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9 Replies Latest reply: Mar 7, 2013 9:23 PM by 995602

# Substring instr issue in obiee

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Hi,

I want to use the INSTR function in OBIEE

POSITION function found in the forum

INSTR function takes four values:
INSTR (string1, string2, number, number)

111.222.333.444

The desired result
SUBSTR (IP_ADD, 1, INSTR (IP_ADD, '.', 1,3) -1)
111.222.333

POSITION function, how should I use?

Thanks,
Mino
• ###### 1. Re: Substring instr issue in obiee
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Mino,

Here you go, Unfortunately, in this case you can't use OBIEE Position string function as the IP has repeated number of decimals(.) in it. Only function you can use here is LOCATEN. But there is a bug in OBIEE 10g with LOCATEN syntax. So, in order to locate the 3rd '.' position, you will have to use below function:

LOCATE(expr1, expr2,int) where int is the no. you want the search to start from. For sure, in any case IPs 3rd decimal will start from position 9. So, in this case you will have to hardcode int 9 here.

So, here is your formula to be used in OBIEE:
SUBSTRING('111.222.333.444' FROM 1 FOR LOCATE('.', '111.222.333.444', 9)-1)
OR
SUBSTRING(IP_ADD FROM 1 FOR LOCATE('.', IP_ADD, 9)-1)

This will give you 111.222.333

Hope its useful...
• ###### 2. Re: Substring instr issue in obiee
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Thanks for the answer.

However, IP number of the information is not uniform.

0.0.0.0
001.02.033.3
03.222.333.343

IP irregular value.

any other solution??
• ###### 3. Re: Substring instr issue in obiee
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Description. position function(BMM layer)

<<expr>> Position (% 1 IN% 2) (<<strExpr>>, <<strExpr>>)

(<<strExpr>>, <<strExpr>>) What is the meaning?
• ###### 4. Re: Substring instr issue in obiee
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1st example '0.0.0.0' will def be a problem here. Unfortunately, I don't think any other OBIEE string function could support this.

I was assuming for min 2 numbers like 00.00.00.00. Do you have any case like the first one in your table ?

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As I said, you can't use POSITION function here..Just do help for String Function in RPD)

Position
Returns the numerical position of the character_expression1 in a character expression. If the character_expression1 is not found, the function returns 0.

Syntax:

POSITION(character_expression1 IN character_expression2)

where:

character_expression1
Any expression that evaluates to a character string. Used to search in the second string.

character_expression2
Any expression that evaluates to a character string.

So, these are the 2 expression, In your case its '.' & IP_ADDR.

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• ###### 5. Re: Substring instr issue in obiee
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830583 wrote:
any other solution??
It's really simple
``SUBSTRING('111.222.333.444' FROM 1 FOR LOCATE('.','111.222.333.444',-1)-1)``
• ###### 6. Re: Substring instr issue in obiee
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I has solved in the following way.

SUBSTRING ('111 .222.333.444 'FROM 1 FOR LOCATE ('.', '111 .222.333.444', LOCATE ('.', '111 .222.333.444 ', LOCATE ('.', '111 .222.333.444') + 1) + 1))

But your suggestion is a better way than my way.

Thanks AnTonic.
Mino
• ###### 7. Re: Substring instr issue in obiee
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Good Learning. Thnx Antonic !!
• ###### 8. Re: Substring instr issue in obiee
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This is great solution. Saved my time. Locate N didnt work in 11g. But this solution works!!!!

Thank You :)
• ###### 9. Re: Substring instr issue in obiee
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if you are facing substring issue in obiee regarding specified string. and u want to string before specified character or words.
i have one solution for this.

you want to string before 'g' character
SUBSTRING("deepak123gupta" FROM 1 FOR LOCATE('g',"deepak123gupta")-1)

this will display deepak123