Removing Trailing Characters from String

Difficulty Level

Beginner

Published on : 08/11/2022

The `mf_trimstr` macro takes two arguments: `basestr` (the string to modify) and `trimstr` (the string to remove). It determines the length of both strings and checks if `basestr` is long enough. If `trimstr` is found at the end of `basestr`, it is removed. Otherwise, the original `basestr` is returned. The macro also handles cases where `basestr` and `trimstr` are identical, or if `basestr` is empty.

Data Analysis

Type : CREATION_INTERNE

The macro operates only on strings provided as parameters. It does not read or write external data or SAS tables. It uses macro functions for string manipulation.

1 Code Block

MACRO

Explanation :
This block defines the `mf_trimstr` macro.
1. **Local variable declaration**: `%local baselen trimlen trimval;` declares local macro variables for string lengths and the value to compare.
2. **Length check**: The lengths of `basestr` and `trimstr` are calculated. If `basestr` is shorter than `trimstr` or is empty, the macro terminates (`%return`).
3. **Substring extraction**: The macro extracts a substring from `basestr` of the same length as `trimstr`, starting from the end of `basestr`. `%superq` is used to handle special characters.
4. **Comparison and removal**:
- If `basestr` is identical to `trimstr`, the macro terminates (implicitly returns an empty string).
- If the extracted substring (`trimval`) matches `trimstr`, then `trimstr` is removed from the end of `basestr` by returning a substring of `basestr` excluding `trimstr`.
- Otherwise (no match), the original `basestr` is returned.

Copied!

1	%macro mf_trimstr(basestr,trimstr);
2	%local baselen trimlen trimval;
3
4	/* return if basestr is shorter than trimstr (or 0) */
5	%let baselen=%LENGTH(%superq(basestr));
6	%let trimlen=%LENGTH(%superq(trimstr));
7	%IF &baselen < &trimlen or &baselen=0 %THEN %return;
8
9	/* obtain the characters from the end of basestr */
10	%let trimval=%qsubstr(%superq(basestr)
11	,%LENGTH(%superq(basestr))-&trimlen+1
12	,&trimlen);
13
14	/* compare and if matching, chop it off! */
15	%IF %superq(basestr)=%superq(trimstr) %THEN %DO;
16	%return;
17	%END;
18	%ELSE %IF %superq(trimval)=%superq(trimstr) %THEN %DO;
19	%qsubstr(%superq(basestr),1,%LENGTH(%superq(basestr))-&trimlen)
20	%END;
21	%ELSE %DO;
22	&basestr
23	%END;
24
25	%mend mf_trimstr;

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Difficulty Level

Published on : 08/11/2022

Data Analysis

1 Code Block

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