case to the "dude" context.Here is class pattern that you will need to define your own CLASS emulation library:var Class = function() {   var _class = function() {      this.init.apply(this, arguments);   };   _class.prototype.init = function(){};   return _class;}var Person a new Class();Person.prototype.init = function() {};var person = new Person;In order for the class emulator to support adding functions and properties to static classes as well as object instances of People, change the emulator:var Class = function() {   var _class = function() {      this.init.apply(this, arguments);   };   _class.prototype.init = function(){};   _class.fn = _class.prototype;   _class.fn.parent = _class;   //adding class properties   _class.extend = function(obj) {      var extended = obj.extended;      for(var i in obj) {         _class[i] = obj[i];      };      if(extended) extended(_class);   };   //adding new instances   _class.include = function(obj) {      var included = obj.included;      for(var i in obj) {         _class.fn[i] = obj[i];      };      if(included) included(_class);   };   return _class;}Now you can use it to create and extend your own object instances://adding static functions to the class Personvar Person = new Class();Person.extend({   find: function(name) {/*....*/},      delete: function(id) {/*....*/},});//calling static function findvar person = Person.find('renso');   //adding properties and functions to the class' prototype so that they are available on instances of the class Personvar Person = new Class;Person.extend({   save: function(name) {/*....*/},   delete: function(id) {/*....*/}});var dude = new Person;//calling instance functiondude.save('renso');

    Read the article

  • How not to suffer from ideologists when you're a pragmatic person?

    - by Lukas Eder
    My story: I'm a pragmatic person. Sometimes, the most simple solution to a problem to get the job done is the one that fits best for me, if its not an utter blasphemy and reproach to any design principles. Check out my answer to this question on stackoverflow. Simple. Works. Was accepted. Could be improved. Is clearly not perfect. And along comes this guy. He downvotes me, comments on the question how his answer is better, more accurate etc and calls me "plain wrong". Reminds me of this comic strip. :-) While on stackoverflow I can laugh at these things because those people are far away, in the real world I'm suffering from ideologies every now and then. Heck, I'm not creating a miracle piece of software, I need to keep that huge legacy thing running, and it's an adventure to me every day. I don't have the time or passion to beautify my code (or other people's code) to that extent. My question(s): How do you deal with ideologies / ideologists, when you're a pragmatic person? How do you deal with pragmatism / pragmatists, when you're an ideologic person? I'm interested in both point of views. Tell me your experience. But please, be fair, somewhat objective, and understand that you may NOT be entirely correct and your opinion is NOT the only true one... :-)

    Read the article

  • How do I create a third Person View using DXUTCamera in DX10?

    - by David
    I am creating a 3d flying game and using DXUTCamera for my view. I can get the camera to take on the characters position, But I would like to view my character in the 3rd person. Here is my code for first person view: //Put the camera on the object. D3DXVECTOR3 viewerPos; D3DXVECTOR3 lookAtThis; D3DXVECTOR3 up ( 5.0f, 1.0f, 0.0f ); D3DXVECTOR3 newUp; D3DXMATRIX matView; //Set the viewer's position to the position of the thing. viewerPos.x = character->x; viewerPos.y = character->y; viewerPos.z = character->z; // Create a new vector for the direction for the viewer to look character->setUpWorldMatrix(); D3DXVECTOR3 newDir, lookAtPoint; D3DXVec3TransformCoord(&newDir, &character->initVecDir, &character->matAllRotations); // set lookatpoint D3DXVec3Normalize(&lookAtPoint, &newDir); lookAtPoint.x += viewerPos.x; lookAtPoint.y += viewerPos.y; lookAtPoint.z += viewerPos.z; g_Camera.SetViewParams(&viewerPos, &lookAtPoint); So does anyone have an ideas how I can move the camera to the third person view? preferably timed so there is a smooth action in the camera movement. (I'm hoping I can just edit this code instead of bringing in another camera class)

    Read the article

  • Explain why folder's permissions differ depending on HOW user is accessing server AFP vs SSH

    - by Meltemi
    Hoping someone can explain what is probably fairly obvious...but confuses me. Imagine two users with admin privileges on our server (Mac OS X Server 10.5). Call them joe & bob. both users are members of these groups: Staff Group ID: 20 Workgroup Group ID: 1025 Shared folder "devfolder" has sharing set as so: POSIX: Owner: joe read & write Group: admin read & write Other no access ACL: Workgroup Allow Read & write Question is why when looking at same folder does the ownership appear to change depending on who's doing the looking?!? Both looking at same folder on the server: From Joe's perspective: xserve:devfolder joe$ ls -l drwxrwxr-x 6 joe workgroup 204 May 20 19:32 app drwxrwxr-x 9 joe workgroup 306 May 20 19:32 config drwxrwxr-x 3 joe workgroup 102 May 20 19:32 db drwxrwxr-x 3 joe workgroup 102 May 20 19:32 doc drwxrwxr-x 3 joe workgroup 102 May 20 19:32 lib And from Bob's perspective (folder mounted on his machine via AFP): bobmac:devfolder bob$ ls -l drwxrwxr-x 6 bob _bob 264 May 20 19:32 app drwxrwxr-x 9 bob _bob 264 May 20 19:32 config drwxrwxr-x 3 bob _bob 264 May 20 19:32 db drwxrwxr-x 3 bob _bob 264 May 20 19:32 doc drwxrwxr-x 3 bob _bob 264 May 20 19:32 lib Now if Bob connects to server via SSH then his output is identical to Joe's, as expected. Can anyone tell me what the client is doing in this case and what should be expected when bob creates or updates files in this folder? What tools do I have to better understand this from the command line? Is this normal? Perhaps a "cleaner" way that wouldn't be confusing with "bob _bob"?!?

    Read the article

  • Using XSD to validate node count

    - by heath
    I don't think this is possible but I thought I'd throw it out there. Given this XML: <people count="3"> <person>Bill</person> <person>Joe</person> <person>Susan</person> </people> Is it possible in an XSD to force the @count attribute value to be the correct count of defined elements (in this case, the person element)? The above example would obviously be correct and the below example would not validate: <people count="5"> <person>Bill</person> <person>Joe</person> <person>Susan</person> </people>

    Read the article

  • Objects with permissions assigned by django-guardian not visible in admin

    - by jul
    I'm using django-guardian in order to manage per object permission. For a given user I give permission all permission on one object: joe = User.objects.get(username="joe") mytask = Task.objects.get(pk=1) assign('add_task', joe, mytask) assign('change_task', joe, mytask) assign('delete_task', joe, mytask) and I get, as expected: In [57]: joe.has_perm("add_task", mytask) Out[57]: True In [58]: joe.has_perm("change_task", mytask) Out[58]: True In [59]: joe.has_perm("delete_task", mytask) Out[59]: True In admin.py I also make TaskAdmin inherit from GuardedModelAdmin instead of admin.ModelAdmin Now when I connect to my site with joe, on the admin I get: You don't have permission to edit anything Am I not supposed to be able to edit the object mytask? Do I have to set some permissions using the built-in model-based permission system? Am I missing anything? Thank you

    Read the article

  • Objective-C: fetchManagedObjectsForEntity problem

    - by Meko
    Hi.I am trying to get value from CoreData entity name Person with predicate and then comparing with new data in dictionary.But it it returns every time 0 .And it creates about 5 person with same name. NSPredicate *predicate = [NSPredicate predicateWithFormat:@"userName == %@",[flickr usernameForUserID:@"owner"]]; peopleList = (NSMutableArray *)[flickr fetchManagedObjectsForEntity:@"Person" withPredicate:predicate]; NSEnumerator *enumerator = [peopleList objectEnumerator]; Person *person; BOOL exists = FALSE; while (person = [enumerator nextObject]) { NSLog(@" Person is: %@ ", person.userName); NSLog(@"Person ID IS %@",person.userID); NSLog(@"Dict ID is %@",[dict objectForKey:@"owner"]); if([person.userID isEqualToString:[dict objectForKey:@"owner"]]) { exists = TRUE; NSLog(@"-- Person Exists : %@--", person.userName); [newPhoto setPerson:person]; } } Here peopleList returns 0 and the enumerator also 0 and it does not use if and not comparing.In my entity I have Person and Photo entities.In Person I have userName and userID attributes and also one-to many relationship with Photo entity. I think problem in predicate but i cant figure out it .

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >