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  • How to enable CDR on AsteriskNow 1.5

    - by Michal Niklas
    I have upgraded PBX to Asterisk 1.6.2.7 and now CDR files are not created. It looks that such logging is disabled: Connected to Asterisk 1.6.2.7 currently running on pbx2 (pid = 5824) Verbosity is at least 3 pbx2*CLI> cdr show status pbx2*CLI> Call Detail Record (CDR) settings ---------------------------------- Logging: Disabled Mode: Simple Asterisk shows that CDR modules are loaded: pbx2*CLI> module show like cd Module Description Use Count cdr_manager.so Asterisk Manager Interface CDR Backend 0 cdr_csv.so Comma Separated Values CDR Backend 0 app_cdr.so Tell Asterisk to not maintain a CDR for 0 app_forkcdr.so Fork The CDR into 2 separate entities 0 func_cdr.so Call Detail Record (CDR) dialplan functi 0 cdr_custom.so Customizable Comma Separated Values CDR 0 6 modules loaded How to enable creating CDR csv files?

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  • How to enable CDR on AsteriskNow 1.5

    - by Michal Niklas
    I have upgraded PBX to Asterisk 1.6.2.7 and now CDR files are not created. It looks that such logging is disabled: Connected to Asterisk 1.6.2.7 currently running on pbx2 (pid = 5824) Verbosity is at least 3 pbx2*CLI> cdr show status pbx2*CLI> Call Detail Record (CDR) settings ---------------------------------- Logging: Disabled Mode: Simple Asterisk shows that CDR modules are loaded: pbx2*CLI> module show like cd Module Description Use Count cdr_manager.so Asterisk Manager Interface CDR Backend 0 cdr_csv.so Comma Separated Values CDR Backend 0 app_cdr.so Tell Asterisk to not maintain a CDR for 0 app_forkcdr.so Fork The CDR into 2 separate entities 0 func_cdr.so Call Detail Record (CDR) dialplan functi 0 cdr_custom.so Customizable Comma Separated Values CDR 0 6 modules loaded How to enable creating CDR csv files?

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  • Inkscape won't open .cdr files

    - by Bora George
    Hello I'm running Kubuntu 12.04 and I've installed Inkscape because I need to edit a .cdr file, Inkscape works like normal but the sole task I need it to do is edit said .cdr file which I understand it should be capable off by default. Whenever I try to open the file though I get the following error: UniConvertor failed: The eror itself is longer but what it seems to be saying that it can't find this application, and I myself haven't found it in the package manager. If someone has experienced something similar or has a simple solution for editing cdr's I'm no artist I just need to change some names, I would be very gratefull. The full error.

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  • elastix cdr stop working

    - by dreddko
    CDR was working before 19 march. Unfortunately i dont remember what kind of changes i made to configuration, but this exactly not changes to CDR config. elastix 2.4.0 asterisk 11.7.0 mysql 5.0.95 elastix*CLI> cdr show status Call Detail Record (CDR) settings ---------------------------------- Logging: Disabled Mode: Simple /etc/asterisk/cdr.conf [general] enable=yes unanswered = yes /etc/asterisk/cdr_mysql.conf [global] hostname = localhost dbname=asteriskcdrdb password = *MYPASSWROD* user = asteriskcdruser userfield=1 ;port=3306 ;sock=/tmp/mysql.sock loguniqueid=yes mysql> SHOW GRANTS FOR 'asteriskcdruser'@'localhost'; +-----------------------------------------------------------------------------------------------+ | Grants for asteriskcdruser@localhost | +-----------------------------------------------------------------------------------------------+ | GRANT USAGE ON *.* TO 'asteriskcdruser'@'localhost' IDENTIFIED BY PASSWORD 'HASHHERE' | | GRANT ALL PRIVILEGES ON `asteriskcdrdb`.* TO 'asteriskcdruser'@'localhost' | +-----------------------------------------------------------------------------------------------+ 2 rows in set (0.00 sec)

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  • Asterisk Outgoing CDR Logging To Mysql

    - by user3295551
    Trying to utilize the cdr logging (to mysql) using custom fields. The problem I am facing is only when an outbound call is placed, during inbound calls the custom field I am able to log no problem. The reason I am having an issue is because the custom cdr field I need is a unique value for each user on the system. sip.conf ... ... [sales_department](!) type=friend host=dynamic context=SalesAgents disallow=all allow=ulaw allow=alaw qualify=yes qualifyfreq=30 ;; company sales agents: [11](sales_agent) secret=xxxxxx callerid="<...>" [12](sales_agent) secret=xxxxxx callerid="<...>" [13](sales_agent) secret=xxxxxx callerid="<...>" [14](sales_agent) secret=xxxxxx callerid="<...>" extensions.conf [SalesAgents] include => Services ; Outbound calls exten=>_1NXXNXXXXXX,1,Dial(SIP/${EXTEN}@myprovider) ; Inbound calls exten=>100,1,NoOp() same => n,Set(CDR(agent_id)=11) same => n,CELGenUserEvent(Custom Event) same => n,Dial(${11_1},25) same => n,GotoIf($["${DIALSTATUS}" = "BUSY"]?busy:unavail) same => n(unavail),VoiceMail(11@asterisk) same => n,Hangup() same => n(busy),VoiceMail(11@asterisk) same => n,Hangup() exten=>101,1,NoOp() same => n,Set(CDR(agent_id)=12) same => n,CELGenUserEvent(Custom Event) same => n,Dial(${12_1},25) same => n,GotoIf($["${DIALSTATUS}" = "BUSY"]?busy:unavail) same => n(unavail),VoiceMail(12@asterisk) same => n,Hangup() same => n(busy),VoiceMail(12@asterisk) same => n,Hangup() ... ... For the inbound section of the dialplan in the above example I am able to insert the custom cdr field (agent_id). But above it you can see for the Oubound section of the dialplan I have been stumped on how I would be able to tell the dialplan which agent_id is making the outbound call. My Question: how to take the agent_id=[11] & agent_id=[12] and agent_id=[13] and agent_id=[14] etc and use that as a custom field for cdr on outbound calls?

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  • Shut Out of XP - No Admin Password or CDR

    - by ashes999
    I inherited an old WinXP/Linux dual-boot machine from the stoneage. Because it has Linux, the regular boot process is replaced with the Fedora boot loader; I cannot, therefore, press F8 strategically to tell my PC to boot from CD. Even if I could, it's a moot point; the CDR doesn't seem to recognize any CDs. To make things worse, there's no option to network boot. The original user is probably long gone; I don't know the password for any of the Administrator group users. I can login using my corp account, but that's unprivileged on this machine. Since I'm not an admin, I can't do crazy things, like looking at boot.ini. Or deleting files. I only have 500MB free on my C drive. I'm pretty sure I can't boot from a USB, since I didn't see any settings for this in my BIOS. How can I get admin access for my user? Edit: Things I've tried: Boot from CD (CD not recognized) Launch CD from XP (CD not recognized) Install Daemon Tools Lite so I can install from an ISO -- don't have admin privileges XP password recovery tool -- requires admin privileges Adding an admin user -- no access to Control Panel Users since I'm not an admin Logging in as both the admin users on the system (trying some standard passwords) Using Fedora to chntpw (the Fedora version installed is ancient -- 2.7)

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  • car and cdr in scheme is driving me crazy ...

    - by kristian Roger
    Hi Im facing a probem with the car and cdr functions for example: first I defined a list caled it x (define x (a (bc) d ( (ef) g ) )) so x now is equal to (a (bc) d ( (ef) g ) now for example I need to get the g from this list using only car and cdr (!! noshortcuts as caddr cddr !!) the correct answer is: (car(cdr(car(cdr(cdr(cdr x)))))) BUT how ? :-( I work according to the rule (the car gives the head of list and cdr gives the tail) and instead of getting the answer above I keep reaching wronge answers can any one help me in understanding this ... give me step or a way to solve it step by step thanx in advance Im really sick of scheme language.

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  • car and cdr in Scheme are driving me crazy ...

    - by kristian Roger
    Hi Im facing a problem with the car and cdr functions for example: first I defined a list called it x (define x (a (bc) d ( (ef) g ) )) so x now is equal to (a (bc) d ( (ef) g ) ) now for example I need to get the g from this list using only car and cdr (!! noshortcuts as caddr cddr !!) the correct answer is: (car(cdr(car(cdr(cdr(cdr x)))))) BUT how ? :-( I work according to the rules (the car gives the head of list and cdr gives the tail) and instead of getting the answer above I keep reaching wrong answers. Can any one help me in understanding this ... give me step or a way to solve it step by step Thanks in advance. I'm really sick of Scheme.

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  • geting the median of 3 values using sheme car & cdr

    - by kristian Roger
    Hi still stuck with the ugly scheme the problem this time is to get the median of three values (easy) I did all these : (define (med x y z) (car(cdr(x y z))) and it was accepted but when testing it (med 3 4 5) I will get this error Error: attempt to call a non-procedure (2 3 4) and when entering letters inetead of number i got (md x y z) Error: undefined varia y (package user) using somthin else than x y z i got (md d l m) Error: undefined variable d (package user) so what is wronge ?!

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  • Getting the median of 3 values using scheme's car & cdr

    - by kristian Roger
    The problem this time is to get the median of three values (easy) I did this: (define (med x y z) (car(cdr(x y z))) and it was accepted but when testing it: (med 3 4 5) I get this error: Error: attempt to call a non-procedure (2 3 4) And when entering letters instead of number i get: (md x y z) Error: undefined varia y (package user) Using something besides x y z I get: (md d l m) Error: undefined variable d (package user) the question was deleted dont know how anyway write a function that return the median of 3 values

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  • How to burn rEfit .cdr or .dmg in ubuntu or windows?

    - by beemzet
    Hi all, I don't have mac installed on my macbook pro, but I have Ubuntu and Windows. I was following this website http://mac.linux.be/content/single-boot-linux-without-delay to boot faster into Ubuntu without a 30 second delay. But I can't even burn a rEfit cdr image. Do you know how to burn it? Thanks

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  • Matrix addition in Scheme

    - by user285012
    I am trying to add a matrix and it is not working... (define (matrix-matrix-add a b) (map (lambda (row) (row-matrix-add row b)) a)) (define (row-matrix-add row matrix) (if (null? (car matrix)) '() (cons (add-m row (map car matrix)) (row-matrix-add row (map cdr matrix))))) (define (add-m row col) (if (null? col) 0 (+ (car row) (car col) (add-m (cdr row) (cdr col)))))

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  • How Can I Improve This Algorithm (LCS)

    - by superguay
    (define (lcs lst1 lst2) (define (except-last-pair list) (if (pair? (cdr list)) (cons (car list) (except-last-pair (cdr list))) '())) (define (car-last-pair list) (if (pair? (cdr list)) (car-last-pair (cdr list)) (car list))) (if (or (null? lst1) (null? lst2)) null (if (= (car-last-pair lst1) (car-last-pair lst2)) (append (lcs (except-last-pair lst1) (except-last-pair lst2)) (cons (car-last-pair lst1) '())) **(if (> (length (lcs lst1 (except-last-pair lst2))) (length (lcs lst2 (except-last-pair lst1)))) (lcs lst1 (except-last-pair lst2)) (lcs lst2 (except-last-pair lst1)))))) I dont want it to run over and over.. Regards, Superguay

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  • Scheme homework Black jack help....

    - by octavio
    So I need to do a game of blackjack simulator, butt can't seem to figure out whats wrong with the shuffle it's suppose to take a card randomly from the pack the put it on top of the pack. The delete it from the rest. so : (ace)(2)(3)(4)(5)...(k) if random card is let say 5 (5)(ace)(2)(3)(4)(5)...(k) then it deletes the 2nd 5 (5)(ace)(2)(3)(4)(6)...(k) here is the code: (define deck '((A . C) (2 . C) (3 . C) (4 . C) (5 . C) (6 . C) (7 . C) (8 . C) (9 . C) (10 . C) (V . C) (Q . C) (K . C))) ;auxilliary function for shuffle let you randomly select a card. (define shuffAux (lambda (t) (define cardR (lambda (t) (list-ref t (random 13)))) (cardR t))) ;auxilliary function used to remove the card after the car to prevent you from removing the randomly selected from the car(begining of the deck). (define (removeDupC card deck) (delete card (cdr deck)) ) (define shuffle2ndtry (lambda (deck seed) (define do-shuffle (lambda (deck seed) (if (> seed 0)( (cons (shuffAux deck) deck) (removeDupC (car deck) deck) (- 1 seed)) (write deck) ) ) ) (do-shuffle deck seed))) (define (shuffle deck seed) (define cards (cons (shuffAux deck) deck)) (write cards) (case (> seed 0) [(#t) (removeDupC (car cards) (cdr cards)) (shuffle cards (- seed 1))] [(#f) (write cards)])) (define random (let ((seed 0) (a 3141592653) (c 2718281829) (m (expt 2 35))) (lambda (limit) (cond ((and (integer? limit)) (set! seed (modulo (+ (* seed a) c) m)) (quotient (* seed limit) m)) (else (/ (* limit (random 34359738368)) 34359738368)))))) ;function in which you can delete an element from the list. (define delete (lambda (item list) (cond ((equal? item (car list)) (cdr list)) (else (cons (car list) (delete item (cdr list))))))) (

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  • Scheme: what are the benefits of letrec?

    - by Ixmatus
    While reading "The Seasoned Schemer" I've begun to learn about letrec. I understand what it does (can be duplicated with a Y-Combinator) but the book is using it in lieu of recurring on the already defined function operating on arguments that remain static. An example of an old function using the defined function recurring on itself (nothing special): (define (substitute new old lat) (cond ((null? l) '()) ((eq? (car l) old) (cons new (substitute new old (cdr l)))) (else (cons (car l) (substitute new old (cdr l)))))) Now for an example of that same function but using letrec: (define (substitute new old lat) (letrec ((replace (lambda (l) (cond ((null? l) '()) ((eq? (car l) old) (cons new (replace (cdr l)))) (else (cons (car l) (replace (cdr l)))))))) (replace lat))) Aside from being slightly longer and more difficult to read I don't know why they are rewriting functions in the book to use letrec. Is there a speed enhancement when recurring over a static variable this way because you don't keep passing it?? Is this standard practice for functions with arguments that remain static but one argument that is reduced (such as recurring down the elements of a list)? Some input from more experienced Schemers/LISPers would help!

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  • vim: sending tab-completion key against a mapped keystroke

    - by CDR
    To switch between buffers without installing any plugins, a good way is to type :b <tab> Which shows all the current buffers names in status bar and you can pick one using cursor keys and enter. But :b <tab> is 5 keystrokes and I would like to map it to a <leader>. But setting the following is not working. :nnoremap <Leader>. :b <Tab> It shows ":b ^I" in status bar and doesn't actually open the buffer names on status bar. Anyone knows why?

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  • Modifying Google Search Plugin

    - by TimeTrap
    I want Google search results to default with a publish date. I'm trying to change the Google search plugin, but I'd be just as happy using a 3rd-party tool that does this. So I'm using Firefox 15.0.1 and I've tried changing {install dir}/searchplugins/google.xml, but Firefox is just ignoring these changes. For example, I'm adding the below in 3 places in google.xml, but nothing is happening: <Param name="cdr" value="cdr:1,cd_min:01/01/1900,cd_"/>

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  • The remote server returned an error: (401) Unauthorized.

    - by Zaidman
    Hi I'm trying to get the html code of certain webpage, I have a username and a password that are correct but i still can't get it to work, this is my code: private void buttondownloadfile_Click(object sender, EventArgs e) { NetworkCredentials nc = new NetworkCredentials("?", "?", "http://cdrs.globalpopsvoip.com/0000069/20091229/20091228_20091228.CDR"); WebClient client = new WebClient(); client.Credentials = nc; String htmlCode = client.DownloadString("http://cdrs.globalpopsvoip.com/0000069/20091229/20091228_20091228.CDR"); MessageBox.Show(htmlCode); } The MessageBox is just to test it, the problem is that every time I get to this line: String htmlCode = client.DownloadString("http://cdrs.globalpopsvoip.com/0000069/20091229/20091228_20091228.CDR"); I get an exception: The remote server returned an error: (401) Unauthorized. How do I fix this?

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  • Traversing Scheme function as a list

    - by csl
    Isn't it possible to treat functions in Scheme as any other list? Basically, what I want do to is something like this: (define (foo) "hello") (cdr foo) ; or similar, should return the list ((foo) "hello") I've found a similar discussion about this, and I feel a bit disappointed if this is not possible with Scheme. If so, why is this impossible? Is it possible in other lisps? EDIT: Changed (cdr 'foo) to (cdr foo) -- it was misleading. I'm asking, why can't I access a function as a list?

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  • Why does using cons to create a pair of two lists produce a list and two elements?

    - by fingerprint211b
    I've started learning Scheme, for fun mostly, and because I've never used a functional language before. I chose Scheme because I wanted to read SICP for a long time. Anyway, I'm currently learning about lists, and before that I learned about cons, car and cdr. And there's an example that creates a list of lists with cons, like this : (cons (list 1 2) (list 3 4)) The resulting list is ((1 2) 3 4), which doesn't make sense to me, I would expect ((1 2)(3 4)) to be the result (a list made out of two lists). Why does it behave like that? I realize that if I were to use car, I would get (1 2), and cdr I'd get (3 4) becaue cdr always returns "the rest", but I don't understand why the list isn't made of two lists?

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  • Generate Permutations of a List

    - by Eric Mercer
    I'm writing a function that takes a list and returns a list of permutations of the argument. I know how to do it by using a function that removes an element and then recursively use that function to generate all permutations. I now have a problem where I want to use the following function: (define (insert-everywhere item lst) (define (helper item L1 L2) (if (null? L2) (cons (append L1 (cons item '())) '()) (cons (append L1 (cons item L2)) (helper item (append L1 (cons (car L2) '())) (cdr L2))))) (helper item '() lst)) This function will insert the item into every possible location of the list, like the following: (insert-everywhere 1 '(a b)) will get: '((1 a b) (a 1 b) (a b 1)) How would I use this function to get all permutations of a list? I now have: (define (permutations lst) (if (null? lst) '() (insert-helper (car lst) (permutations (cdr lst))))) (define (insert-helper item lst) (cond ((null? lst) '()) (else (append (insert-everywhere item (car lst)) (insert-helper item (cdr lst)))))) but doing (permutations '(1 2 3)) just returns the empty list '().

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  • How do I play flash files in pure C?

    - by CDR
    As part of a customized media player written in C (Win32), I need to enable my app play flash movies (.swf files) inside the player window. Can someone please indicate the most C compatible low level way to achieve this, giving me highest control? Specially control on display window and network access. I am looking for reference to a Windows DLL like flash_player.dll (if there is such thing) with documentation. Or at least a COM/ActiveX controls. Note that ActiveX is harder to use in C, than in say VB.

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  • Scheme. Tail recursive ?

    - by n00b
    Hi guys, any tail-recursive version for the below mentioned pseudocode ? Thanks ! (define (min list) (cond ((null? list '()) ((null? (cdr list)) (car list)) (#t (let ((a (car list)) (b (min (cdr list))) ) (if (< b a) b a) ) ) ) )

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