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  • Sorting in Hash Maps in Java

    - by Crystal
    I'm trying to get familiar with Collections. I have a String which is my key, email address, and a Person object (firstName, lastName, telephone, email). I read in the Java collections chapter on Sun's webpages that if you had a HashMap and wanted it sorted, you could use a TreeMap. How does this sort work? Is it based on the compareTo() method you have in your Person class? I overrode the compareTo() method in my Person class to sort by lastName. But it isn't working properly and was wondering if I have the right idea or not. getSortedListByLastName at the bottom of this code is where I try to convert to a TreeMap. Also, if this is the correct way to do it, or one of the correct ways to do it, how do I then sort by firstName since my compareTo() is comparing by lastName. import java.util.*; public class OrganizeThis { /** Add a person to the organizer @param p A person object */ public void add(Person p) { staff.put(p.getEmail(), p); //System.out.println("Person " + p + "added"); } /** * Remove a Person from the organizer. * * @param email The email of the person to be removed. */ public void remove(String email) { staff.remove(email); } /** * Remove all contacts from the organizer. * */ public void empty() { staff.clear(); } /** * Find the person stored in the organizer with the email address. * Note, each person will have a unique email address. * * @param email The person email address you are looking for. * */ public Person findByEmail(String email) { Person aPerson = staff.get(email); return aPerson; } /** * Find all persons stored in the organizer with the same last name. * Note, there can be multiple persons with the same last name. * * @param lastName The last name of the persons your are looking for. * */ public Person[] find(String lastName) { ArrayList<Person> names = new ArrayList<Person>(); for (Person s : staff.values()) { if (s.getLastName() == lastName) { names.add(s); } } // Convert ArrayList back to Array Person nameArray[] = new Person[names.size()]; names.toArray(nameArray); return nameArray; } /** * Return all the contact from the orgnizer in * an array sorted by last name. * * @return An array of Person objects. * */ public Person[] getSortedListByLastName() { Map<String, Person> sorted = new TreeMap<String, Person>(staff); ArrayList<Person> sortedArrayList = new ArrayList<Person>(); for (Person s: sorted.values()) { sortedArrayList.add(s); } Person sortedArray[] = new Person[sortedArrayList.size()]; sortedArrayList.toArray(sortedArray); return sortedArray; } private Map<String, Person> staff = new HashMap<String, Person>(); public static void main(String[] args) { OrganizeThis testObj = new OrganizeThis(); Person person1 = new Person("J", "W", "111-222-3333", "[email protected]"); Person person2 = new Person("K", "W", "345-678-9999", "[email protected]"); Person person3 = new Person("Phoebe", "Wang", "322-111-3333", "[email protected]"); Person person4 = new Person("Nermal", "Johnson", "322-342-5555", "[email protected]"); Person person5 = new Person("Apple", "Banana", "123-456-1111", "[email protected]"); testObj.add(person1); testObj.add(person2); testObj.add(person3); testObj.add(person4); testObj.add(person5); System.out.println(testObj.findByEmail("[email protected]")); System.out.println("------------" + '\n'); Person a[] = testObj.find("W"); for (Person p : a) System.out.println(p); System.out.println("------------" + '\n'); a = testObj.find("W"); for (Person p : a) System.out.println(p); System.out.println("SORTED" + '\n'); a = testObj.getSortedListByLastName(); for (Person b : a) { System.out.println(b); } } } Person class: public class Person implements Comparable { String firstName; String lastName; String telephone; String email; public Person() { firstName = ""; lastName = ""; telephone = ""; email = ""; } public Person(String firstName) { this.firstName = firstName; } public Person(String firstName, String lastName, String telephone, String email) { this.firstName = firstName; this.lastName = lastName; this.telephone = telephone; this.email = email; } public String getFirstName() { return firstName; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getLastName() { return lastName; } public void setLastName(String lastName) { this.lastName = lastName; } public String getTelephone() { return telephone; } public void setTelephone(String telephone) { this.telephone = telephone; } public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } public int compareTo(Object o) { String s1 = this.lastName + this.firstName; String s2 = ((Person) o).lastName + ((Person) o).firstName; return s1.compareTo(s2); } public boolean equals(Object otherObject) { // a quick test to see if the objects are identical if (this == otherObject) { return true; } // must return false if the explicit parameter is null if (otherObject == null) { return false; } if (!(otherObject instanceof Person)) { return false; } Person other = (Person) otherObject; return firstName.equals(other.firstName) && lastName.equals(other.lastName) && telephone.equals(other.telephone) && email.equals(other.email); } public int hashCode() { return this.email.toLowerCase().hashCode(); } public String toString() { return getClass().getName() + "[firstName = " + firstName + '\n' + "lastName = " + lastName + '\n' + "telephone = " + telephone + '\n' + "email = " + email + "]"; } }

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  • What would your three most-telling interview questions be for a new hire?

    - by Phil.Wheeler
    I've been asked to interview my company's next junior developer candidate and I want to come up with a couple of questions that will challenge him / her. What are some of the best interview questions you asked a developer candidate that revealed the most about the person's character, ability or nature? These do not necessarily have to be technical questions, but I am after some insight into the person's ability to reason or think fast under pressure or when faced with an unusual problem.

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  • http_post() variables to php script from SMS message but only append data to text file if certain variable (telephone number) equals a known

    - by user1592162
    All, I have an SMS gateway application running on my server that passes incoming SMS messages to a php script. That script writes the data to a text file. Certain aspects of the message can be extracted as variables - for example: Message content: $msgdata Sender: $originator Time: $receivedtime etc... Using the script below I can easily write the message to a text file but it will write the data of every SMS that comes in. I only want to write to a text file if the $originator variable is a certain telephone number (e.g 07123456321) $date = date("d.m.Y H:i:s"); $msgdata = $_REQUEST["msgdata"]; $myFile3 = "test.txt"; $fh = fopen($myFile3, 'a') or die("can't open file"); $stringData3 = "$date - $msgdata\n"; fwrite($fh, $stringData3); fclose($fh); Your assistance is bery much appreciated. Many thanks in advance.

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  • add presence for a remote user to a legacy telephone system?

    - by niko
    we have a small call center that uses an old nortel phone system with analog lines. one of our sales people works from home so her calls do not go through the phone system. this creates a problem at time as the receptionist does not know if she is on the phone or not. we can easily get around this by using instant messenger status but i wanted to ask if there is another way that we can do it so that calls can also be forwarded to her when she is not on the phone. i realize that we can do this with a voip system but we're not planning on upgrading to voip until next year. does anyone know if there is an inexpensive way to add this capability today?

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  • Easiest way to find out if user has either Windows 7 or Vista (through telephone support)?

    - by Rabarberski
    If you have to provide some initial troubleshooting support by phone [or email], and you don't have access to the PC itself, what is the easiest and most foolproof question to find out if the 'dumb' user is using either Windows 7 or Windows Vista? For example: determining if the user has either Windows XP or Windows Vista/7 is easy. Just ask the user if the button at the left bottom corner is (a) either square with the word 'Start' on it, or (b) it is a round button. But how to determine the difference between Vista and 7? Edit: For all the existing answers the user has to type something, and do it correctly. Sometimes even that is already hard for a computer illiterate user. My XP example just requires looking. If it exists (although I am afraid it doesn't), I think a solution that is just based on something this is visually different between Vista and 7 would stand above all others. (Which makes Dan's suggestion to turn over the box and look at the label" not so stupid). Perhaps the small 'show desktop' rectangle at the right side of the task bar (was that present in Vista)?

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  • Podcast Show Notes: Red Room Interview &ndash; Part 3: Ninja BPM

    - by Bob Rhubart
    The third and final segment of my conversation with Red Room bloggers Sean Boiling, Richard Ward, and Mervin Chaing is now available. Listen to Part 1 Listen to Part 2 Listen to Part 3 As you’ll hear, this segment gets its title from another example of Mervin’s tactic for tweaking terminology to make it easier to sell stakeholders on certain SOA concepts. These are some very bright, very knowledgeable guys, so I encourage you to connect with them via the links below to pick their brains on any SOA or related issues that might have you reaching for the aspirin bottle. Sean Boiling - Sales Consulting Manager for Oracle Fusion Middleware LinkedIn | Twitter | Blog Richard Ward - SOA Channel Development Manager at Oracle LinkedIn | Blog Mervin Chiang - Consulting Principal at Leonardo Consulting LinkedIn | Twitter | Blog Once again, you’ll find the complete list of Red Room SOA Best Practice Posts in here. Up Next Next week’s program features another panel discussion recorded during a virtual min meet-up. The panel includes Oracle ACE Directors Mike van Alst (IT-Eye) and Jordan Braunstein (TUSC) along with The Definitive Guide to SOA: Oracle Service Bus author Jeff Davies. Stay tuned: RSS   Technorati Tags: oracle technology network,oracle,archbeat,podcast. arch2arch,soa,bpm del.icio.us Tags: oracle technology network,oracle,archbeat,podcast. arch2arch,soa,bpm

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  • What should I wear to a job interview with a game development company?

    - by Bill
    Many game development companies are less formal in terms of workplace attire than other types of software development houses. For example, I know that one place at which I will be interviewing soon has a predominant workplace culture of jeans and polos or t-shirts. Should I wear a suit? Shirt and tie? Shirt and sport jacket, with or without tie? I want to show that I'm serious about the job, but that I understand the culture, too.

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  • Interview: Nina Paley (author of "Sita Sings the Blues" and the two "Minute ...

    <b>Free Software Magazine:</b> "Nina Paley is a cartoonist and animator who in recent years discovered the serious faults of our long-standing copyright and licensing system through her experiences in making and releasing &#8220;Sita Sings the Blues.&#8221; After considerable trouble involved in fulfilling all of the licensing obligations for the film, she became a serious advocate for both free culture and copyright..."

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  • I wired up a z 80 using telephone wire and put a jump to 0000 0000 0000 0000

    - by john
    I put 1100 0011 0000 0000 0000 0000 in the 2764 eprom --- this is supposed to test the z80 -- I have a 555 timer running at 500 khz. Can this small program work with the z80 ? I looked at the address pins on a m465 oscilloscope. The address shows highs up to 0100 0000. I think it should only count to 0000 0000 0000 0011. Can the z80 be tested? The Santa Clara Valley also made the lm1871 radio control chip that could not show a high or a low without completing the entire rc loop.

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  • how to effectively keep/update postal and telephone code format for each country?

    - by melaos
    hi there, currently we have a table for regex format for phone and postal code for countries that we use to validate when the user register through our forms. but the problem remains on the maintenance on the correctness of these format, thus what's a good way to ensure that we always have the latest copy of this information? is there a web service/etc that i can use to get this? or does it even make sense to keep all these format but instead use a relaxed method to ensure that the user just keys in something which roughly matches the format? the information is used solely for shipping and billing address. we're using asp.net 2.0 btw. thanks ~steve

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  • Can you suggest some UI related flex 3 interview questions for a senior pos?

    - by mohan talluri
    Our Company is looking for a sr.flex developer. As part of interview process customized UI understanding and implementation is also included. I am Usability&design Lead for the same product team have some understanding of flex 3 but am not sure if pure UI/usability questions can be answered by flex developer. So can you suggest some UI related questions to see if he/she has competency to refer a prototype(html/mockup's) and build the same UI in flex.

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  • BlackBerry 10 en images (8/9) : un téléphone ? Non, six ! A tous les prix, avec et sans clavier

    BlackBerry 10 en images (1/9) : BlackBerry Flow RIM dévoile les nouveautés au compte-goutte et promet de très grosses surprises Deux jours après les annonces officielles du PDG de RIM, la filiale Française nous a conviés à une démonstration pour nous dévoiler « en vrai » quelques nouveautés supplémentaires de son prochain BlackBerry 10. « Son plus gros lancement de tous les temps », selon David Derrida, le responsable produit. Les voici en images au moment où le code est officiellement gelé. BlackBerry Flow C'est la nouvelle manière d'interagir avec l'OS. ...

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  • T-SQL Right Joins to ALL Entries inc Selected Column

    - by Pace
    Hi Experts, I have the following Query which produces the output below; SELECT TBLUSERS.USERID, TBLUSERS.ADusername, TBLACCESSLEVELS.ACCESSLEVELID, TBLACCESSLEVELS.AccessLevelName FROM TBLACCESSLEVELS INNER JOIN TBLACCESSRIGHTS ON TBLACCESSLEVELS.ACCESSLEVELID = TBLACCESSRIGHTS.ACCESSLEVELID INNER JOIN TBLUSERS ON TBLACCESSRIGHTS.USERID = TBLUSERS.USERID The output is this; 29 administrator 1 AllUsers 29 administrator 2 JobQueue 29 administrator 3 Telephone Directory Admin 29 administrator 4 Jobqueueadmin 29 administrator 5 UserAdmin 29 administrator 6 Product System 27 alan 1 AllUsers 97 andy 1 AllUsers 26 barry 1 AllUsers 26 barry 2 JobQueue 26 barry 3 Telephone Directory Admin 26 barry 4 Jobqueueadmin 26 barry 5 UserAdmin 26 barry 6 Product System 26 barry 7 Newseditor 26 barry 8 GreetingBoard What I would like to do is modify the query so I get all Access Levels regardless of weather there is an entry for that user. What I would also like to do is some sort of exist case so that I get output like the following; 29 administrator 1 AllUsers True 29 administrator 2 JobQueue True 29 administrator 3 Telephone Directory Admin True 29 administrator 4 Jobqueueadmin True 29 administrator 5 UserAdmin True 29 administrator 6 Product System True 29 administrator 7 Newseditor False 29 administrator 8 GreetingBoard False 27 alan 1 AllUsers True 27 alan 2 JobQueue False 27 alan 3 Telephone Directory Admin False 27 alan 4 Jobqueueadmin False 27 alan 5 UserAdmin False 27 alan 6 Product System False 27 alan 7 Newseditor False 27 alan 8 GreetingBoard False 97 andy 1 AllUsers True 97 andy 2 JobQueue False 97 andy 3 Telephone Directory Admin False 97 andy 4 Jobqueueadmin False 97 andy 5 UserAdmin False 97 andy 6 Product System False 97 andy 7 Newseditor False 97 andy 8 GreetingBoard False 26 Barry 1 AllUsers True 26 Barry 2 JobQueue True 26 Barry 3 Telephone Directory Admin True 26 Barry 4 Jobqueueadmin True 26 Barry 5 UserAdmin True 26 Barry 6 Product System True 26 Barry 7 Newseditor True 26 Barry 8 GreetingBoard True ......................................... So the rules are ALWAYS show ALL Entries for ACCESSLEVELS and where EXISTS in ACCESSRIGHTS produce a true / false to show this. I hope this makes sense and hopefully you dont need the table definitions as everything I need to work with is in the original Query. I just need a way of manipulating it slightly and getting the join in the right place. Thank you. Pace

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  • You Are Hiring But Do Candidate&rsquo;s Want to Work For You

    - by david.talamelli
    So here you are – it has happened, you are now interviewing for that position that you have either applied for or maybe were called about. Whether you are an “active” candidate looking for a job or a “passive” candidate who was contacted about the opportunity, it doesn’t matter now. Regardless of the circumstances of how you got to the interview stage, how you and your new potential manager connect with each other at interview will play a part in whether you are successful in landing that job. The best manager/employee relationships I think tend to be the ones where both the manager and employee have a common goal that they are both working towards and they work together in unison to achieve these goals. Candidates – when you are interviewing for a role, remember that an interview is a two way process. An interview shouldn’t be just a case of a company interviewing you to see if you are a good fit for a certain role. Don’t forget in an interview process it is equally important that you take the opportunity to similarly interview the company to see if that role/company are the right place for you to move to as the next step in your career. I think an interview should not only be a chance for a Hiring Manager to get to better know a candidate and asses his capability and cultural fit for a team/company but it should also be a chance for the candidate to similarly assess a company or manager about whether they are someone that they want to work with. Managers – I know Recruiters have been talking about the “war for talent” since before many of you were managers, but there is no denying it – it exists. You are not only competing with other companies for talented individuals but you are also competing with the existing companies that those talented individuals are working at. Companies are not going to let the people they have identified as superstars resign without a fight (this is the classic Counter Offer scenario which may be another blog post in itself). So how do we get these great people – their current employer will do all they can to keep them, everyone else wants them – does this mean all hope is lost? No, absolutely not. The same reasons that have always existed on why candidates are interested in other opportunities is still there: it could be that someone is looking for career advancement, or they want the chance to work with new technology or maybe you have an opportunity that is exactly what that person is looking to do. As a Hiring Manager don’t just conduct your interviews in question/answer mode. You should talk to that individual to work out what it is they are looking for and you can then relate how your role addresses that. It is potentially going to be the two of you working together so you two are the ones who have to be most comfortable with each other. Don’t oversell the role – set realistic expectations of what that candidate can expect working in your team – give them the good, the bad and the ugly so they can make an informed decision. Manager’s think back to when you last were looking for a job and put yourself in the candidate’s shoes. When you were looking for a job, what was it that you wanted to know about Oracle, or what was it that you wanted more information about. There are some great Business Leaders that work here at Oracle – if you are one of them it is likely that you already are doing all these things anyway. The good news for you is that you are also likely raising yourself head and shoulders above what many interviewers do – that in itself gives you a competitive advantage in this ‘war for talent’ but as a great Business Leader you already know that

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  • Online Classes on ASP .Net for FREE

    Here are some of the FREE upcoming ASP .Net classes at WiZiQ. 1. 5 imp ASP.NET interview questions by Questpond    Sunday, April 11, 2010 12:30 AM (EST)5 imp .NET interview questions by Questpond     Sunday, April 18, 2010 12:30 AM (EST)3. 5 imp ADO.NET interview questions by Questpond     Saturday, April 24, 2010 12:30 AM (EST)4. 5 important .NET interview questions by ...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Pos receipt printer

    - by unknown (google)
    Is it possible to connect a receipt printer to a telephone for printing similar to how a Credit card terminal works when connected to a telephone line. Our clients donot have internet connections where we can connect the printer over Ethernet, so was thinking if it was possible to do the same via telephone line.

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  • Broad Band LAN connection through existing 6 wire phone line ( no phone connected )

    - by Paul Taylor
    I have an (up to but never achieved ) 10 mb broadband signal coming into my house along the telephone line. The modem then connects the broadband signal via a LAN connection and Wi-Fi signal to my computers and iPad. My workshop desk is 54 yards (approx. 50m) downhill from the modem (part of a separate building) – too far to give a good direct signal. The inside corner of the workshop (by a window) receives a weak signal. I have a disconnected telephone cable consisting of 6 wires going from the house to the workshop. The telephone is no longer used in the workshop - we use our mobile number for business calls. A broad band signal using the cable would not have to share with a phone. What would be the most economic price/efficient way to get the broadband signal to the workshop? I am writing in hope that since the telephone didn't need the six wires, an Ethernet connection might be similar and not need all the eight wires for a LAN connection. In theory could use the telephone wire to pull an Ethernet cable trough the underground pipe but I doubt if the cable would survive the strain. The broad band connection in the workshop does not have to have network facility. My skill level is high enough to do house wiring, plumbing and gas fitting; so given any sound advice and a wiring diagram or instructions I can probably work out the rest myself.

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  • Linq To Xml problems using XElement's method Elements(XName)

    - by Dorian McHensie
    Hello everyone. I have a problem using Linq To Xml. A simple code. I have this XML: <?xml version="1.0" encoding="utf-8" ?> <data xmlns="http://www.xxx.com" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.xxx.com/directory file.xsd"> <contact> <name>aaa</name> <email>[email protected]</email> <birthdate>2002-09-22</birthdate> <telephone>000:000000</telephone> <description>Description for this contact</description> </contact> <contact> <name>sss</name> <email>[email protected]</email> <birthdate>2002-09-22</birthdate> <telephone>000:000000</telephone> <description>Description for this contact</description> </contact> <contact> <name>bbb</name> <email>[email protected]</email> <birthdate>2002-09-22</birthdate> <telephone>000:000000</telephone> <description>Description for this contact</description> </contact> <contact> <name>ccc</name> <email>[email protected]</email> <birthdate>2002-09-22</birthdate> <telephone>000:000000</telephone> <description>Description for this contact</description> </contact> I want to get every contact mapping it on an object Contact. To do this I use this fragment of code: XDocument XDoc = XDocument.Load(System.Web.HttpRuntime.AppDomainAppPath + this.filesource); XElement XRoot = XDoc.Root; //XElement XEl = XElement.Load(this.filesource); var results = from e in XRoot.Elements("contact") select new Contact((string)e.Element("name"), (string)e.Element("email"), "1-1-1", null, null); List<Contact> cntcts = new List<Contact>(); foreach (Contact cntct in results) { cntcts.Add(cntct); } Contact[] c = cntcts.ToArray(); // Encapsulating element Elements<Contact> final = new Elements<Contact>(c); Ok just don't mind that all: focus on this: When I get the root node, it is all right, I get it correctly. When I use the select directive I try to get every node saying: from e in XRoot.Elements("contact") OK here's the problem: if I use: from e in XRoot.Elements() I get all contact nodes, but if I use: from e in XRoot.Elements("contact") I GET NOTHING: Empty SET. OK you tell me: Use the other one: OK I DO SO, let's use: from e in XRoot.Elements(), I get all nodes anyway, THAT's RIGHT BUT HERE COMES THE OTHER PROBLEM: When Saying: select new Contact((string)e.Element("name"), (string)e.Element("email"), "1-1-1", null, null); I Try to access <name>, <email>... I HAVE TO USE .Element("name") AND IT DOES NOT WORK TOO!!!!!!!!WHAT THE HELL IS THIS????????????? IT SEEMS THAT I DOES NOT MATCH THE NAME I PASS But how is it possible. I know that Elements() function takes, overloaded, one argument that is an XName which is mapped onto a string. Please consider that the code I wrote come from an example, It should work. Please somebody help me. THANKS IN ADVANCE

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  • Regular Expression Longes Possible Matching

    - by syker
    So I have an input string which is a directory addres: Example: ProgramFiles/Micro/Telephone And I want to match it against a list of words very strictly: Example: Tel|Tele|Telephone I want to match against Telephone and not Tel. Right now my reg looks like this: my( $output ) = ( $input =~ m/($list)/o ); The regex above will match against Tel. What can I do to fix it?

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  • I can't create a view in oracle database using sqlplus (insufficient privileges)

    - by Nubkadiya
    I'm running this SQL: CREATE VIEW showMembersInfo(MemberID,Fname,Lname,Address,DOB,Telephone,NIC,Email,WorkplaceID,WorkName,WorkAddress,WorkTelephone,StartingDate,ExpiryDate,Amount,WitnessID,WitName,WitAddress,WitNIC,WitEmail,WitTelephone) AS SELECT mem.MemberID,mem.FirstName,mem.LastName,mem.Address,mem.DOB,mem.Telephone,mem.NIC,mem.Email, wrk.WorkPlaceID,wrk.Name,wrk.Address,wrk.Telephone, anl.StartingDate,anl.ExpiryDate,anl.Amount, wit.WitnessID,wit.Name,wit.Address,wit.NIC,wit.Email,wit.Telephone FROM Member mem, WorkPlace wrk, AnnualFees anl, Witness wit WHERE mem.MemberID = anl.MemberID AND mem.WorkPlaceID = work.WorkPlaceID AND mem.WitnessID = wit.WitnessID When I try to create the view I get this error: ERROR at line 1: ORA-01031: insufficient privileges Why is that? I'm logged in to sqlplus using sysman

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