вторник, 20 ноября 2007 г.
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| Don't know if this is still an issue. But I ran into the... Guest 15:04:40 |
| | Don't know if this is still an issue. But I ran into the same problem.
Found the following solution @ http://forums.devshed.com/perl-programming-6/net-ssh-perl-won-t-perform-agent-authentication-191670.html
Protocol must be explicitly defined in params string. (protocol=>'2,1')
Posting this for public knowledge.
Will be sending this as a bug report to developer of Net::SSH::Perl. |
| |  2 answer | Add comment |
воскресенье, 30 сентября 2007 г.
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| Hithe end of the format(dot) should be in starting of the... Guest 12:16:40 |
| | Hi
the end of the format(dot) should be in starting of the line .
Thanks
sabari
sabari.us@gmail.com
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| | Add comment |
пятница, 28 сентября 2007 г.
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| ewrer:-* Guest 13:33:59 |
| | ewrer |
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четверг, 13 сентября 2007 г.
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| If you solved this problem please post it because I have... Guest 13:48:59 |
| | If you solved this problem please post it because I have the same problem!!
Request must have exactly one security token |
| | 2 answer | Add comment |
вторник, 31 июля 2007 г.
четверг, 12 июля 2007 г.
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| On Linux: echo 64 > /proc/sys/net/ipv4/ip_default_ttlwhere... Guest 07:46:39 |
| | On Linux:
echo 64 > /proc/sys/net/ipv4/ip_default_ttl
where 64 is the new TTL you want. |
| | 1 answer | Add comment |
воскресенье, 8 июля 2007 г.
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| Can i know , Window VISTA supports PERL?Please reply on... Guest 10:09:15 |
| | Can i know , Window VISTA supports PERL?
Please reply on nishant.talegaonkar@gmail.com
Best Regards,
Nishant |
| | Add comment |
четверг, 5 июля 2007 г.
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| ]:-) X-( ]:-) where are the answersssssss]:-) ]:-) ]:-)... Guest 21:24:39 |
| | ![]:-)](http://qaix.com/i/smiles/devil.png "]:-)")  where are the answersssssss  ![:-\](http://qaix.com/i/smiles/confused.png ":-\")  |
| | Add comment |
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| see the value of __PACKAGE__ variable. Guest 16:57:02 |
| | see the value of __PACKAGE__ variable. |
| | Add comment |
пятница, 22 июня 2007 г.
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| did you ever solve this, I have the same issue-A Guest 16:02:52 |
| | did you ever solve this, I have the same issue
-A |
| | Add comment |
вторник, 22 мая 2007 г.
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| Here ya go. an actual copywrite symbol. Just copy & paste... Guest 06:31:15 |
| | Here ya go. an actual copywrite symbol. Just copy & paste. Should work with most word processors. Save in word file for copy and paste any time. Hope this helps. |
| | 3 answer | Add comment |
вторник, 15 мая 2007 г.
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| the whole world seems to be getting this error, and it... Guest 11:02:17 |
| | the whole world seems to be getting this error, and it seems the solutions are not present, Most of them end with the suggestion to check ORACLE_HOME and NLS settings.
I am also still to get anything that works for UNKNOWN OCI STATUS 1804) OCIInitialize
The db comes up fine when started with sqlplus, but when using DBI, it just throws this error, and does not even say to check ORACLE_HOME and NLS settings as many others get.
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| | Add comment |
пятница, 27 апреля 2007 г.
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| Thank you guys!I had a same problem, and it has really... Guest 07:58:31 |
| | Thank you guys!
I had a same problem, and it has really helped me. |
| | 1 answer | Add comment |
пятница, 23 марта 2007 г.
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| New formula for the number PiThe method is to circumscribe... Guest 10:58:28 |
| | New formula for the number Pi
The method is to circumscribe a polygon around the circumference
[of a circle] and calculate the perimeter of this polygon.
A polygon with infinitely many sides results in the number Pi.
The method is that of Archimedes for polygons who are
outside the circumference of those (in other words).
The formula is the following iterative one:
2 * A[n] * B[n]
A[n+1] = ------------------
1 + B[n]
1 + B[n]
B[n+1] = sqrt ( ----------- )
2
The initial values are A[0] = 4 and B[0] = sqrt( 1 / 2 )
The value 4 of A[0] equals 4 * tan[Pi/4]
considered as a circumference of radius 1 of the square
around the unit circle.
the value sqrt(1/2) of B[0] equals cos(Pi/4)
The value of A[n] ( lim(A[n]) for n--> infinity ) equals Pi.
Proof
we consider A[n] the perimeter of the polygon of 2^q sides
and A[n + 1] the polygon of 2^(q + 1) sides , the proof of
the relation between A[n] and A[n + 1] is the following:
sin [ pi/n ]
A[n] = 2^q * tan[ pi/n ] = 2^q * ------------------
cos [ pi/n ]
If we divide by the inverse cosine of Pi/n
sin [ pi/n ]
2^q * ----------------
cos [ pi/n ]
----------------------------
1
------------------
cos [ pi/n ]
the cosines cancel.The sine of Pi/n is equal to
2 * sin [ pi/(2*n) ] * cos [ pi/(2*n) ]
with this we divide between
2^q * 2 * sin [ pi/(2*n) ] * cos [ pi/(2*n) ]
---------------------------------------------------
cos [ pi/(2*n) ] * cos [ pi/(2*n) ]
the cosine cancels and it remains
2^q * 2 * sin [ pi/(2*n) ]
------------------------------
cos [ pi/(2*n) ]
therefore A[n] = 2^q * tan[ pi/n ]
and A[n + 1] = 2^(q + 1) * tan[ pi/(2*n) ]
therefore the final formula is
sin [ pi/n ]
2^q * -------------------
cos [ pi/n ]
-------------------------------------------------------------------
1
----------------------- * cos [ pi/(2*n) ] * cos [ pi/(2*n) ]
cos [ pi/n ]
which is equal to
sin [ pi/n ]
2^q * -------------------
cos [ pi/n ]
----------------------------------------------------
1 1 + cos [ pi/n ]
------------------------ * ------------------------
cos [ pi/n ] 2
this leaves us finally with an identity that we did write in the
beginning
2 * A[n] * B[n]
A[n+1] = ------------------
1 + B[n]
for which A[n] > A[n + 1] > PI
and consequentially the computation of A[n] for n-->infinity
is greater of equal to Pi. A[infinity] == PI.
The computation of B[n] is the computation of the successive
cosines for lengths of the sides in the preceding steps.
para cualquier duda o consulta contactar con la direccion
MSN MESSENGER oteropera@hotmail.com
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| | Add comment |
пятница, 9 февраля 2007 г.
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| Textpad - has syntax highlighting and a lot of other free... Guest 19:56:15 |
| | Textpad - has syntax highlighting and a lot of other free add-ons for perl developing |
| | 5 answers | Add comment |
четверг, 9 ноября 2006 г.
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| on 10/27 billing i was billed for $l8.59, I did not... Guest 13:22:26 |
| | on 10/27 billing i was billed for $l8.59, I did not authorize to be advanced to Turbo. My usual fee thru Visa was $9.99. Please corect my account. |
| | 3 answer | Add comment |
пятница, 6 октября 2006 г.
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| Weird reading my own words after years gone by. Guest 03:40:17 |
| | Weird reading my own words after years gone by. |
| | 1 answer | Add comment |
четверг, 31 августа 2006 г.
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| Did you ever get this sorted?I've got the exact same issue. Guest 21:13:02 |
| | Did you ever get this sorted?
I've got the exact same issue. |
| | Add comment |
пятница, 25 августа 2006 г.
вторник, 8 августа 2006 г.
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| :-* Guest 12:20:25 |
| | |
| | Add comment |
понедельник, 24 июля 2006 г.
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| When I migrate from Sun Solaris 8 to Linux Red Hat... Guest 10:24:04 |
| | When I migrate from Sun Solaris 8 to Linux Red Hat Enterprise, my old perl script written in Sun can not be used. Why? |
| | Add comment |
пятница, 7 июля 2006 г.
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Can't install DBD-Oracle- Error 405 method not found  karthikpr 07:38:38 |
| | Hi all
I have installed Oracle 9i, Activeperl 5.8.8 build and DBI on my windows XP machine.
But when I try to install DBD-Oracle.ppd on my machine I got the following error.
*******************************************************************ppm> install DBD-Oracle
Error: No valid repositories: Error: 405 Method not allowedppm>
********************************************************************
Can someone help me please.
karthik |
| | Add comment |
четверг, 22 июня 2006 г.
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| Interrupt PerlSAX Guest 20:40:51 |
| | I would like to know if anyone has any ideas about how to interrupt, or even stop, the PerlSAX parser after let s say 1000 elements or so. From where it's done doesn't really matter, as long as it really interrupts the parsing-process and is performed in a fairly clean way.
Any ideas are welcome!
//Matt |
| | Add comment |
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