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.

  1. The first character of string is at start_position 1. start_position is defaulted as 1. If start_position is set to 0, 0 will always be returned, and thus, is not a useful value. If start_position is negative, searching for substring will begin at start_position characters counted from the end (right) of string and searching will be conducted towards the start (right-to-left) of string.
    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
  2. substring may exist within string more than once. The ocurrence attribute 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 of substring within string, you should search using a start_position of -1 to indicate that searching should be conducted backwards, leaving occurrence at 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.

Bookmark and Share

Related Information:

  1. Oracle SUBSTR Function – SQL Syntax Examples
    The Oracle SUBSTR SQL Function...
  2. Oracle SUBSTR with INTSR Function – SQL Syntax Examples
    Oracle SUBSTR and INSTR SQL...
  3. Oracle TO_DATE Function – SQL Syntax Examples
    The Oracle TO_DATE SQL function...
  4. Oracle TO_CHAR Function – SQL Syntax Examples (Most With Dates, TO_DATE)
    This article provides common examples...
  5. SQL LIKE – Oracle LIKE Operator – SQL Syntax Examples
    The reserved word LIKE in...
  6. First and Last Day of Month – Oracle SQL Syntax Examples
    First Day of Current Month...
  7. XMLFOREST vs XMLELEMENT – Missing vs Empty in SQLX
    Like me, some people may...
  8. Oracle SQL Subselect Statements
    It is becoming more common...

5 Comments

  1. Mukesh Pandey

    Thanks for above examples. It really help me lot.

  2. ashok.aggarwal

    You’re welcome, Mukesh. Glad this helped.

  3. Souvik Ghosh

    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

    • Brad Gallien

      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.

  4. Brad Gallien

    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

Leave a Reply