substr clob to varchar2 in the insertion
HelloI try inserting several varchar2 values that make up a unique clob in a table.
Version is;
SQL> select * from v$version;
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - Production
I get the following error message when you try to insert a value in addition to 8,000 characters;SQL> begin
2 xddq_ins(rpad('x',8001,'x'));
3 end;
4 /
begin
*
ERROR at line 1:
ORA-06502: PL/SQL: numeric or value error: character string buffer too small
But works very well for values less than 8000 characters, for example;SQL> begin
2 xddq_ins(rpad('x',4001,'x'));
3 end;
4 /
PL/SQL procedure successfully completed.
SQL> select length(str) str_len, pos1,pos2,chunk_cnt
2 from xddq;
STR_LEN POS1 POS2 CHUNK_CNT
---------- ---------- ---------- ----------
4000 1 4000 1
1 4001 8000 2
The definition of table and the procedure I use is the following;create table xddq (Str varchar2(4000)
,t timestamp default current_timestamp
,Pos1 integer
,Pos2 integer
,Chunk_Cnt integer);
create or replace procedure XDDQ_Ins (p_Str in varchar2) as
STR_EMPTY exception;
l_Clob clob := to_clob(p_Str); -- first convert to a clob
l_Len pls_integer := dbms_lob.getlength(l_Clob); -- clob length
l_Chunk_Cnt pls_integer := ceil(l_Len / 4000); -- number of chunks
begin
if (l_Len < 1) then
raise STR_EMPTY;
end if;
insert into xddq (t
,str
,Pos1
,Pos2
,Chunk_Cnt)
(
select current_timestamp
,dbms_lob.substr(l_Clob, level*4000, ((level - 1) * 4000) + 1)
,((level - 1) * 4000) + 1
,level*4000
,level
from dual
connect by level <= l_Chunk_Cnt
);
exception
when STR_EMPTY then
raise_application_error('No string to insert', -20088);
end XDDQ_Ins;
/
The instance character sets are the following;SQL> select value
2 from nls_database_parameters
3 where parameter in ('NLS_CHARACTERSET','NLS_NCHAR_CHARACTERSET')
4 /
VALUE
----------------------------------------
WE8MSWIN1252
AL16UTF16
Could anyone help regarding how I can fix this problem?I need perform an INSERTION of a CLOB in chunks of 4,000 characters.
The size of the of the varchar2 parameter in the procedure is never more than 10000 characters.
Thank you
Hello
The amount should be not a constant (= 4000)?
dbms_lob.substr(l_Clob, 4000, ((level - 1) * 4000) + 1)
Tags: Database
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Hello
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SQL * more: Production of liberation 12.1.0.1.0 kills him Sep 10 16:21:39 2013
Copyright (c) 1982, 2013, Oracle. All rights reserved.
Enter the password:
Last successful login time: kills Sep 10-2013 16:18:26 + 02:00
Connected to:
Database Oracle 12 c Enterprise Edition Release 12.1.0.1.0 - 64 bit Production
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CON_NAME
------------------------------
PDBORCL
C##SA@pdborcl 10.09.2013 > see the con_id
CON_ID
------------------------------
3
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...
IX
IX
IX
SH
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PLAN_TABLE_OUTPUT
------------------------------------------------------------------------------------------------------------
SQL_ID, aj3vkggtvv9d9, number of children 0
-------------------------------------
INSERT INTO SA_VIEWS_V (OWNER, VIEW_NAME) SELECT VIEW_NAME, MASTER OF
DBA_VIEWS
Hash value of plan: 1585970530
-----------------------------------------------------------------------------------------------------
| ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time |
-----------------------------------------------------------------------------------------------------
| 0 | INSERT STATEMENT. | | | 136 (100) | |
| 1. LOAD TABLE CLASSIC | | | | | |
|* 2 | FILTER | | | | | |
|* 3 | HASH JOIN | | 65. 6045. 136 (0) | 00:00:01 |
|* 4 | HASH JOIN | | 65. 4875. 132 (0) | 00:00:01 |
| 5. NESTED LOOPS | | | | | |
| 6. NESTED LOOPS | | 65. 3315 | 131 (0) | 00:00:01 |
| 7. INDEX SCAN FULL | I_VIEW1 | 65. 325. 1 (0) | 00:00:01 |
|* 8 | INDEX RANGE SCAN | I_OBJ1 | 1. | 1 (0) | 00:00:01 |
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| 10. INDEX SCAN FULL | I_USER2 | 131. 3144 | 1 (0) | 00:00:01 |
| 11. TABLE ACCESS FULL | USER$ | 131. 2358. 4 (0) | 00:00:01 |
| * 12 | TABLE ACCESS FULL | USER_EDITIONING$ | 1. 6. 2 (0) | 00:00:01 |
| 13. SEMI NESTED LOOPS. | 1. 29. 2 (0) | 00:00:01 |
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| * 15 | INDEX RANGE SCAN | I_OBJ4 | 1. 9. 1 (0) | 00:00:01 |
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-----------------------------------------------------------------------------------------------------
Information of predicates (identified by the operation identity card):
---------------------------------------------------
2 filter (((IS NULL AND "O". ("TYPE #" <>88) OR BITAND ("O". ("" FLAGS ", 1048576) = 1048576 OR
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PLAN_TABLE_OUTPUT
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-------------------------------------
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Hash value of plan: 1508506130
--------------------------------------------------------------------------------------------
| ID | Operation | Name | Lines | Bytes | Cost (% CPU). Pstart. Pstop |
--------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | | 1 (100) | | |
| 1. PARTITION LIST ALL | | 10000 | 1289K | 0 (0) | 1. 2.
| 2. TABLE FIXED FULL | X$ COMVW$ 5885ef62 | 10000 | 1289K | 0 (0) | | |
--------------------------------------------------------------------------------------------
14 selected lines.
You have an idea about this result?
Thank you in advance,
Arnaud.
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Hello
I was wondering if someone could help write me a sql statement.
Here is my table:
Here are the instructions for correct insertion. This time, I posted 2 examples with 2 numbers different otherwise.CREATE TABLE "TEMP_INVOICE" ("INVOICE" VARCHAR2(100 BYTE), "DATE_OF_DOCUMENT" DATE, "DATE_OF_PAY_DAY" DATE, "D" NUMBER, "K" NUMBER );
I want to do is make an insert allows you to table another call is MADE:insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (1000,to_date('01.02.2012','dd.mm.yyyy'),to_date('01.03.2012','dd.mm.yyyy'),5000,0); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (1000,to_date('01.04.2012','dd.mm.yyyy'),'','',1000); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (1000,to_date('01.05.2012','dd.mm.yyyy'),'','',3000); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (1000,to_date('01.06.2012','dd.mm.yyyy'),'','',1000); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (2000,to_date('01.07.2012','dd.mm.yyyy'),to_date('01.09.2012','dd.mm.yyyy'),8000,0); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (2000,to_date('01.10.2012','dd.mm.yyyy'),'','',5000); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (2000,to_date('01.11.2012','dd.mm.yyyy'),'','',2000); insert into temp_invoice (invoice,DATE_OF_DOCUMENT,DATE_OF_PAY_DAY,d,k) values (2000,to_date('01.12.2012','dd.mm.yyyy'),'','',1000);
Statements in the INVOICE table should be like this:CREATE TABLE "INVOICE" ("INVOICE" VARCHAR2(100 BYTE), "DATE_OF_DOCUMENT" DATE, "DATE_OF_PAY_DAY" DATE, "DATE_OF_PAYMENT_REC" DATE, "VALUE" NUMBER, "VALUE_DEDUCT" NUMBER, "DATE_FROM" DATE, "DATE_TO" DATE );
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1...... 1000...............1.1.2012.................1.3.2012................NULL............................ 5000... NULL... 1.3.2012...1.4.2012
2...... 1000...............1.4.2012.................NULL..................... 1.4.2012...1000... 4000... 2.4.2012... 1.5.2012
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I would try to explain.
(1) the first statement that is to be inserted is original imply that at which is different DATE_OF_PAY_DAY to NULL.
To this inserted negative of the original imply, we must add date_from that is exatly the same date_of_pay_day and date_to which is exactly the same as the date of the first payment. Payment which came first!
(2) we have now in the second insert statement. It will be the first payment of lease with date_of_document and date_pf_payment_rec, which is the same as date_fo_document. Value field will be populated with the amount of payment received and value_deduct field will be the value of the original imply - value of the first payment. Date is date_of_document + 1 and date_to is the date of the next payment.
(3) Insert us the next installment. Date_od_payment_rec is the same as date_of_document... value is the amount of the second payment and value_deduct's previous value_deduct which was of 4000 - value of this second payment. date_from date_of_payment_rec + 1 and date_to is the date of the next payment
So we continue this same pattern until we reached the final payment when we finish insert with the statement:
Date of the document (date of the last payment received) and even for date_of_payment_rec and field value with the amount of the payment receieved. The rest (value_deduct, date_from, date_to) is null.
I really hope you understand what I'm trying to do here.
If you have any other questions please.
Published by: user13071990 on November 22, 2012 04:16
Published by: user13071990 on November 22, 2012 04:16Hello
user13071990 wrote:
... Here are the instructions for correct insertion. This time, I posted 2 examples with 2 numbers different otherwise.Ok!
Be sure to post the results you want new data.You probably need to add "PARTITION BY the Bill" to all analytical clauses in my solution:
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Try to run this Fixit: http://support.microsoft.com/mats/Program_Install_and_uninstall
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normally on the insertion of the sd chip a pop-up dialog box appears asking what route you want that information to sd etc. This pop up does not work and the only way I can get into the sd information is to search on D:, then click on 'Open' please notify if MS can restore the pop up facillity?
Hi JohnWack,
Welcome to the Microsoft Community Forums.
We're here to help and guide you in the right direction.
According to the description of the problem, I understand that you do not get the AutoPlay option when you insert the SD chip into the Windows XP computer.
-What do you mean by "what route you want to than the sd information etc."? Are you referring to the auto play feature?
Try these methods.
Method 1
Double-click on "My Computer", right-click on the drive you want to Auto play and click 'properties '.
Click the 'Auto play '.
Method 2
Steps to enable AutoPlay.
1. start, click run, type Gpedit.msc in the Open box, and then click OK.
2. under Computer Configuration, expand models of administration and then click System.
3. in the settings pane, do a right click Enable Autoplay, and then click Properties.For your reference.
Windows XP does not recognize a secure digital card when you reinsert it after you remove it
http://support.Microsoft.com/kb/812661It will be useful.
Let us know if you encounter problems under windows in the future. We will be happy to help you. -
my Media Player shows no music, or rip music of the inserted disc. What can solve this problem
Hey Lucky,
You can try the steps in the follwoign and check if it helps.
Method 1:
You can try to rebuild the media player library, and then try to extract again
This problem can occur if the media library is corrupt. Rebuild the library and check if this solves the problem.
Follow these steps:
a. exit Windows Media Player.
b. Click Start, run, type %LOCALAPPDATA%\Microsoft\Media Player and then click OK.
c. Select all files in the folder and then click on Delete on the file menu.
Note: You don't have to remove the folders that are in this folder.
d. restart Windows Media Player.
Note: Windows Media Player automatically rebuilds the database.
If this does not resolve the problem, disable the Windows Media Player database cache files. To do this, follow these steps:
a. exit Windows Media Player.
b. Click Start, click Run, type % LOCALAPPDATA%\Microsoft and then click OK.
c. Select the Media Player folder, and then click on Delete on the file menu.
d. restart Windows Media Player.
Method 2:
You can also try to change rip settings and check.
For more information, you can consult the following article:
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