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  • Convince developer to use IDE

    - by artjom
    There is a developer, lets call him John (currently on probationary period) in company(pretty small company approx. 10 persons, 3 developers, one of them works long in this company know business process around and can be consider as Team leader) who didn't want to use any IDE at all(he is using some text editor). Application this team working on is medium size Java application with Spring Hibernate technology stack and refactoring/adding new features to launch new version of that application in near future. John performance working without IDE on this application is lower then desirable, team leader's (lets call him Bill) assumption is this happens because John is not using IDE. Bill try to persuade John to use IDE, but this idea meets a lot of resistance and main reason is "I want to be in total control of what I am doing, so I need to write all code by myself". How can Bill convince John to try to use IDE? (considering the fact what Bill already protected John from company owner several complaints about John performance)

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  • DOS batch command to read some info from text file

    - by Ray
    Hello All, I am trying to read some info from a text file by using windows command line, and save it to a variable just like "set info =1234" Below is the content of the txt file, actually I just need the revision number, and the location of it is always the same line 5, and from column 11 to 15. In the sample it's 1234, and I am wondering is there a way to save it to a variable in Dos command line. Thanks a lot! svninfo.txt: Path: . URL: https://www.abc.com Repository Root: https://www.abc.com/svn Repository UUID: 12345678-8b61-fa43-97dc-123456789 Revision: 1234 Node Kind: directory Schedule: normal Last Changed Author: abc Last Changed Rev: 1234 Last Changed Date: 2010-04-01 18:19:54 -0700 (Thu, 01 Apr 2010)

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  • Search 2 Columns with 1 Input Field

    - by Norbert
    I have a db with two columns: first name and last name. The first name can have multiple words. Last name can contain hyphenated words. Is there a way to search both columns with only one input box? Database ID `First Name` `Last Name` 1 John Peter Doe 2 John Fubar 3 Michael Doe Search john peter returns id 1 john returns id 1,2 doe returns id 1,3 john doe returns id 1 peter john returns id 1 peter doe returns id 1 doe john returns id 1 I previously tried the following. Searching for John Doe: SELECT * FROM names WHERE ( `first` LIKE '%john%' OR `first` LIKE '%doe%' OR `last` LIKE '%john%' OR `last` LIKE '%doe%' ) which returns both 1 and 3

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  • Need help tuning a SQL statement

    - by jeffself
    I've got a table that has two fields (custno and custno2) that need to be searched from a query. I didn't design this table, so don't scream at me. :-) I need to find all records where either the custno or custno2 matches the value returned from a query on the same table based on a titleno. In other words, the user types in 1234 for the titleno. My query searches the table to find the custno associated with the titleno. It also looks for the custno2 for that titleno. Then it needs to do a search on the same table for all other records that have either the custno or custno2 returned in the previous search in the custno or custno2 fields for those other records. Here is what I've come up with: SELECT BILLYR, BILLNO, TITLENO, VINID, TAXPAID, DUEDATE, DATEPIF, PROPDESC FROM TRCDBA.BILLSPAID WHERE CUSTNO IN (select custno from trcdba.billspaid where titleno = '1234' union select custno2 from trcdba.billspaid where titleno = '1234' and custno2 != '') OR CUSTNO2 IN (select custno from trcdba.billspaid where titleno = '1234' union select custno2 from trcdba.billspaid where titleno = '1234' and custno2 != '') The query takes about 5-10 seconds to return data. Can it be rewritten to work faster?

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  • sql query need a help.

    - by benjamin button
    If i have a table with two fields.customer id and order. let's say i have in total order ID 1,2,3,4 all the customer can have all the four orders.like below 1234 1 1234 2 1234 3 1234 4 3245 3 3245 4 5436 2 5436 4 you can see above that 3245 customer doesnt have order id 1 and 2. how could i print in the query output like 3245 1 3245 2 5436 1 5436 3 EDIT: i dont order table but i have list of order's like we can hard code it in the query(1,2,3,4) i dont have an orders table.

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  • Socket left in TIME_WAIT after file transfer via netcat

    - by com
    Using Copying by NetCat I am trying to copy files throught network by NetCat. From console it work pretty well. First I run listening netcat on the destination machine and after I run sending on source machine. The problem is it's doen't work from script from the source machine: ssh -f user@$desthost 'nc -l 1234 | tar xvf - /dev/null &' #listening on destination host tar cv /tmp/file | nc $desthost 1234 #sending to destination host I saw that after running port 1234 is still was open and status of the socket was TIME_WAIT. If you know what's the problem, please, help me out. And by the way, after copying how can I validate that the content is identical? Thanks! Addendum: I found one very strange thing, the same implementation with screen on destination work works, but not stable, sometimes it doesn't copy a file. ssh user@$desthost screen -dm -S test 'nc -l 1234 | tar xvf - ' #listening on destination host Maybe there is an issue with timeout?

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  • rails 3, active record: any way to tell how many unique values match a "x LIKE ?" query

    - by jpwynn
    I have a query to find all the phone numbers that match a partial expression such as "ends with 234" @matchingphones = Calls.find :all, :conditions => [ "(thephonenumber LIKE ?)", "%234"] The same phone number might be in the database several times, and so might be returned multiple times by this query if it matches. What I need is to know is UNIQUE phone numbers the query returns. For example if the database contains 000-111-1234 * 000-111-3333 000-111-2234 * 000-111-1234 * 000-111-4444 the existing query will return the 3 records marked with * (eg returns one phone number -1234 twice since it's in the database twice) what I need is a query that returns just once instance of each match, in this case 000-111-1234 * 000-111-2234 *

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  • Fill in missing values in a SELECT statement

    - by benjamin button
    If i have a table with two fields.customer id and order. let's say i have in total order ID 1,2,3,4 all the customer can have all the four orders.like below 1234 1 1234 2 1234 3 1234 4 3245 3 3245 4 5436 2 5436 4 you can see above that 3245 customer doesnt have order id 1 and 2. how could i print in the query output like 3245 1 3245 2 5436 1 5436 3 EDIT: i dont have order table but i have list of order's like we can hard code it in the query(1,2,3,4) i dont have an orders table.

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  • TPROXY Not working with HAProxy, Ubuntu 14.04

    - by Nyxynyx
    I'm trying to use HAProxy as a fully transparent proxy using TPROXY in Ubuntu 14.04. HAProxy will be setup on the first server with eth1 111.111.250.250 and eth0 10.111.128.134. The single balanced server has eth1 and eth0 as well. eth1 is the public facing network interface while eth0 is for the private network which both servers are in. Problem: I'm able to connect to the balanced server's port 1234 directly (via eth1) but am not able to reach the balanced server via Haproxy port 1234 (which redirects to 1234 via eth0). Am I missing out something in this configuration? On the HAProxy server The current kernel is: Linux extremehash-lb2 3.13.0-24-generic #46-Ubuntu SMP Thu Apr 10 19:11:08 UTC 2014 x86_64 x86_64 x86_64 GNU/Linux The kernel appears to have TPROXY support: # grep TPROXY /boot/config-3.13.0-24-generic CONFIG_NETFILTER_XT_TARGET_TPROXY=m HAProxy was compiled with TPROXY support: haproxy -vv HA-Proxy version 1.5.3 2014/07/25 Copyright 2000-2014 Willy Tarreau <[email protected]> Build options : TARGET = linux26 CPU = x86_64 CC = gcc CFLAGS = -g -fno-strict-aliasing OPTIONS = USE_LINUX_TPROXY=1 USE_LIBCRYPT=1 USE_STATIC_PCRE=1 Default settings : maxconn = 2000, bufsize = 16384, maxrewrite = 8192, maxpollevents = 200 Encrypted password support via crypt(3): yes Built without zlib support (USE_ZLIB not set) Compression algorithms supported : identity Built without OpenSSL support (USE_OPENSSL not set) Built with PCRE version : 8.31 2012-07-06 PCRE library supports JIT : no (USE_PCRE_JIT not set) Built with transparent proxy support using: IP_TRANSPARENT IPV6_TRANSPARENT IP_FREEBIND Available polling systems : epoll : pref=300, test result OK poll : pref=200, test result OK select : pref=150, test result OK Total: 3 (3 usable), will use epoll. In /etc/haproxy/haproxy.cfg, I've configured a port to have the following options: listen test1235 :1234 mode tcp option tcplog balance leastconn source 0.0.0.0 usesrc clientip server balanced1 10.111.163.76:1234 check inter 5s rise 2 fall 4 weight 4 On the balanced server In /etc/networking/interfaces I've set the gateway for eth0 to be the HAProxy box 10.111.128.134 and restarted networking. auto eth0 eth1 iface eth0 inet static address 111.111.250.250 netmask 255.255.224.0 gateway 111.131.224.1 dns-nameservers 8.8.4.4 8.8.8.8 209.244.0.3 iface eth1 inet static address 10.111.163.76 netmask 255.255.0.0 gateway 10.111.128.134 ip route gives: default via 111.111.224.1 dev eth0 10.111.0.0/16 dev eth1 proto kernel scope link src 10.111.163.76 111.111.224.0/19 dev eth0 proto kernel scope link src 111.111.250.250

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  • Encrypting peer-to-peer application with iptables and stunnel

    - by Jonathan Oliver
    I'm running legacy applications in which I do not have access to the source code. These components talk to each other using plaintext on a particular port. I would like to be able to secure the communications between the two or more nodes using something like stunnel to facilitate peer-to-peer communication rather than using a more traditional (and centralized) VPN package like OpenVPN, etc. Ideally, the traffic flow would go like this: app@hostA:1234 tries to open a TCP connection to app@hostB:1234. iptables captures and redirects the traffic on port 1234 to stunnel running on hostA at port 5678. stunnel@hostA negotiates and establishes a connection with stunnel@hostB:4567. stunnel@hostB forwards any decrypted traffic to app@hostB:1234. In essence, I'm trying to set this up to where any outbound traffic (generated on the local machine) to port N forwards through stunnel to port N+1, and the receiving side receives on port N+1, decrypts, and forwards to the local application at port N. I'm not particularly concerned about losing the hostA origin IP address/machine identity when stunnel@hostB forwards to app@hostB because the communications payload contains identifying information. The other trick in this is that normally with stunnel you have a client/server architecture. But this application is much more P2P because nodes can come and go dynamically and hard-coding some kind of "connection = hostN:port" in the stunnel configuration won't work.

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  • How to get apache to look for files in different subfolders folders?

    - by prb
    I am definitely new to mod-rewrite stuff. Note:- here the URL is common, and all the folders and subfolders on same host. The url a user uses to access their page is http://myurl.com/1234/filename.jpg Here the name of the subfolder is an integer is unique and generated dynamically by another application. The subfolder stores images specific to an individual user. So the folder structure is as follows main1 = document root main2 is another folder within main1 or document root. /main1/1234/filename.jpg /main1/5678/filename.jpg /main1/2345/filename.jpg /main1/1212/filename.jpg /main1/main2/2367/filename.jpg /main1/main2/8790/filename.jpg /main1/main2/9966/filename.jpg So, I want to write a rewrite a rule so that if a user tries to type in http://myurl.com/1234/filename.jpg, the rewrite rule will need to look where the file is and serve the request; so for request http:/myurl.com/1234/filename.jpg the actual page is located at /main1/1234/filename.jpg and then need to serve that page from that folder. So, if another users makes a request as http://myurl.com/9966/filename.jpg, it should serve the page from the following destination /main1/main2/9966/filename.jpg Please let me know if the question is still not clear. This is what i have done so far and does not work at all. RewriteCond {DOCUMENT_ROOT}/%{REQUEST_FILENAME} -f RewriteRule ^(.*)$ {DOCUMENT_ROOT}/$1 [L] RewriteCond {DOCUMENT_ROOT}/main2/%{REQUEST_FILENAME} -f RewriteRule ^(.*)$ {DOCUMENT_ROOT}/main2/$1 [L] any help is really grateful

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  • "ldap_add: Naming violation (64)" error when configuring OpenLDAP

    - by user3215
    I am following the Ubuntu server guide to configure OpenLDAP on an Ubuntu 10.04 server, but can not get it to work. When I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file: # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: john[email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Can anyone help me?

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  • How to Configure OpenLDAP on Ubuntu 10.04 Server

    - by user3215
    I am following the Ubuntu server guide to configure OpenLDAP on an Ubuntu 10.04 server, but can not get it to work. When I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file: # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: john[email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Can anyone help me?

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  • How to configure ldap on ubuntu 10.04 server

    - by user3215
    I am following the link to configure ldap on ubuntu 10.04 server but could not. when I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: john[email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Anybody could help me?

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  • Select From MySQL PHP

    - by Liju
    Sir, I have one Database Table named "table1" with 8 column, that is Date, Time, Name, t1, t2, t3, t4, t5. I want to update the same table like the following... my existing table:- Date Time Name t1 t2 t3 t4 t5 10/11/2010 08:00 bob 10/11/2010 09:00 bob 10/11/2010 10:00 bob 10/11/2010 13:00 bob 10/11/2010 10:00 john 10/11/2010 12:00 john 10/11/2010 14:00 john 12/11/2010 08:00 bob 12/11/2010 09:00 bob 12/11/2010 10:00 bob 12/11/2010 13:00 bob 12/11/2010 10:00 john 12/11/2010 12:00 john 12/11/2010 14:00 john 12/11/2010 16:00 john I want to update this as follows :- Date Time Name t1 t2 t3 t4 t5 10/11/2010 08:00 bob 08:00 09:00 10:00 13:00 10/11/2010 10:00 john 10:00 12:00 14:00 12/11/2010 08:00 bob 08:00 09:00 10:00 13:00 12/11/2010 10:00 john 10:00 12:00 14:00 16:00 is it posible to update like this please help me.. Liju

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  • Select and copy to MySQL table PHP

    - by Liju
    Can insert the table1 value to Table2 like the follows.. based on Name Date. Table1 Id Date Name time 1 20/11/2010 Tom 08:00 2 20/11/2010 Tom 08:30 3 20/11/2010 Tom 09:00 4 20/11/2010 Tom 09:30 5 20/11/2010 Tom 10:00 6 20/11/2010 Tom 10:30 7 20/11/2010 Tom 11:30 8 20/11/2010 Tom 14:30 9 20/11/2010 John 08:10 10 20/11/2010 John 09:30 11 20/11/2010 John 11:00 12 20/11/2010 John 13:00 13 20/11/2010 John 14:30 14 20/11/2010 John 16:00 15 20/11/2010 John 17:30 16 20/11/2010 John 19:00 17 20/11/2010 Ram 08:05 18 20/11/2010 Ram 08:30 19 20/11/2010 Ram 09:00 20 20/11/2010 Ram 09:45 21 20/11/2010 Ram 12:00 22 20/11/2010 Ram 13:30 23 20/11/2010 Ram 15:00 Table2 Id Date Name Time In1 Time Out1 Time In1 Time Out1 Time In1 Time Out1 Time In4 Time Out4 1 20/11/2010 Tom 08:00 08:30 09:00 09:30 10:00 10:30 11:30 14:30 2 20/11/2011 John 08:10 09:30 11:00 13:00 14:30 16:00 17:30 19:00 3 20/11/2012 Ram 08:05 08:30 09:00 09:45 12:00 13:30 15:00 Null Help me Please... Liju

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  • How to tell endianness from this output?

    - by Nick Rosencrantz
    I'm running this example program and I'm suppossed to be able to tell from the output what machine type it is. I'm certain it's from inspecting one or two values but how should I perform this inspection? /* pointers.c - Test pointers * Written 2012 by F Lundevall * Copyright abandoned. This file is in the public domain. * * To make this program work on as many systems as possible, * addresses are converted to unsigned long when printed. * The 'l' in formatting-codes %ld and %lx means a long operand. */ #include <stdio.h> #include <stdlib.h> int * ip; /* Declare a pointer to int, a.k.a. int pointer. */ char * cp; /* Pointer to char, a.k.a. char pointer. */ /* Declare fp as a pointer to function, where that function * has one parameter of type int and returns an int. * Use cdecl to get the syntax right, http://cdecl.org/ */ int ( *fp )( int ); int val1 = 111111; int val2 = 222222; int ia[ 17 ]; /* Declare an array of 17 ints, numbered 0 through 16. */ char ca[ 17 ]; /* Declare an array of 17 chars. */ int fun( int parm ) { printf( "Function fun called with parameter %d\n", parm ); return( parm + 1 ); } /* Main function. */ int main() { printf( "Message PT.01 from pointers.c: Hello, pointy World!\n" ); /* Do some assignments. */ ip = &val1; cp = &val2; /* The compiler should warn you about this. */ fp = fun; ia[ 0 ] = 11; /* First element. */ ia[ 1 ] = 17; ia[ 2 ] = 3; ia[ 16 ] = 58; /* Last element. */ ca[ 0 ] = 11; /* First element. */ ca[ 1 ] = 17; ca[ 2 ] = 3; ca[ 16 ] = 58; /* Last element. */ printf( "PT.02: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); printf( "PT.03: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); printf( "PT.04: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.05: Dereference pointer ip and we find: %d \n", *ip ); printf( "PT.06: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.07: Dereference pointer cp and we find: %d \n", *cp ); *ip = 1234; printf( "\nPT.08: Executed *ip = 1234; \n" ); printf( "PT.09: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); printf( "PT.10: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.11: Dereference pointer ip and we find: %d \n", *ip ); printf( "PT.12: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); *cp = 1234; /* The compiler should warn you about this. */ printf( "\nPT.13: Executed *cp = 1234; \n" ); printf( "PT.14: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); printf( "PT.15: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.16: Dereference pointer cp and we find: %d \n", *cp ); printf( "PT.17: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); ip = ia; printf( "\nPT.18: Executed ip = ia; \n" ); printf( "PT.19: ia[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ia[0], ia[0], ia[0] ); printf( "PT.20: ia[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ia[1], ia[1], ia[1] ); printf( "PT.21: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.22: Dereference pointer ip and we find: %d \n", *ip ); ip = ip + 1; /* add 1 to pointer */ printf( "\nPT.23: Executed ip = ip + 1; \n" ); printf( "PT.24: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.25: Dereference pointer ip and we find: %d \n", *ip ); cp = ca; printf( "\nPT.26: Executed cp = ca; \n" ); printf( "PT.27: ca[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[0], ca[0], ca[0] ); printf( "PT.28: ca[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[1], ca[1], ca[1] ); printf( "PT.29: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.30: Dereference pointer cp and we find: %d \n", *cp ); cp = cp + 1; /* add 1 to pointer */ printf( "\nPT.31: Executed cp = cp + 1; \n" ); printf( "PT.32: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.33: Dereference pointer cp and we find: %d \n", *cp ); ip = ca; /* The compiler should warn you about this. */ printf( "\nPT.34: Executed ip = ca; \n" ); printf( "PT.35: ca[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[0], ca[0], ca[0] ); printf( "PT.36: ca[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[1], ca[1], ca[1] ); printf( "PT.37: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.38: Dereference pointer ip and we find: %d \n", *ip ); cp = ia; /* The compiler should warn you about this. */ printf( "\nPT.39: Executed cp = ia; \n" ); printf( "PT.40: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.41: Dereference pointer cp and we find: %d \n", *cp ); printf( "\nPT.42: fp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &fp, (long) fp, (long) fp ); printf( "PT.43: Dereference fp and see what happens.\n" ); val1 = (*fp)(42); printf( "PT.44: Executed val1 = (*fp)(42); \n" ); printf( "PT.45: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); return( 0 ); } Output Message PT.01 from pointers.c: Hello, pointy World! PT.02: val1: stored at 21e50 (hex); value is 111111 (dec), 1b207 (hex) PT.03: val2: stored at 21e54 (hex); value is 222222 (dec), 3640e (hex) PT.04: ip: stored at 21eb8 (hex); value is 138832 (dec), 21e50 (hex) PT.05: Dereference pointer ip and we find: 111111 PT.06: cp: stored at 21e6c (hex); value is 138836 (dec), 21e54 (hex) PT.07: Dereference pointer cp and we find: 0 PT.08: Executed *ip = 1234; PT.09: val1: stored at 21e50 (hex); value is 1234 (dec), 4d2 (hex) PT.10: ip: stored at 21eb8 (hex); value is 138832 (dec), 21e50 (hex) PT.11: Dereference pointer ip and we find: 1234 PT.12: val1: stored at 21e50 (hex); value is 1234 (dec), 4d2 (hex) PT.13: Executed *cp = 1234; PT.14: val2: stored at 21e54 (hex); value is -771529714 (dec), d203640e (hex) PT.15: cp: stored at 21e6c (hex); value is 138836 (dec), 21e54 (hex) PT.16: Dereference pointer cp and we find: -46 PT.17: val2: stored at 21e54 (hex); value is -771529714 (dec), d203640e (hex) PT.18: Executed ip = ia; PT.19: ia[0]: stored at 21e74 (hex); value is 11 (dec), b (hex) PT.20: ia[1]: stored at 21e78 (hex); value is 17 (dec), 11 (hex) PT.21: ip: stored at 21eb8 (hex); value is 138868 (dec), 21e74 (hex) PT.22: Dereference pointer ip and we find: 11 PT.23: Executed ip = ip + 1; PT.24: ip: stored at 21eb8 (hex); value is 138872 (dec), 21e78 (hex) PT.25: Dereference pointer ip and we find: 17 PT.26: Executed cp = ca; PT.27: ca[0]: stored at 21e58 (hex); value is 11 (dec), b (hex) PT.28: ca[1]: stored at 21e59 (hex); value is 17 (dec), 11 (hex) PT.29: cp: stored at 21e6c (hex); value is 138840 (dec), 21e58 (hex) PT.30: Dereference pointer cp and we find: 11 PT.31: Executed cp = cp + 1; PT.32: cp: stored at 21e6c (hex); value is 138841 (dec), 21e59 (hex) PT.33: Dereference pointer cp and we find: 17 PT.34: Executed ip = ca; PT.35: ca[0]: stored at 21e58 (hex); value is 11 (dec), b (hex) PT.36: ca[1]: stored at 21e59 (hex); value is 17 (dec), 11 (hex) PT.37: ip: stored at 21eb8 (hex); value is 138840 (dec), 21e58 (hex) PT.38: Dereference pointer ip and we find: 185664256 PT.39: Executed cp = ia; PT.40: cp: stored at 21e6c (hex); value is 138868 (dec), 21e74 (hex) PT.41: Dereference pointer cp and we find: 0 PT.42: fp: stored at 21e70 (hex); value is 69288 (dec), 10ea8 (hex) PT.43: Dereference fp and see what happens. Function fun called with parameter 42 PT.44: Executed val1 = (*fp)(42); PT.45: val1: stored at 21e50 (hex); value is 43 (dec), 2b (hex)

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  • SQL Server: What locale should be used to format numeric values into SQL Server format?

    - by Ian Boyd
    It seems that SQL Server does not accept numbers formatted using any particular locale. It also doesn't support locales that have digits other than 0-9. For example, if the current locale is bengali, then the number 123456789 would come out as "?????????". And that's just the digits, nevermind what the digit grouping would be. But the same problem happens for numbers in the Invariant locale, which formats numbers as "123,456,789", which SQL Server won't accept. Is there a culture that matches what SQL Server accepts for numeric values? Or will i have to create some custom "sql server" culture, generating rules for that culture myself from lower level formatting routines? If i was in .NET (which i'm not), i could peruse the Standard Numeric Format strings. Of the format codes available in .NET: c (Currency): $123.46 d (Decimal): 1234 e (Exponentional): 1.052033E+003 f (Fixed Point): 1234.57 g (General): 123.456 n (Number): 1,234.57 p (Percent): 100.00 % r (Round Trip): 123456789.12345678 x (Hexadecimal): FF Only 6 accept all numeric types: c (Currency): $123.46 d (Decimal): 1234 e (Exponentional): 1.052033E+003 f (Fixed Point): 1234.57 g (General): 123.456 n (Number): 1,234.57 p (Percent): 100.00 % r (Round Trip): 123456789.12345678 x (Hexadecimal): FF And of those only 2 generate string representations, in the en-US locale anyway, that would be accepted by SQL Server: c (Currency): $123.46 d (Decimal): 1234 e (Exponentional): 1.052033E+003 f (Fixed Point): 1234.57 g (General): 123.456 n (Number): 1,234.57 p (Percent): 100.00 % r (Round Trip): 123456789.12345678 x (Hexadecimal): FF Of the remaining two, fixed is dependant on the locale's digits, rather than the number being used, leaving General g format: c (Currency): $123.46 d (Decimal): 1234 e (Exponentional): 1.052033E+003 f (Fixed Point): 1234.57 g (General): 123.456 n (Number): 1,234.57 p (Percent): 100.00 % r (Round Trip): 123456789.12345678 x (Hexadecimal): FF And i can't even say for certain that the g format won't add digit groupings (e.g. 1,234). Is there a locale that formats numbers in the way SQL Server expects? Is there a .NET format code? A java format code? A Delphi format code? A VB format code? A stdio format code? latin-numeral-digits

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  • MS Word 2007 Mail Merge fails on ZIP codes with leading Zeros (eg. 01234)

    - by Pretzel
    I have an Excel Spreadsheet with a ZIP code column. For some dumb reason the original spreadsheet I got had all the zip codes stored as numbers, so a ZIP code like 01234 was stored as 1234. Easy to fix with "Format Column" as "Special = ZIP Code". All values like 1234, show up as 01234. Great! When I import it into Word via Mail Merge (to print address labels), the ZIP codes on all the addresses starting with a leading zero (like 01234) revert to their old form (1234). How do I fix this?

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  • using Linux vncviewer

    - by Darkoni
    Hi ! when i am connecting to VNC server using wine on linux $ wine vncviewer.exe i have to enter: VNC Server: 1.1.1.21 Proxy/Reapeter: 195.29.18.33:1234 and then, when i connect, on top there is txt: 1.1.1.21:5900 (195.29.18.33:1234) mine question is: how to connect using vncviewer ? what to put in VNC_VIA_CMD ? $ export xlocalPort=1234 $ export xremoteHost=1.1.1.21 $ export xremotePort=5900 $ export xgateway=195.29.18.33 $ export VNC_VIA_CMD="/usr/bin/ssh -f -L $xlocalPort:$xremoteHost:$xremotePort $xgateway sleep 20" $ vncviewer $xremoteHost -via $xgateway and i get error: unable connect to socket: Connection refused (111) i was trying to help myself with page http://www.tightvnc.com/vncviewer.1.php Please help, couse i need to use "native" linux vncviewer installed by $ yum install tigervnc tigervnc.i686 0:1.0.90-0.13.20100420svn4030.fc13 Thnx

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  • Standardize Outlook Contacts Format

    - by Nathan DeWitt
    I have Outlook syncing with my cell phone (WinMo 6.1) and vice versa. Everything works fine, but my numbers are all in different formats. I have some contacts with numbers of 5555551234, some are 555.555.1234, some are (555) 555-1234, some are 555-555-1234, etc. I don't really care how it's displayed, because they all dial fine on my phone. But I want to clean them up so they all look the same. Any easy ways to do this?

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  • using Linux vncviewer

    - by Darkoni
    when i am connecting to VNC server using wine on linux $ wine vncviewer.exe i have to enter: VNC Server: 1.1.1.21 Proxy/Reapeter: 195.29.18.33:1234 and then, when i connect, on top there is txt: 1.1.1.21:5900 (195.29.18.33:1234) mine question is: how to connect using vncviewer ? what to put in VNC_VIA_CMD ? $ export xlocalPort=1234 $ export xremoteHost=1.1.1.21 $ export xremotePort=5900 $ export xgateway=195.29.18.33 $ export VNC_VIA_CMD="/usr/bin/ssh -f -L $xlocalPort:$xremoteHost:$xremotePort $xgateway sleep 20" $ vncviewer $xremoteHost -via $xgateway and i get error: unable connect to socket: Connection refused (111) i was trying to help myself with page http://www.tightvnc.com/vncviewer.1.php Please help, couse i need to use "native" linux vncviewer installed by $ yum install tigervnc tigervnc.i686 0:1.0.90-0.13.20100420svn4030.fc13

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  • Tunneling over HTTP

    - by Morgan
    Hello, I have a network at work that is locked behind a firewall and Internet connection is available only by using a proxy server. At work, I can connect to databases that are distributed across the network. However, at home, I cannot connect to the proxy server or the databases. How can this be done? I can access my workstation via LogMeIn, so I can install anything on it. I thought of installing some kind of tunneling mechanism in my workstation. Then, at home, I could connect to this mechanism, which would in turn do the required connections. So essentially, what I'd like to do can be represented by the following diagram: Home = Workstation = Database. For example, whenever I connect to, say, 10.140.0.1:1234 at home, this would be redirected to 10.140.0.1:1234 of my Workstation, because 10.140.0.1:1234 is only available through the corporate network. NOTE: I'm using Windows XP.

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