# How do you remove spaces in the middle of a string in Cobol?

## How do you remove spaces in the middle of a string in Cobol?

CALL “STR\$TRIM” USING BY DESCRIPTOR user_output, user_input, BY REFERENCE WS-STRING-LENGTH. a more general solution: 01 length pic 99. perform varying length from 1 by 1 until length > 30 or user-input[length] = space end-perform.

How do I remove special characters in Cobol?

PROCEDURE DIVISION. PERFORM VARYING WS-I FROM 1 BY 1 UNTIL WS-I > FUNCTION LENGTH(WS-STR) IF WS-STR(WS-I:1) = ‘*’ THEN CONTINUE ELSE MOVE WS-STR(WS-I:1) TO WS-LETTER(WS-J) ADD 1 TO WS-J ADD 1 TO WS-CNT END-IF END-PERFORM DISPLAY WS-CHAR STOP RUN.

How do you get rid of extra spaces in Cobol?

You can simply use the Function UNSTRING and remove the spaces. Code: 01 WS-STR PIC X(20) VALUE ‘ BBXXXYYYY’. 01 WS-SPACE PIC X(20).

### How do you remove trailing spaces in Cobol string?

MOVE 1 TO W-IX1. COMPUTE W-ROUNDED-AMNT ROUNDED = W-INPUT-AMNT * 1. MOVE W-ROUNDED-AMNT TO W-TEMP-AMNT. INSPECT W-TEMP-AMNT TALLYING W-IX1 FOR LEADING SPACES.

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How do you remove leading zeros using inspect?

Try this:

1. DEFINE a tally-field with a PIC of S9(2) COMP.
2. REDEFINE source-field as source-field-X with a PIC of X(17).
3. MOVE SPACES TO target-field.
4. MOVE ZERO TO tally-field.
5. INSPECT source-field-X TALLYING tally-field FOR LEADING ZEROS.

How can I remove multiple special characters from a string in SQL?

How To Remove Characters & Special Symbols From String Using SQL Function

1. Create function [dbo].[RemoveCharSpecialSymbolValue](@str varchar(500))
2. returns varchar(500)
3. begin.
4. declare @startingIndex int.
5. set @startingIndex=0.
6. while 1=1.
7. begin.
8. set @startingIndex= patindex(‘\%[^0-9. ]\%’,@str)

#### How do you inspect in Cobol?

INSPECT Statements

1. INSPECT statement with TALLYING phrase (Key Phrases to use: BEFORE/AFTER, CHARACTERS, ALL, LEADING and FIRST)
2. INSPECT statement with REPLACING phrase (Key Phrases to use: BEFORE/AFTER, CHARACTERS BY, ALL, LEADING and FIRST)
3. INSPECT statement with TALLYING and REPLACING phrases.

How do you remove leading and trailing spaces from string in Cobol?

INSPECT FUNCTION REVERSE (W-OUTPUT-AMNT) TALLYING W-TRAIL FOR LEADING SPACE. SUBTRACT W-TRAIL FROM LENGTH OF W-OUTPUT-AMNT GIVING W-LENGTH. DISPLAY “RESULT:” W-OUTPUT-AMNT(1:W-LENGTH) “:”. MOVE SPACE TO W-OUTPUT-AMNT.

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How do you trim a space in Cobol?

Couple of ways to trim the trailing spaces. MOVE FUNCTION REVERSE(INPUT) TO INPUT-R. INSPECT INPUT-R TALLYING CNTR FOR LEADING SPACES. COMPUTE INP-LEN = LENGTH OF INPUT-R – CNTR.

## Which function used to remove all leading and trailing whitespaces in string?

The STRIP function is similar to the TRIM function. It removes both the leading and trailing spaces from a character string.

How do you remove spaces from a string in Swift?

To remove all leading whitespaces, use the following code: var filtered = “” var isLeading = true for character in string { if character. isWhitespace && isLeading { continue } else { isLeading = false filtered.

How do I remove leading spaces in COBOL?

If you want to remove leading spaces, a similar approach can be used starting at the front of USER-INPUT or you could try playing with the INSPECT verb with TALLYING and LEADING modifiers. Sorry, but sting management is not one of COBOL’s strong points. – NealB Jan 12 ’10 at 17:43

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### What are the string handling statements in COBOL?

String handling statements in COBOL are used to do multiple functional operations on strings. Following are the string handling statements −. Inspect; String; Unstring; Inspect. Inspect verb is used to count or replace the characters in a string. String operations can be performed on alphanumeric, numeric, or alphabetic values.

How to replace string characters with JCL in COBOL?

JCL to execute the above COBOL program. When you compile and execute the above program, it produces the following result − Replacing option is used to replace the string characters. INSPECT input-string REPLACING ALL char1 BY char2. input-string − The string whose characters are to be replaced from char1 to char2.

How to move 0 to w-ix2 for characters before space?

MOVE 0 TO W-IX2 INSPECT W-OUTPUT-AMNT TALLYING W-IX2 FOR CHARACTERS BEFORE SPACE. DISPLAY “RESULT:” W-OUTPUT-AMNT (1:W-IX2) “:”. The easy way to do this is to count backwards from the end of the field to the first non-space character.