ENCRYPTION and DECRYPTION of a FILE BLOB

Similar Questions

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  I suggest you to ask your question in the below link:

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  Hello

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• 15 - r036tu: pci encryption and decryption hp 15-r038tu

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• Packet encryption and decryption by IPSEC Tunnel

  Hello world

  You must confirm if Site A has VPN IPSEC to Site B through Public network.

  PCs on site say a package is sent in clear text to the switch, then switch sends to the VPN router.

  Router VPN to site A will make encryption and send over a WAN link encrypted.

  When the packet reaches a router B Site it will decrypt the packet and send clear text to PC right of the site B?

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  Hello Manu,

  Yes, you are right.

  Encryption and decryption will perform VPN closing devices.

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• Packages that do not receive encryption and decrypt IPSEC

  Hello world

  I have 2691 conencted to the Internet router and it does NAT.

  This connects to the 3550A shift that has the connection to the router 1811W.

  I have VPN installation between 1811W and 3550.

  3550 has connection to 2691 via ospf.

  OSPF is running between 1811w and 3550.

  1811

  1811w # sh crypto isakmp his

  IPv4 Crypto ISAKMP Security Association

  DST CBC conn-State id

  192.168.99.2 192.168.99.1 QM_IDLE 2005 ASSETS

  IPv6 Crypto ISAKMP Security Association

  1811w # sh crypto ipsec his

  Interface: FastEthernet0

  Tag crypto map: VPN_MAP, local addr 192.168.99.1

  protégé of the vrf: (none)

  local ident (addr, mask, prot, port): (192.168.0.0/255.255.0.0/0/0)

  Remote ident (addr, mask, prot, port): (192.168.99.0/255.255.255.0/0/0)

  current_peer 192.168.99.2 port 500

  LICENCE, flags is {origin_is_acl},

  #pkts program: encrypt 0, #pkts: 0, #pkts digest: 0

  #pkts decaps: 0, #pkts decrypt: 0, #pkts check: 0

  compressed #pkts: 0, unzipped #pkts: 0

  #pkts uncompressed: 0, #pkts compr. has failed: 0

  #pkts not unpacked: 0, #pkts decompress failed: 0

  #send 30, #recv errors 0

  local crypto endpt. : 192.168.99.1, remote Start crypto. : 192.168.99.2

  Path mtu 1500, mtu 1500 ip, ip mtu IDB FastEthernet0

  current outbound SPI: 0x0 (0)

  PFS (Y/N): N, Diffie-Hellman group: no

  SAS of the esp on arrival:

  the arrival ah sas:

  SAS of the CFP on arrival:

  outgoing esp sas:

  outgoing ah sas:

  outgoing CFP sas:

  3550A

  3550SMIA # sh crypto isakmp his

  IPv4 Crypto ISAKMP Security Association

  status of DST CBC State conn-id slot

  192.168.99.2 192.168.99.1 QM_IDLE 1001 ACTIVE

  IPv6 Crypto ISAKMP Security Association

  3550SMIA #sh cry

  3550SMIA #sh crypto ipsec his

  Interface: FastEthernet0/8

  Tag crypto map: VPN_MAP, local addr 192.168.99.2

  protégé of the vrf: (none)

  local ident (addr, mask, prot, port): (192.168.0.0/255.255.0.0/0/0)

  Remote ident (addr, mask, prot, port): (192.168.99.0/255.255.255.0/0/0)

  current_peer 192.168.99.1 port 500

  LICENCE, flags is {origin_is_acl},

  #pkts program: encrypt 0, #pkts: 0, #pkts digest: 0

  #pkts decaps: 0, #pkts decrypt: 0, #pkts check: 0

  compressed #pkts: 0, unzipped #pkts: 0

  #pkts uncompressed: 0, #pkts compr. has failed: 0

  #pkts not unpacked: 0, #pkts decompress failed: 0

  #send 15, #recv errors 0

  local crypto endpt. : 192.168.99.2, remote Start crypto. : 192.168.99.1

  Path mtu 1500, ip mtu 1500

  current outbound SPI: 0x0 (0)

  SAS of the esp on arrival:

  the arrival ah sas:

  SAS of the CFP on arrival:

  outgoing esp sas:

  As we have seen more top packets are not encrypted between 1811w and 3550.

  I used the same ACLs on 1811W and 3550A

  INTERESTING_TRAFFIC extended IP access list

  IP 192.168.0.0 allow 0.0.255.255 192.168.99.0 0.0.0.255 connect

  Reasons why packages do not encrypt and decrypt?

  Thank you

  MAhesh

  Hello

  Access-list for interesting traffic should be mirrored.

  Best regards

  Eugene

• Encrypt and decrypt using the key of the table

  Dear Experts,

  Here is my package to encrypt and decrypt with triples in oracle 11 g.

  In my package, I used encryption_key (3FECCDC7D348A85B096F0B43C4C6A38DBBD369DB37FEA435) according to this key we are encryption and decryption.

  My requirement is now, I don't want to spend the encryption_key in my code.we key stored in a table (key_details) and using the key column, we her encrypt and decrypt the data.

  -----------------------------------

  create the table key_details

  (

  Identification number,

  VARCHAR2 (48) key

  );

  insert into key_details values(1,'3FECCDC7D348A85B096F0B43C4C6A38DBBD369DB37FEA435');

  -------------------------------------

  CREATE or REPLACE PACKAGE encr_decr

  AS

  FUNCTION encrypt (p_plainText VARCHAR2) RETURN RAW DETERMINISTIC;

  FUNCTION decrypt (p_encryptedText RAW) RETURN VARCHAR2 DETERMINISTIC;

  END;

  /

  CREATE or REPLACE PACKAGE encr_decr BODY

  AS

  encryption_type PLS_INTEGER: = DBMS_CRYPTO. ENCRYPT_3DES

  + DBMS_CRYPTO. CHAIN_ECB

  + DBMS_CRYPTO. PAD_PKCS5;

  encryption_key RAW (48): = UTL_RAW.cast_to_raw ('3FECCDC7D348A85B096F0B43C4C6A38DBBD369DB37FEA435');

  FUNCTION encrypt (p_plainText VARCHAR2) RETURN RAW DETERMINISTIC

  IS

  encrypted_raw RAW (2000);

  BEGIN

  encrypted_raw: = DBMS_CRYPTO. ENCRYPT

  (

  SRC = > UTL_RAW. CAST_TO_RAW (p_plainText),

  Typ = > encryption_type,.

  key = > encryption_key

  );

  RETURN encrypted_raw;

  END encrypt;

  FUNCTION decrypt (p_encryptedText RAW) RETURN VARCHAR2 DETERMINISTIC

  IS

  decrypted_raw RAW (2000);

  BEGIN

  decrypted_raw: = DBMS_CRYPTO. DECRYPT

  (

  SRC = > p_encryptedText,

  Typ = > encryption_type,.

  key = > encryption_key

  );

  RETURN (UTL_RAW. CAST_TO_VARCHAR2 (decrypted_raw));

  END decrypt;

  END;

  /

  Help, please.

  create or replace package body encr_decr

  as

  encryption_type pls_integer: = dbms_crypto.encrypt_3des

  + dbms_crypto.chain_ecb

  + dbms_crypto.pad_pkcs5;

  gross encryption_key (48);

  --

  function encrypt (p_plaintext varchar2)

  gross return deterministic

  is

  Start

  Return dbms_crypto.encrypt (CBC-online utl_i18n.string_to_raw (p_plaintext)

  typ-online encryption_type

  key-online encryption_key

  );

  end encrypt;

  --

  function decrypt (gross p_encryptedtext) return varchar2 deterministic

  is

  Start

  Return utl_i18n.raw_to_char (dbms_crypto.decrypt (src-online p_encryptedtext

  typ-online encryption_type

  key-online encryption_key

  )

  );

  put an end to decrypt;

  Start

  Select the key in encryption_key

  of key_details

  where id = 1;

  end;

  /

• Error DBMS_CRYPTO in encrypt and decrypt

  Hello
  
  CREATE or REPLACE PACKAGE enc_dec
  AS
  FUNCTION encrypt (p_plainText VARCHAR2) RETURN RAW DETERMINISTIC;
  FUNCTION decrypt (p_encryptedText RAW) RETURN VARCHAR2 DETERMINISTIC;
  END;
  /
  
  CREATE or REPLACE PACKAGE enc_dec BODY
  AS
  encryption_type PLS_INTEGER: = DBMS_CRYPTO. ENCRYPT_DES
  + DBMS_CRYPTO. CHAIN_CBC
  + DBMS_CRYPTO. PAD_PKCS5;
  encryption_key RAW (32): = UTL_RAW.cast_to_raw ('MyEncryptionKey');
  FUNCTION encrypt (p_plainText VARCHAR2) RETURN RAW DETERMINISTIC
  IS
  encrypted_raw RAW (2000);
  BEGIN
  encrypted_raw: = DBMS_CRYPTO. ENCRYPT
  (
  SRC = > UTL_RAW. CAST_TO_RAW (p_plainText),
  Typ = > encryption_type,.
  key = > encryption_key
  );
  RETURN encrypted_raw;
  END encrypt;
  FUNCTION decrypt (p_encryptedText RAW) RETURN VARCHAR2 DETERMINISTIC
  IS
  decrypted_raw RAW (2000);
  BEGIN
  decrypted_raw: = DBMS_CRYPTO. DECRYPT
  (
  SRC = > p_encryptedText,
  Typ = > encryption_type,.
  key = > encryption_key
  );
  RETURN (UTL_RAW. CAST_TO_VARCHAR2 (decrypted_raw));
  END decrypt;
  END;
  /
  
  I used the script above to encrypt a column of data. But I get the error below
  
  ERROR on line 1:
  ORA-12899: value too large for column
  "TEST1". «TESTS ".» "" SECURE_ID "(real: 32,)
  maximum: 12)
  
  The column data type is NOT NULL VARCHAR2 (12 CHAR). Where I have to change my script to encrypt this column.
  
  Kind regards
  007

  >
  Where I have to change my script to encrypt this column.
  >
  Oracle knows nothing about your data is encrypted. So like all other columns, the column must be defined with a length that will contain the largest value that it must take. Redefine the column to make it longer.

  Have you considered using the Oracle Transparent data encryption?

  For examples, see this AskTom blog
  http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:44742967463133

  And the Doc for the characteristics
  http://docs.Oracle.com/CD/B19306_01/network.102/b14268/asotrans.htm

• Encrypt and decrypt any

  Please help... I have this...

  main.cfm (test only)

  < cfset txtuserid = 123 >
  < p > < A HREF = ' test.cfm? txtuserid = < cfoutput > #URLEncodedFormat (Encrypt (txtuserid, "#txtuserid #")) # < / cfoutput > "> click me < /A >". "

  Once you click on the click me

  the url becomes: http://localhost/Newl/Main2.cfm?txtuserid= 25% 2F5% 3BW 5th % 3 C 4% 20% 0% > > > for his work

  but once the test.cfm comes

  the output of the txtuserid instead of 123

  txtuserid is e P1

  I need the txtuserid being 123

  Here is my sample code at main2.cfm

  < cfset txtuserid1 = #Decrypt(txtuserid, "#txtuserid#") # >

  < cfoutput > #txtuserid1 # < / cfoutput >

  Thank you!!!

  In the encryption

  Encrypt (txtuserid, "#txtuserid #")

  1st parameter is the string to encrypt.

  2Md parameter is key or seed used to encrypt the string.

  In decription

  encrypted_string	String or a variable that contains a. String to decipher
  seeds	Required string. The 32-bit key that was used to encrypt the string.

  main.cfm (test only)

  
  

  #URLEncodedFormat(Encrypt(txtuserid, "> ' > Click me.

  Here the key to encryption is 'txtuserid' (123)

  for decription same key must be used for encryption

  Main2.cfm

  #txtuserid1 #.

  Here the value of txtuserid IE the 2nd parameter is the decryption key that wasn't the same key you used to 4encryption u have to also use the same key for decryption. If you change

  Main2.cfm

  #txtuserid1 #.

  you will get the response as 123.

  Please try the example below... Then, you'll get the IDEA

  main.cfm (test only)

  
  
  

  #URLEncodedFormat(Encrypt(txtuserid, "> ' > Click me.

  Main2.cfm

  
  
  #txtuserid1 #.

• How to read the data of BLOB and save to the file system

  Hi friends.
  
  I have called DEMO table there image stored as a BLOB. now my requirement is to know how to read the data of BLOB and saved in the file system. is there a procedure
  to read the blob object and stored in the file system. pls suggest me with an example.
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  HI - don't know what version you have, but have a read through this:

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  I use the PGP 10.0 Beta for the last two weeks (it is now been shipped and is the current version), and one of the new features is Whole Disk Encryption on Boot Camp partitions. It works fine - except for the fact that VMWare Fusion 3.0.1 can not load the Boot Camp partition is encrypted (it starts initially, but does not advance beyond what I suppose is the PGP bootloader).

  I know that I can create a virtual machine and run Windows to work around the problem - I'm just curious to know if this is something that VMWare are willing to study and operate now PGP 10.0 has been shipped.

  It seems to be fixed/supported in a future version?

  Thank you very much .

  This looked interesting so I tried myself. I downloaded a demo of PGP 10.0 and encrypted to my hard drive which has OS X 10.5.8 and a Windows 7 Boot Camp partition.

  After starting the machine virtual Boot Camp in Fusion, I see a black screen that says bootguard _

  

  I guess that's also what you see?

  You're right - it's the PGP bootloader which replaced the Windows loader in the MBR of the physical disk. Fusion accesses the data in the partition through Mac OS X, which already has access to the decrypted data. The boot loader assumes that the drive is always encrypted and fails when he tries to decrypt again.

  Fortunately, Boot Camp VMs include a copy of the MBR (with the boot code and the partition table) in a separate file, so it's very easy to fix:

  #. Start the machine virtual of Boot Camp (not native!) with a Windows CD/DVD installation (or ISO image).

  #. Go to the repair command prompt.

  #. Run the fixmbr command in Windows XP or bootrec /fixmbr in Windows Vista or Windows 7 to replace the MBR code.

  In this way, your Boot Camp virtual machine will use the default Windows boot loader to access the already decrypted data, while the native Boot Camp startup will use code PGP MBR (which is always the physical disk) to access the prompt for the PGP password and decrypt the drive.

• processing order of encryption and ACLs

  Hi people,

  I am preparing to a test lab and have the following scenario:

  R6---172.16.50/24---PIX---172.16.10/24--R1

  R6 I have two interfaces:

  lo0 6.6.6.6/24

  FA0/1 172.16.50.50/24

  R1 two int:

  lo0 1.1.1.1/24

  E0 172.16.10.1/24

  I want to protect all traffic between the 6.6.6.0 and network 1.1.1.0 with IPSec. I use ESP to protect traffic.

  Another condition is that I want to put an ACL to e0 allowing IPSec traffic.

  I created an ACL named ACL_E0_IN that is applied on e0 for inbound traffic.

  R1 #sh of access lists

  Expand the IP ACL_E0_IN access list

  esp permits 172.16.50.50 host 172.16.10.1 (15 matches)

  permit udp host 172.16.50.50 host 172.16.10.1 eq isakmp (116 matches)

  refuse the log host 1.1.1.1 icmp host 6.6.6.6 (5 matches)

  Ping of R6 R1 does not work:

  R6 #p 1.1.1.1 source lo 0

  Type to abort escape sequence.

  Send 5, echoes ICMP 100 bytes of 1.1.1.1, time-out is 2 seconds:

  Packet sent with the address source 6.6.6.6

  .....

  Success rate is 0% (0/5)

  R6 #.

  On the R1, I get the following message:

  * 8 Mar 03:58:37.917: % s-6-IPACCESSLOGDP: ACL_E0_IN denied icmp 6.6.6.6 - list

  > 1.1.1.1 (8/0), 4 packs

  This scenario works ONLY when I allow ICMP of R6 and ESP traffic.

  I wonder why the decrypted packets are denied by the ACL. I expect that the ACL is processed BEFORE the packet is decrypted. When I look at the meter on the hit of the ACL, it seems that the ACL is checked twice.

  Someone at - it an idea on the exact order of encryption and the treatment of the ACL?

  Thank you

  Michael

  Attached you will find the configs of the R1 and R6

  Michael

  I recently saw an explanation of encryption and ACLs that indicates there has recently been a change in behavior. In most versions of IOS, the behavior is as you describe, the package is evaluated by the ACL twice. The explanation is that the package is evaluated first in his State encrypted to check that it was something that must be dealt with. After the package has been decrypted, the IOS necessary to assess the package decrypted to see if things like the quality of Service necessary to apply. So basically the decrypted packet passed through the interface again and the ACL again. In recent versions of the code (12.3 (4) T, if memory serves) a change has been made and the package will now go through the ACL only once.

  HTH

  Rick

• Encryption and Protection of SQLite database - App of signature with the Signature of Blackberry tool

  Hello

  I am trying to encrypt and protect my SQLite database. To do this, I have generated a pair of public/private key using the administration Blackberry signature tool. Then, I added the .key file to my project and written the code for encryption and data protection. Then, I signed the app with Blackberry signing authority tool such as mentioned in the documentation. It was duly signed. But my application does not work on the device. It shows that my application tries to access a secure API which means that it is not properly signed.

  What could be the problem?

  And also I want to know what is the difference between the signing of the "Tool of signed" app within the eclipse and sign the application with "Blackberry signing authority tool '?

  I sign my application with eclipse itself "tool of signed" before adding the public .key file and it worked very well.

  Please suggest.

  Thank you.

  You must sign with both.

• A rESTful Service: display file BLOB in DB as a RESTful service?

  Hello
  
  With Oracle Cloud we can expose the data as RESTful Web Service, but how about exposing BLOB... we need to generate a link to this file that is stored in the DB.
  
  To make it clearer:
  
  Two Tables:

  Properties (Property_ID number PK, Project varchar2(40), Property_number varchar2(20) ...)

  Images  (ID number PK, Property_ID number FK, content blob, 
  mimetype varchar2(200), filename varchar2 (200), filesize varchar2 (200))

  And I want to expose this request as repairman:

  Select p.property_id, p.project, p.property_number, i.content 
  from properties p inner join images i on (p.property_id = i.property_id)

  Usually we use: [url http://docs.oracle.com/cd/E14373_01/apirefs.32/e13369/apex_util.htm#CHDICGDA] apex_util.get_blob_file_src, but that one must be called from a session valid apex! ("i.content" in this query should give a link).
  
  How can do us?
  
  One other point, because we expose the data in JSON format:
  Suppose I have a property with 10 images, so it is meaningless to get 10 rows for the same property just to give a different image. I think we need another (nested) level for images!
  
  I hope my point is clear,
  
  Thanks in advance...
  Fateh

  Hi Fateh-

  Good question. You identify the source as a multimedia resource type and use a SQL statement with the primary key and the BLOB column. When you use multimedia resource, you are basically saying your Cloud Service database not to marshall the data, simply send it - which is exactly what you are looking for.

  Thanks to this application you will need a separate SQL call for each retrieval of BLOB. However, you may be able to use a PL/SQL block as the endpoint for the RESTful Service and take care of several BLOB in the processing block.

  I hope this helps.

  -Rick Greenwald

• Universal Clipboard is encrypted and how the data is stored in iCloud

  Hello

  You need a deep understanding of the works of the new, universal Clipboard feature safety? The feature is nice, but do you copy your password by company store system administrator password... This password is moving all over the world and also through Apple servers... Don't make me happy.

  Who can guarantee that the communication is encrypted and the data stored in iCloud are not used for other purpose?

  Thanks and greetings

  Security and privacy - Apple Support Overview iCloud

  Although not listed here, all data transmitted to the Apple servers is encrypted in transit and on the server.

  I can't offer any advice on the question of whether your transmissions could be intercepted and decoded.

  They say they also unequivocally that they never provide encryption keys to any third party. Which has been further strengthened with their recent dealings with the FBI.

  You'll have to decide for yourself if you agree with what Apple says.