What does $$ mean when using DO, CREATE FUNCTION, etc? [duplicate] - sql

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What are '$$' used for in PL/pgSQL
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I have seen the double dollar sign ($$) being used often in FUNCTION declarations and while doing declare statements. I acknowledge that it is probably some syntax related feature, but I am not too clear how/why to use it and when.
Code examples of usage below:
CREATE OR REPLACE FUNCTION BuildFunction(IN prefix TEXT, IN func_id INTEGER) RETURNS BOOLEAN AS $$
BEGIN
...
END; $$
Point of confusion, how do you return a boolean as a $$, is it some special type or something?
DO $$
DECLARE
a integer := 10;
b integer := 20;
c integer;
BEGIN
c := a + b;
RAISE NOTICE'Value of c: %', c;
END $$;
This code snippet is from How do you use variables in a simple PostgreSQL script?

From the fine manual:
4.1.2.4. Dollar-Quoted String Constants
While the standard syntax for specifying string constants is usually convenient, it can be difficult to understand when the desired string contains many single quotes or backslashes, since each of those must be doubled. To allow more readable queries in such situations, PostgreSQL provides another way, called “dollar quoting”, to write string constants. A dollar-quoted string constant consists of a dollar sign ($), an optional “tag” of zero or more characters, another dollar sign, an arbitrary sequence of characters that makes up the string content, a dollar sign, the same tag that began this dollar quote, and a dollar sign.
The dollar quoting is a PostgreSQL-specific way of quoting strings that will contain a lot of internal single quotes without forcing you to escape quotes and make an unreadable mess. So these are equivalent string literals:
'R''hllor'
$$R'hllor$$
Dollar-quoting isn't specific to functions, you can use it for any string literals, but it is most commonly seen in function definitions as in your two examples.

Related

Remove accents from string in Oracle

When trying to remove all accents from a string in Oracle using the techniques described in this stackoverflow answer: how replace accented letter in a varchar2 column in oracle I’m getting mixed results.
select CONVERT('JUAN ROMÄN', 'US7ASCII') from dual;
Returns the original string but replaces characters with for example ñ by a question mark (probably because of the chosen charset - tests with different charsets led to different results).
Using the following technique:
select utl_raw.cast_to_varchar2(nlssort(NAME_USER, 'nls_sort=binary_ai')) from YOUR_TABLE;
Returns the complete string but also places a NUL value at the end of the string.
Is there a characterset that I can use with Spanish accents to get a correct result (the original string with the different accents removed); is there a way to avoid the NUL value in the utl_raw.cast_to_varchar2 technique?
Based on the comments the the replace char(0) seems to remove the NUL value. For example
select
upper(utl_raw.cast_to_varchar2((nlssort('this is áà ñew test','nls_sort=binary_ai')))) as test,
replace(upper(utl_raw.cast_to_varchar2((nlssort('this is áà ñew test','nls_sort=binary_ai')))),chr(0),'') as test2
from dual;
If possible I would however to have a more 'straightforward/simpler' solution.
You can use TRANSLATE(your_string, from_chars, to_chars) https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions196.htm
Just put all chars with accents in from_chars string and their corresponding replacement chars in to_chars.

What regular expression characters have to be escaped in SQL?

To prevent SQL injection attack, the book "Building Scalable Web Sites" has a function to replace regular expression characters with escaped version:
function db_escape_str_rlike($string) {
preg_replace("/([().\[\]*^\$])/", '\\\$1', $string);
}
Does this function escape ( ) . [ ] * ^ $? Why are only those characters escaped in SQL?
I found an excerpt from the book you mention, and found that the function is not for escaping to protect against SQL injection vulnerabilities. I assumed it was, and temporarily answered your question with that in mind. I think other commenters are making the same assumption.
The function is actually about escaping characters that you want to use in regular expressions. There are several characters that have special meaning in regular expressions, so if you want to search for those literal characters, you need to escape them (precede with a backslash).
This has little to do with SQL. You would need to escape the same characters if you wanted to search for them literally using grep, sed, perl, vim, or any other program that uses regular expression searches.
Unfortunately, active characters in sql databases is an open issue. Each database vendor uses their own (mainly oracle's mysql, that uses \ escape sequences)
The official SQL way to escape a ', which is the string delimiter used for values is to double the ', as in ''.
That should be the only way to ensure transparency in SQL statements, and the only way to introduce a proper ' into a string. As soon as any vendor admits \' as a synonim of a quote, you are open to support all the extra escape sequences to delimit strings. Suppose you have:
'Mac O''Connor' (should go into "Mac O'Connor" string)
and assume the only way to escape a ' is that... then you have to check the next char when you see a ' for a '' sequence and:
you get '' that you change into '.
you get another, and you terminate the string literal and process the char as the first of the next token.
But if you admit \ as escape also, then you have to check for \' and for \\', and \\\' (this last one should be converted to \' on input) etc. You can run into trouble if you don't detect special cases as
\'' (should the '' be processed as SQL mandates, or the first \' is escaping the first ' and the second is the string end quote?)
\\'' (should the \\ be converted into a single \ then the ' should be the string terminator, or do we have to switch to SQL way of encoding and consider '' as a single quote?)
etc.
You have to check your database documentation to see if \ as escape characters affect only the encoding of special characters (like control characters or the like) and also affects the interpretation of the quote character or simply doesn't, and you have to escape ' the other way.
That is the reason for the vendors to include functions to do the escape/unescape of character literals into values to be embedded in a SQL statement. The idea of the attackers is to include (if you don't properly do) escape sequences into the data they post to you to see if that allows them to modify the text of the sql command to simply add a semicolon ; and write a complete sql statement that allows them to access freely your database.

Round A Variable Autohotkey

I've been trying to round a variable for the past hour and cannot figure out how to do so.
average_load = 1.234
average_load := Round(%average_load%)
It keeps saying that it contains an illegal character.
Functions are always expression capable and thus variables do not need to be quoted.
average_load = 1.234
average_load := Round(average_load)
What is expression capable? See examples of Variables and Expressions.
It is same difference as := is expression ready so variables do not need to be declared whereas for = the expressions must be declared with surrounding % or a leading %.

What is significance of trailing "$" in some function names?

I recently looked at some vba source at Microsoft: [Convert Fractions to Decimal Values][1]
[1]: https://support.microsoft.com/en-us/kb/185424 and I noticed that several functions had a trailing "$", specifically trim$(), left$(), and mid$(). My question is: what does the "$" signify?
I downloaded the microsoft function and it ran correctly under Excel 2007.
Since VBA trim() works differently from the worksheet function trim(), I wrote a small program to compare the operation of the 3 possible trim() calls. I found that trim() and trim$() produced identical output. worksheetfunction.trim(), of course, produces output that has extraneous space characters removed from inside the string.
I am very curious about the trailing "$", and will be grateful for enlightenment!
Thank you,
Dave
To quote from https://bytes.com/topic/access/answers/196893-difference-between-left-left-function
Allen Browne
The trailing $ is a type declaration character for the String data type in
VBA.
The result returned from Left$() is a string, whereas Left() returns a
Variant.
You must use Left(), not Left$() if there is any chance of Null values,
since the Variant can be Null but the String cannot.
That post has a full worked example
The syntax is a left-over habit from ancient history. In early versions of Basic variables did not have to be declared but data types were implied by the name of the variable. Any variable ending with $ was a string and any variable ending with % was an integer.
FORTRAN had a similar convention: any variable starting with the letters I, J, K, L, M or N were integers, all others were real.

How do you escape this regular expression?

I'm looking for
"House M.D." (2004)
with anything after it. I've tried where id~'"House M\.D\." \(2004\).*'; and there's no matches
This works id~'.*House M.D..*2004.*'; but is a little slow.
I suspect you're on an older PostgreSQL version that interprets strings in a non standards-compliant C-escape-like mode by default, so the backslashes are being treated as escapes and consumed. Try SET standard_conforming_strings = 'on';.
As per the lexical structure documentation on string constants, you can either:
Ensure that standard_conforming_strings is on, in which case you must double any single quotes (ie ' becomes '') but backslashes aren't treated as escapes:
id ~ '"House M\.D\." \(2004\)'
Use the non-standard, PostgreSQL-specific E'' syntax and double your backslashes:
id ~ E'"House M\\.D\\." \\(2004\\)'
PostgreSQL versions 9.1 and above set standard_conforming_strings to on by default; see the documentation.
You should turn it on in older versions after testing your code, because it'll make updating later much easier. You can turn it on globally in postgresql.conf, on a per-user level with ALTER ROLE ... SET, on a per-database level with ALTER DATABASE ... SET or on a session level with SET standard_conforming_strings = on. Use SET LOCAL to set it within a transaction scope.
Looks that your regexp is ok
http://sqlfiddle.com/#!12/d41d8/113
CREATE OR REPLACE FUNCTION public.regexp_quote(IN TEXT)
RETURNS TEXT
LANGUAGE plpgsql
STABLE
AS $$
/*******************************************************************************
* Function Name: regexp_quote
* In-coming Param:
* The string to decoded and convert into a set of text arrays.
* Returns:
* This function produces a TEXT that can be used as a regular expression
* pattern that would match the input as if it were a literal pattern.
* Description:
* Takes in a TEXT in and escapes all of the necessary characters so that
* the output can be used as a regular expression to match the input as if
* it were a literal pattern.
******************************************************************************/
BEGIN
RETURN REGEXP_REPLACE($1, '([[\\](){}.+*^$|\\\\?-])', '\\\\\\1', 'g');
END;
$$
Test:
SELECT regexp_quote('"House M.D." (2004)'); -- produces: "House M\\.D\\." \\(2004\\)