Find Charactors in String

Need to know how to only show certain numbers in a string.
Example: NDC#49281-392-15
I need to only see the numbers and drop the NDC#
Example: Patient tolerated well. NDC#66521-116-11. This was given at 11/18/13 office visit, could not document. Computers were down. Sandy Wiseman CMA 11/19/2013
I need to drop all characters except 66521-116-11
Trouble is each set of numbers changes and may be a different length.
Some number sets may have "-" while others don't.
Thanks for any help.

DATA code TYPE string VALUE 'NDC#66521-116-11'
DATA first TYPE string.
DATA second TYPE string.
SPLIT code AT '#' INTO first second.
This code will return the following (drops the # symbol):
first = NDC
second = 66521-116-11
Hope this helps.

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Find out varchar2 string NULL columns and Empty columns

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SET ECHO OFF;
SET FEEDBACK OFF;
SET SERVEROUTPUT ON;
SET VERIFY OFF;
SET PAGES 0;
SET HEAD OFF;
spool D:\stringnull.csv
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v_tab_indent NUMBER(5);
v_col_indent NUMBER(5);
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v_val2       VARCHAR2(20);
v_query1     VARCHAR(500);
v_query2     VARCHAR(500);
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--DBMS_OUTPUT.ENABLE(100000);
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INTO v_tab_indent,v_col_indent
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Spool OFF
Set echo on
Set feedback onERROR:
Error report:
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ORA-06512: at line 31
01403. 00000 - "no data found"
*Cause:
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Example:
SQL> DECLARE
2    v_val       VARCHAR2(20);
3    v_query1     VARCHAR(32767);
4 BEGIN
5   FOR i IN (SELECT table_name, column_name FROM user_tab_columns
6             WHERE data_type IN ('NVARCHAR2', 'CHAR', 'NCHAR', 'VARCHAR2')
7             ORDER BY table_name, column_name
8            )
9   LOOP
10     v_query1 := 'SELECT count(*) FROM '||i.table_name||' where '||i.column_name||' IS NULL';
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13   END LOOP;
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15 /
CHILD_TAB                      : DESCRIPTION                    : 0
DEPT                           : DNAME                          : 0
DEPT                           : LOC                            : 0
EMP                            : ENAME                          : 0
EMP                            : JOB                            : 0
MYEMP_OLD                      : ENAME                          : 0
MYEMP_OLD                      : JOB                            : 0
MYNULLS                        : ENAME                          : 0
MYNULLS                        : JOB                            : 4
PARENT_TAB                     : DESCRIPTION                    : 0
T                              : CHAR_VALUE                     : 0
TABLE1                         : COL1_DESC                      : 0
PL/SQL procedure successfully completed.

Use REGEXP_INSTR to find a text string with space(s) in it

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Let's say ParmIn_Look_For has a value of 'black dog'. I want to see if
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test@ORA10G>
test@ORA10G>pratz
Also, 'x' ignores whitespace characters. Link to doc:
http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/conditions007.htm#i1048942
Message was edited by:
pratz

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Hi All,
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You may also try the below :
http://gallery.technet.microsoft.com/c0c57332-8624-48c0-b4c3-5b31fe641c58

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hiii
use following code
data: wa_str(100) type c.
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data: len type i.
data: ofset type i.
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do.
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else.
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endif.
ofset = ofset + 1.
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statement have added .i hope this will solve your problem.
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Hi,
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For oracle try using oracle function.
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I'm don't write scripts but do 'enjoy' copying, pasting and changing existing code/txt files.
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seb400 napisał(-a):
What do you mean by I would mark "changing" in "copying, pasting and changing"?
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Is it always true, as in your example, that the replacement is the same length as the existing string? If so, this one's easy.
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Solved!
Go to Solution.

The Find function evaluates to a number:
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CTA, CLA
teststandhelp.com
~Will work for kudos and/or BBQ~

Find Charactors in String

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