How do I add colon as a syntactic construct in my language? I've added it to the syntax, but when I try to match a colon inside a rule it gives me a parse error.
K is getting confused when you surround it with parenthesis. Delete the surrounding parenthesis and the parse error should be removed.
Related
I am getting a syntax error in a PostgreSQL query. I am working on a project developed in YII1, I am getting an error
CDbCommand failed to execute the SQL statement: SQLSTATE[42601]:
Syntax error: 7 ERROR: syntax error at or near "s" LINE 1: ...OT NULL
AND sub_heading like '%Women and Children's Voices%'.
As you can see above, I am using the like operator in single quotes, and in the string there is another single quote (Children's). So PostgreSQL is throwing me an error. Please provide me a solution to escape the string.
You can escape a single quote in a string by using another single quote (i.e., '' instead of '. Note that these are two ' characters, not a single " character):
sub_heading LIKE '%Women and Children''s Voices%'
-- Here -----------------------------^
You should use the format function to construct the SQL statement, using the %L placeholder for the pattern.
I solved this problem by replacing the single quote with double quotes using PHP. Here is the code
There is a variable $var with value Women and Children's Voices. I replace that single quote using the str_replace() function.
$var = str_replace("'", "''", $var);
I am using ANTLR to parse a language which uses the colon for both a comment indicator and as part of a 'becomes equal to' assignment. So for example in the line
Index := 2 :Set Index
I need to recognize the first part as an assignment statement and the text after the second colon as a comment. Currently I do this using the rule:
COMMENT : ':'+ ~[:='\r\n']*;
This seems to work OK apart from when the colon is immediately followed by a new line. e.g. in the line
Index := 2 :
the newline occurs immediately after the second colon. In this case the comment is not recognized and the rest of the code is not parsed in the correct context. If there is a single space after the second colon the line is parsed correctly.
I expected the '\r'\n' to cope with this but it only seems to work if there is at least one character after the comment symbol - have I missed something from the command?
The braces denote a collection of characters without any quotes. Hence your '\r\n' literal doesn't work there (you should have got a warning that the apostrophe is included more than once in the char range.
Define the comment like this instead:
COMMENT: ':'+ ~[:=\n\r]*;
I'm trying to use REGEXP_REPLACE to remove all punctuation from a varchar. I'm using the following:
regexp_replace(d.NAME, [.,\/#!$%\^&\*;:{}=\-_`~()])
But it gives me an error, saying:
Statement 1 is not valid. ERROR: syntax error at or near "."
How can I fix this to remove all punctuation?
Firstly, the dash in a character class means a range, except when it's first or last... so put it there:
[.,\/#!$%\^&\*;:{}=\_`~()-]
And, you have to put it in quotes, and most characters don't need escaping:
regexp_replace(d.NAME, '[.,/#!$%^&*;:{}=_`~()-]')
I'm trying to do something like this
SELECT epi_week {week}, state
FROM
lookerdata:cdc.project_tycho_reports
LIMIT 10
Error: Encountered " "{" "{ "" at line 1, column 17. Was expecting: EOF>
It seems that curly braces are not legal syntax. I've tried escaping or using quotes without success.
Is there a way around this? We use the braces as an indication for post-processing string replacement to support multiple languages.
Is there a way around this?
No way unfortunatelly.
Field name must contain only letters, numbers, and underscores, start with a letter or underscore, and be at most 128 characters long.
As an option - you might want to come up with another name convention for post-processing
I'm writing a regular expression in Objective-C.
The escape sequence \w is illegal and emits a warning, so the regular expression /\w/ must be written as #"\\w"; the escape sequence \? is valid, apparently, and doesn't emit a warning, so the regular expression /\?/ must be written as #"\\?" (i.e., the backslash must be escaped).
Question marks aren't invisible like \t or \n, so why is \? a valid escape sequence?
Edit: To clarify, I'm not asking about the quantifier, I'm asking about a string escape sequence. That is, this doesn't emit a warning:
NSString *valid = #"\?";
By contrast, this does emit a warning ("Unknown escape sequence '\w'"):
NSString *invalid = #"\w";
It specifies a literal question mark. It is needed because of a little-known feature called trigraphs, where you can write a three-character sequence starting with question marks to substitute another character. If you have trigraphs enabled, in order to write "??" in a string, you need to write it as "?\?" in order to prevent the preprocessor from trying to read it as the beginning of a trigraph.
(If you're wondering "Why would anybody introduce a feature like this?": Some keyboards or character sets didn't include commonly used symbols like {. so they introduced trigraphs so you could write ??< instead.)
? in regex is a quantifier, it means 0 or 1 occurences. When appended to the + or * quantifiers, it makes it "lazy".
For example, applying the regex o? to the string foo? would match o.
However, the regex o\? in foo? would match o?, because it is searching for a literal question mark in the string, instead of an arbitrary quantifier.
Applying the regex o*? to foo? would match oo.
More info on quantifiers here.