Search Results

Search found 1775 results on 71 pages for 'aaron smith'.

Page 3/71 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Rewriting from headers in Postfix

    - by inxilpro
    I want to configure Postfix to replace the 'From' header in all forwarded/aliased messages with a custom email address, and the 'Reply-To' header with the original sender's address. Is that something that can be done with a simple configuration change, or am I looking at a more complex problem? For example: Original Message: From: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Would get translated to: From: "My Email Forwarding Service" <[email protected]> Reply-To: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Ideally, I would also have it rewrite the message body (adding something about how the message was forwarded for them), but I know that's much more difficult. We have a number of email aliases, and everytime someone reports spam they received through their alias, our server gets flagged. I'm trying to minimize that damage as much as possible. Any help is greatly appreciated!

    Read the article

  • Resize Debian in VirtualBox

    - by Poni
    I have a VM with one HD of size 3GB and I'd like to enlarge its HD to 7GB. So I execute this command on the host (while guest is shutdown): VBoxManage modifyhd debian.vdi --resize 7168 Then I run the guest, Debian 6, and then: smith@debian6:~$ df -h Filesystem Size Used Avail Use% Mounted on /dev/sda1 2.8G 2.6G 60M 98% / tmpfs 61M 0 61M 0% /lib/init/rw udev 57M 160K 57M 1% /dev tmpfs 61M 0 61M 0% /dev/shm smith@debian6:~$ sudo parted /dev/sda print Model: ATA VBOX HARDDISK (scsi) Disk /dev/sda: 3221MB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 3035MB 3034MB primary ext3 boot 2 3036MB 3220MB 185MB extended 5 3036MB 3220MB 185MB logical linux-swap(v1) smith@debian6:~$ cat /proc/partitions major minor #blocks name 8 0 3145728 sda 8 1 2962432 sda1 8 2 1 sda2 8 5 180224 sda5 So, no automatic resizing (detection) of the HD/partition (while VirtualBox, in the host, shows it's 7GB now). Ok... Then I do: smith@debian6:~$ sudo resize2fs /dev/sda1 resize2fs 1.41.12 (17-May-2010) The filesystem is already 740608 blocks long. Nothing to do! smith@debian6:~$ sudo parted GNU Parted 2.3 Using /dev/sda Welcome to GNU Parted! Type 'help' to view a list of commands. (parted) select /dev/sda1 Using /dev/sda1 (parted) resize WARNING: you are attempting to use parted to operate on (resize) a file system. parted's file system manipulation code is not as robust as what you'll find in dedicated, file-system-specific packages like e2fsprogs. We recommend you use parted only to manipulate partition tables, whenever possible. Support for performing most operations on most types of file systems will be removed in an upcoming release. Partition number? 1 Start? 0 End? [3034MB]? Here I'm stuck. At the above parted it asks me to resize to 3GB. No point in that, right.. What should I do in order to enlarge this partition?

    Read the article

  • Script to determine if you should update Build version

    - by NeilHambly
    Aaron Betrand has posted a great article on the Patch Tuesday Security Bulletin and I have quickly translated that into a SQL script to check your version and advise what you should be doing http://www.microsoft.com/technet/security/Bulletin/MS11-049.mspx Aaron's article: http://sqlblog.com/blogs/aaron_bertrand/archive/2011/06/14/security-updates-for-all-supported-versions-of-sql-server.aspx#comments Naturally ANY Script needs to be carefully vetted before it is used in your own environments;...(read more)

    Read the article

  • OpenLDAP Authentication UID vs CN issues

    - by user145457
    I'm having trouble authenticating services using uid for authentication, which I thought was the standard method for authentication on the user. So basically, my users are added in ldap like this: # jsmith, Users, example.com dn: uid=jsmith,ou=Users,dc=example,dc=com uidNumber: 10003 loginShell: /bin/bash sn: Smith mail: [email protected] homeDirectory: /home/jsmith displayName: John Smith givenName: John uid: jsmith gecos: John Smith gidNumber: 10000 cn: John Smith title: System Administrator But when I try to authenticate using typical webapps or services like this: jsmith password I get: ldapsearch -x -h ldap.example.com -D "cn=jsmith,ou=Users,dc=example,dc=com" -W -b "dc=example,dc=com" Enter LDAP Password: ldap_bind: Invalid credentials (49) But if I use: ldapsearch -x -h ldap.example.com -D "uid=jsmith,ou=Users,dc=example,dc=com" -W -b "dc=example,dc=com" It works. HOWEVER...most webapps and authentication methods seem to use another method. So on a webapp I'm using, unless I specify the user as: uid=smith,ou=users,dc=example,dc=com Nothing works. In the webapp I just need users to put: jsmith in the user field. Keep in mind my ldap is using the "new" cn=config method of storing settings. So if someone has an obvious ldif I'm missing please provide. Let me know if you need further info. This is openldap on ubuntu 12.04. Thanks, Dave

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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!

    Read the article

  • 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 :-)

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • Apache mod_rewrite for multiple domains to SSL

    - by Aaron Vegh
    Hi there, I'm running a web service that will allow people to create their own "instances" of my application, running under their own domain. These people will create an A record to forward a subdomain of their main domain to my server. The problem is that my server runs everything under SSL. So in my configuration for port 80, I have the following: <VirtualHost *:80> ServerName mydomain.com ServerAlias www.mydomain.com RewriteEngine On RewriteCond %{HTTPS} !=on RewriteRule /(.*) https://mydomain.com/$1 [R=301] </VirtualHost> This has worked well to forward all requests from the http: to https: domain. But as I said, I now need to let any domain automatically forward to the secure version of itself. Is there a rewrite rule that will let me take the incoming domain and rewrite it to the https version of same? So that the following matches would occur: http://some.otherdomain.com -> https://some.otherdomain.com http://evenanotherdomain.com -> https://evenanotherdomain.com Thanks for your help! Apache mod_rewrite makes my brain hurt. Aaron.

    Read the article

  • Help with setting up AIM account

    - by Aaron
    Can anyone help me configure my AIM account? I have Oneiric and I want to set up my AIM account with what I believe is a chat client (the envelope icon on the top right). It is called Empathy. \n I tried to do this using Broadcast Account. It let me choose what account (Jabber, Pidgin, Aim, etc...). Once I entered my password for my AIM account, I got a pop-up asking for the master password for my Keyring. I didn't know what that was at the time so I closed that window after trying to enter my account password. Keyring apparently asks for a master password and it holds any keys you want to remember in the future. It gave me an error so I couldn't set up my AIM account completely. Now I'm trying to get back to that screen but I can only set up a Twitter or Facebook. Thanks, Aaron P.S. Can anyone tell me how to break the message up so it doesn't appear all on one line? I tried 'coding' a \n...seemed to work.

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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!

    Read the article

  • 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.

    Read the article

  • 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.)

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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 :)

    Read the article

  • 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.

    Read the article

  • UIViewerTableViewController.m:15: error: expected identifier before '*' token

    - by Aaron Levin
    I am new to objective-c programming. I come from a C# background. I am having problems with the following code I am writing for a proof of concept for an iPhone App: I am getting a number of compile errors but I think they are all due to the first error (could be wrong) - error: expected identifier before '*' token (@synthesize *lists; in the .m file) I'm not sure why my code is showing up the way it is in the view below the editor..hmm any way, any help would be appreciated. .m file // // Created by Aaron Levin on 4/19/10. // Copyright 2010 RonStan. All rights reserved. // import "UIViewerTableViewController.h" @implementation UIViewerTableViewController @synthesize *lists; @synthesize *icon; (void)dealloc { [Lists release]; [super dealloc]; } pragma mark Table View Methods //Customize number of rows in table view (NSInteger)tableView:(UITableView *) tableView numberOfRowsInSection: (NSInteger) section{ return self.Lists.Count; } //Customize the appearence of table view cells (UITableViewCell *) tableView(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *) indexPath{ static NSString *CellIdentifier = @"Cell"; UITableView *Cell = [tablevView dequeueReusableCellWithIdentifier:CellIdentifier]; if(cell == nil){ cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease]; } cell.textLabel.text = [[self.Lists objectAtIndex:indexPath.row] retain]; cell.imageView = self.Icon; cell.accessoryType = UITableViewCellAccessoryDisclosureIndicator; } @end .h file // // UIMyCardsTableViewController.h // MCS ProtoType v0.1 // // Created by Aaron Levin on 4/19/10. // Copyright 2010 RonStan. All rights reserved. // import @interface UIViewerTableViewController : UITableViewController { NSArray *lists; UIImage *icon; } @property (nonatomic,retain) NSArray *lists; @property (nonatomic,retain) UIImage *icon; @end

    Read the article

  • Using only alphanumeric characters(a-z) inside toCharArray

    - by Aaron
    Below you will find me using toCharArray in order to send a string to array. I then MOVE the value of the letter using a for statement... for(i = 0; i < letter.length; i++){ letter[i] += (shiftCode); System.out.print(letter[i]); } However, when I use shiftCode to move the value such as... a shifted by -1; I get a symbol @. Is there a way to send the string to shiftCode or tell shiftCode to ONLY use letters? I need it to see my text, like "aaron", and when I use the for statement iterate through a-z only and ignore all symbols and numbers. I THINK it is as simple as... letter=codeWord.toCharArray(a,z); But trying different forms of that and googling it didn't give me any results. Perhaps it has to do with regex or something? Below you will find a complete copy of my program; it works exactly how I want it to do; but it iterates through letters and symbols. I also tried finding instructions online for toCharArray but if there exists any arguments I can't locate them. My program... import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /* * Aaron L. Jones * CS219 * AaronJonesProg3 * * This program is designed to - * Work as a Ceasar Cipher */ /** * * Aaron Jones */ public class AaronJonesProg3 { static String codeWord; static int shiftCode; static int i; static char[] letter; /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { // Instantiating that Buffer Class // We are going to use this to read data from the user; in buffer // For performance related reasons BufferedReader reader; // Building the reader variable here // Just a basic input buffer (Holds things for us) reader = new BufferedReader(new InputStreamReader(System.in)); // Java speaks to us here / We get it to query our user System.out.print("Please enter text to encrypt: "); // Try to get their input here try { // Get their codeword using the reader codeWord = reader.readLine(); // Make that input upper case codeWord = codeWord.toUpperCase(); // Cut the white space out codeWord = codeWord.replaceAll("\\s",""); // Make it all a character array letter = codeWord.toCharArray(); } // If they messed up the input we let them know here and end the prog. catch(Throwable t) { System.out.println(t.toString()); System.out.println("You broke it. But you impressed me because" + "I don't know how you did it!"); } // Java Speaks / Lets get their desired shift value System.out.print("Please enter the shift value: "); // Try for their input try { // We get their number here shiftCode = Integer.parseInt(reader.readLine()); } // Again; if the user broke it. We let them know. catch(java.lang.NumberFormatException ioe) { System.out.println(ioe.toString()); System.out.println("How did you break this? Use a number next time!"); } for(i = 0; i < letter.length; i++){ letter[i] += (shiftCode); System.out.print(letter[i]); } System.out.println(); /**************************************************************** **************************************************************** ***************************************************************/ // Java speaks to us here / We get it to query our user System.out.print("Please enter text to decrypt: "); // Try to get their input here try { // Get their codeword using the reader codeWord = reader.readLine(); // Make that input upper case codeWord = codeWord.toUpperCase(); // Cut the white space out codeWord = codeWord.replaceAll("\\s",""); // Make it all a character array letter = codeWord.toCharArray(); } // If they messed up the input we let them know here and end the prog. catch(Throwable t) { System.out.println(t.toString()); System.out.println("You broke it. But you impressed me because" + "I don't know how you did it!"); } // Java Speaks / Lets get their desired shift value System.out.print("Please enter the shift value: "); // Try for their input try { // We get their number here shiftCode = Integer.parseInt(reader.readLine()); } // Again; if the user broke it. We let them know. catch(java.lang.NumberFormatException ioe) { System.out.println(ioe.toString()); System.out.println("How did you break this? Use a number next time!"); } for(i = 0; i < letter.length; i++){ letter[i] += (shiftCode); System.out.print(letter[i]); } System.out.println(); } }

    Read the article

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