Advanced String Manipulation Using REGEXP

Hello all,
I've been trying to split a string using either REGEXP_SUBSTR and REGEXP_REPLACE functions with much success so I wonder whether anyone here can give me a hand. Basically the string I want to split in to 3 is in this format:
'instanceno;partno;serialno'
I have read numerous articles within last few days but I still can't get it to working (I'm sure due to my lack of regular expression handling knowledge in OBIEE). I found two articles discussing similar scenarios. One in here and the other one is here. Both of those are good, but I cannot apply those principles to my query as I'm keep getting issues with the use of semicolon (;).
Could someone explain to me how to split this string so I can have 3 separate columns, please?
Thanks and regards,

Hi Anuhas,
you have to create 3 columns in the report on 3 attribute columns inside le Logical Model with this syntax for the the string str = 'instanceno;partno;serialno'
COLUMN1 = evaluate('regexp_substr(%1,''[^\;]+'', 1,1)',REPLACE(*str*,';','\;'))
COLUMN2 = evaluate('regexp_substr(%1,''[^\;]+'', 1,2)',REPLACE(*str*,';','\;'))
COLUMN3 = evaluate('regexp_substr(%1,''[^\;]+'', 1,3)',REPLACE(*str*,';','\;'))
You have to use the REPLACE function (; --> \;) because the ; is e special character
In this way you have:
COLUMN1 = instanceno
COLUMN2 = partno
COLUMN3 = serialno
The regex_substr(a,b,c) contains:
a = str the string
b = [^\;]+ find for the beginning of the string the character ;
c = number of occurrence
I hope it hepls.
Regards,
Gianluca

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String Match using RegExp

Hi ,
I am new to this:
I have the following requirement :
String1: '1,12,4'
String2: '2'
[both strings could be comma separated]
I need to check if evey member of String2 exists in String1. I was able to write the PL/SQL code for looping but when i use : select REGEXP_SUBSTR ('1,12,4','2') from dual, I get 2 as the result which is incorrect. Can someone please help me with this ?
Thanks in advance

Hi,
Ssharma21 wrote:
Will there always be exactly 1 string1 and 1 string2, or will you be getting.multiple values (from a table, for example)?
Ans
Both Strings have comma separated values.
Ex: String1 ='1,2,3'
String ='2,3'
Are both strings ordered the same way? That is, given that '12' comes before '4' in string1, can we be sure that '12' will come before '4' when both appear in string2 ?
Ans
What if there are duplicates in either string?
Ans: No duplicates
I wrote the following piece of code :
BEGIN
 v_levelIncludedFlag := FALSE;
 v_tempFlag := TRUE;
 v_array := APEX_UTIL.string_to_table (v_String2, ',');
 FOR i IN 1 .. v_array.COUNT
 LOOP
 IF REGEXP_SUBSTR (v_String1, v_array (i)) IS NOT NULL
 THEN
 v_levelIncludedFlag := TRUE;
 ELSE
 v_levelIncludedFlag := FALSE;
 END IF;
 v_levelIncludedFlag := v_levelIncludedFlag AND v_tempFlag;
 v_tempFlag := v_levelIncludedFlag;
 END LOOP;
END
Just the REGEXP_SUBSTR is not giving me the correct result due when String1='12' and String2='2' . I should be getting NULL from REGEXP_SUBSTR (v_String1, v_array (i))
The 1
st question is: Will you be comparing exactly 1 comma-delimited string against exactly 1 comma-delimited string?
It looks like you forgot to post the answer to the 2nd question.
Depending on your answers, you might want something like this:
VARIABLE string1 VARCHAR2 (100)
VARIABLE string2 VARCHAR2 (100)
EXEC :string1 := '1,12,4';
EXEC :string2 := '12,1';
WITH s1 AS
 SELECT REGEXP_SUBSTR ( :string1
 , '[^,]+'
 , 1
 , LEVEL
 ) AS item
 FROM dual
 CONNECT BY LEVEL <= REGEXP_COUNT (:string1, '[^,]+')
, s2 AS
 SELECT REGEXP_SUBSTR ( :string2
 , '[^,]+'
 , 1
 , LEVEL
 ) AS item
 FROM dual
 CONNECT BY LEVEL <= REGEXP_COUNT (:string1, '[^,]+')
SELECT CASE
 WHEN COUNT (s2.item) = 0
 THEN 'Everything in ' || :string2 || ' is also in ' || :string1
 ELSE 'Something in ' || :string2 || ' is missing from ' || :string1
 END AS output
FROM s2
LEFT OUTER JOIN s1 ON s1.item = s2.item
WHERE s1.item IS NULL
Whatever your requirements are, you probably don't need PL/SQL, though you may want to use PL/SQL for several reasons.

String manipulation function (regexp?)

Hi,
I am trying to extract several items from one string and concatenate them into another string with a separator.
Example
input: UV23_I2_P0_NUM > UV23_I2_P1_NUM AND UV23_I2_P1_NUM > UV23_I2_P2_NUM AND UV24_I4_P0_TXT = 'YES'
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Hi,
Merijn8106997 wrote:
... I'll be checking those regexp functions more closely from now on...seeing them in action explains a lot more than just looking at manuals....Yes, the manual is a valuable reference, but it's just a reference. A dictionary is a reference; when you're learning a foreign language, it's extremely valuable to have a dictionary, but that's not the best way to learn the language. I should have thought about that earlier and posted a little bit of an explanation.
Here's one rather subtle point, concerning the '?' in the middle of the pattern, which means the expression right before it is Non-Greedy :
SELECT     TRIM ( BOTH ';'
            FROM REGEXP_REPLACE ( str
                                     , '(^|NUM|TXT)(.*?)(UV|$)'
--                                   |
--                                This ? means "non-greedy"
                        , '\1;\3'
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FROM    table_x
;By default regular expressions are Greedy , that is, when there is a choice about how much a given pattern will match, it it will match as much as possile.
In the example above, \2 is the 0 or more characters that comes after either the beginning of the string, or NUM or TXT, and comes before either UV or the end of the string. Where is the first occurrence of that pattern in the example you gave:
input:         UV23_I2_P0_NUM > UV23_I2_P1_NUM AND UV23_I2_P1_NUM > UV23_I2_P2_NUM AND UV24_I4_P0_TXT = 'YES'
position: 1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234
                    1         2         3         4         5         6         7         8         9Where is the beginning of the 1st instance of the pattern, that is, the first beginning of string, NUM or TXT? Clearly, it's the beginning of the string, before position 1.
Where is the ending of that pattern, that is, UV or end-of-string? There are occurrences of UV at positions 1, 18, 37, 54 and 73, and an end-of-string after position 94. Which one do we consider to be the end of the pattern? The default is to be greedy, and take as much as possible, so that the first (and only) occurrence of the pattern would start before postion 1 and end after position 94. The ? mark tells REGEXP_REPLACE not to be greedy, to take as little as possiible. That means the first UV, at position 1. So the first occurrence of the pattern is not the whole string,l but only the part up through the first UV, at positions 1 and 2.

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Hello Guys,
I stuck in a situation wherein I want to extract specific data from a column of the table .
Below are the values for a particular column wherein I want to ignore values along with bracket which are in bracket and which are like .pdf,.doc .
Tris(dibenzylideneacetone)dipalladium (0) 451CDHA.pdf
AM57001A(ASRM549CDH).DOC
AM23021A Identity of sulfate (draft)
PG-1183.E.2 (0.25 mg FCT)
AS149656A (DEV AERO APPL HFA WHT PROVENTIL)
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PS21001A_1H-NMR.doc (PN 332-00)
AM68311A (Q-One CP 33021.02) Attachment
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Like this?
SQL> with t
2 as
3 (
4 select 'Tris(dibenzylideneacetone)dipalladium (0) 451CDHA.pdf' str from dual
5 union all
6 select 'AM57001A(ASRM549CDH).DOC' str from dual
7 union all
8 select 'AM23021A Identity of sulfate (draft)' str from dual
9 union all
10 select 'PG-1183.E.2 (0.25 mg FCT)' str from dual
11 union all
12 select 'AS149656A (DEV AERO APPL HFA WHT PROVENTIL)' str from dual
13 union all
14 select 'Stability report (RSR) Annex2 semi-solid form (internal information)' str from dual
15 union all
16 select 'TSE(Batch#USLF000332)-242CDH, Lancaster synthesis.pdf' str from dual
17 union all
18 select 'TR3018520A Addendum 1 (PN 3018520)' str from dual
19 union all
20 select 'AM10311A Particle size air-jet sieving (constant sieving) (draft)' str from dual
21 union all
22 select 'ASE00099B Addendum (PN E000099) 90 mesh' str from dual
23 union all
24 select 'AM37101_312-99 (Z11c) Palladium by DCP.doc' str from dual
25 union all
26 select 'PS21001A_1H-NMR.doc (PN 332-00)' str from dual
27 union all
28 select 'AM68311A (Q-One CP 33021.02) Attachment' str from dual
29 union all
30 select 'AM68202-1A (BioReliance no. 02.102006) Attachment' str from dual
31 )
32 select str
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34    from t;
STR                                                                    STR_NEW
Tris(dibenzylideneacetone)dipalladium (0) 451CDHA.pdf                  Trisdipalladium 451CDHA
AM57001A(ASRM549CDH).DOC                                              AM57001A
AM23021A Identity of sulfate (draft)                                  AM23021A Identity of sulfate
PG-1183.E.2 (0.25 mg FCT)                                              PG-1183
AS149656A (DEV AERO APPL HFA WHT PROVENTIL)                            AS149656A
Stability report (RSR) Annex2 semi-solid form (internal information) Stability report Annex2 semi-solid form
TSE(Batch#USLF000332)-242CDH, Lancaster synthesis.pdf                  TSE-242CDH, Lancaster synthesis
TR3018520A Addendum 1 (PN 3018520)                                    TR3018520A Addendum 1
AM10311A Particle size air-jet sieving (constant sieving) (draft)      AM10311A Particle size air-jet sieving
ASE00099B Addendum (PN E000099) 90 mesh                                ASE00099B Addendum 90 mesh
AM37101_312-99 (Z11c) Palladium by DCP.doc                            AM37101_312-99 Palladium by DCP
PS21001A_1H-NMR.doc (PN 332-00)                                        PS21001A_1H-NMR
AM68311A (Q-One CP 33021.02) Attachment                                AM68311A Attachment
AM68202-1A (BioReliance no. 02.102006) Attachment                      AM68202-1A Attachment
14 rows selected.

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