Oracle INSTR Function – SQL Syntax Examples
The Oracle INSTR SQL function is popular and performs materially the same operation as instr functions in many other programming languages.
Below shows the INSTR function along with the arguments it takes:
1 | instr(string, SUBSTRING) |
1 | instr(string, SUBSTRING, start_position) |
1 | instr(string, SUBSTRING, start_position, occurrence) |
The Oracle INSTR function returns the position (an integer) within string of the first character in substring that was found while using the corresponding start_position and occurrence.
Following are important rules to follow along with syntax exemplifying the implications of the rules.
- The first character of
stringis atstart_position1.start_positionis defaulted as 1. Ifstart_positionis set to 0, 0 will always be returned, and thus, is not a useful value. Ifstart_positionis negative, searching forsubstringwill begin atstart_positioncharacters counted from the end (right) ofstringand searching will be conducted towards the start (right-to-left) ofstring.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
SELECT instr('abcdabcd','b') FROM dual --returns: 2 SELECT instr('abcdabcd','b',1) FROM dual --returns: 2 SELECT instr('abcdabcd','b',4) FROM dual --returns: 6 SELECT instr('abcdabcd','b',0) FROM dual --returns: 0 SELECT instr('abcdabcd','b',-1) FROM dual --returns: 6 SELECT instr('abcdabcd','b',-4) FROM dual --returns: 2
substringmay exist withinstringmore than once. Theocurrenceattribute is a positive integer that allows you specify which occurrence you are searching for. Remember, if you would like to find the 2cnd-to-last occurrence ofsubstringwithinstring, you should search using astart_positionof -1 to indicate that searching should be conducted backwards, leavingoccurrenceat a value of 2.1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
SELECT instr('abcdabcd','b',1,1) FROM dual --returns: 2 SELECT instr('abcdabcd','b',1,2) FROM dual --returns: 6 SELECT instr('abcdabcd','b',1,0) FROM dual --returns: ORA-01428: argument '0' is out of range SELECT instr('abcdabcd','b',1,-1) FROM dual --returns: ORA-01428: argument '-1' is out of range SELECT instr('abcdabcd','b',-1,1) FROM dual --returns: 6 SELECT instr('abcdabcd','b',-1,2) FROM dual --returns: 2
string and substringcan be any of the datatypes CHAR, VARCHAR2, NCHAR, NVARCHAR2, CLOB, or NCLOB. Both start_position and ocurrence must be of datatype NUMBER, or any datatype that can be implicitly converted to NUMBER, and must resolve to an integer (remember that start_position should not be 0 and ocurrence should be greater than or equal to 1). The return value is a NUMBER, and will return 0 when nothing is found in the search.
INSTR is most powerful and often used in practice with the Oracle SUBSTR SQL function. Here are some examples of how to combine SUBSTR with INSTR.
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Thanks for above examples. It really help me lot.
You’re welcome, Mukesh. Glad this helped.
Hello Team,
I need some more explaination on how the below syntax is working
“SELECT instr(‘abcdabcd’,'b’,1,1) FROM dual”
Thanks & regards,
Souvik Ghosh
this syntax finds the first occurrence of “b” starting front he first character on the left, hence it will return the value 2. The syntax instr(‘abcdabcd’,’b’,1,2) finds the second occurrence and would return the value 6. you can also start from the right and search left. so instr(‘abcdabcd’,’a’,-1,1) would return 5, instr(‘abcdabcd’,’a’,-1,2) would return 1.
Thanks for this. Got me headed in the right direction. I needed to parse out the second to last level in a URL where the “root” could vary (parsing Sharepoint folders). So, here is what I came up with:
substr(doc_url,instr(document_url,’/',-1,2)+1,instr(doc_url,’/',-1,1)-instr(doc_url,’/',-1,2)-1)
returns hello for
http://a.com/b/hello/world.doc and for
http://a.com/hello/world.doc