I have found a CSV parsing issue with FasterCSV (1.5.0) which seems like a genuine bug, but which I'm hoping there's a workaround for.
Basically, adding a space after the separator (in my case a comma) when the fields are enclosed in quotes generates a MalformedCSVError.
Here's a simple example:
# No quotes on fields -- works fine
FasterCSV.parse_line("one,two,three")
=> ["one", "two", "three"]
# Quotes around fields with no spaces after separators -- works fine
FasterCSV.parse_line("\"one\",\"two\",\"three\"")
=> ["one", "two", "three"]
# Quotes around fields but with a space after the first separator -- fails!
FasterCSV.parse_line("\"one\", \"two\",\"three\"")
=> FasterCSV::MalformedCSVError: Illegal quoting on line 1.
Am I going mad, or is this a bug in FasterCSV?
The MalformedCSVError is correct here.
Leading/trailing spaces in CSV format are not ignored, they are considered part of a field. So this means you have started a field with a space, and then included unescaped double quotes in that field, which would cause the illegal quoting error.
Maybe this library is just more strict than others you have used.
Maybe you could set the :col_sep: option to ', ' to make it parse files like that.
I had hoped that the :col_sep option might allow a regular expression, but it seems to be used for both reading and writing, which is a shame. The documentation doesn't hold out much hope and your need is probably more immediate than could be satisfied by requesting a change or submitting a patch ;-)
If you're calling #parse_line explicitly, then you could always call
gsub(/,\s*/, ',')
on your input line. That regular expression might need to change significantly if you anticipate the possibility of comma-space within quoted strings. (I'd suggest reposting such a question here with a suitable tag and let the RegEx mavens loose on it should that be the case).
Related
I have a string:
"hello\n\nsomeletters\t\nmoreletters\n"
What I want:
"hello\\n\\nsomeletters\\t\\nmoreletters\\n"
How to do it?
I know a gsub way. But it sounds very simple and seems to be a common problem therefore I am sure that Ruby Gods have already sent us a solution.
There are different possibilities. The closest to what you want would be Regexp#escape:
Regexp.escape "hello\n\nsomeletters\t\nmoreletters\n"
#⇒ "hello\\n\\nsomeletters\\t\\nmoreletters\\n"
But be aware it will escape some other symbols having a special meaning in regular expressions.
Also, we have Shellwords#escape, which is probably not what you want here.
For escaping backslashes only there is no dedicated method because this operation basically has a little sense and it is not worth it to call it instead of:
"hello\n\nsomeletters\t\nmoreletters\n".gsub(
/\n|\t/, {"\n" => "\\n", "\t" => "\\t"}
)
Please note, there are no slash characters in the initial string, hence you are to match all the expected sequences.
How can I escape an apostrophe in golang?
I have a string
s = "I've this book"
and I want to make it
s = "I\'ve this book"
How to achieve this?
Thanks in advance.
Escaping a character is only necessary if it can be interpreted in two or more ways. The apostrophe in your string can only be interpreted as an apostrophe, escaping is therefore not necessary as such. This is probably why you see the error message unknown escape sequence: '.
If you need to escape the apostrophe because it is inserted into a database, first consider using library functions for escaping or inserting data directly. Correct escaping has been the culprit of many security problems in the last decades. You will almost certainly do it wrong.
Having said that, you have to escape \ to do what you want (click to play):
fmt.Println("\\'") # outputs \'
As you're using cassandra, you can use packages like gocql which provide you with parametrized queries:
session.Query(`INSERT INTO sometable (text) VALUES (?)`, "'escaping'").Exec();
How can I match a balanced pair of delimiters not escaped by backslash (that is itself not escaped by a backslash) (without the need to consider nesting)? For example with backticks, I tried this, but the escaped backtick is not working as escaped.
regex = /(?!<\\)`(.*?)(?!<\\)`/
"hello `how\` are` you"
# => $1: "how\\"
# expected "how\\` are"
And the regex above does not consider a backslash that is escaped by a backslash and is in front of a backtick, but I would like to.
How does StackOverflow do this?
The purpose of this is not much complicated. I have documentation texts, which include the backtick notation for inline code just like StackOverflow, and I want to display that in an HTML file with the inline code decorated with some span material. There would be no nesting, but escaped backticks or escaped backslashes may appear anywhere.
Lookbehind is the first thing everyone thinks of for this kind of problem, but it's the wrong tool, even in flavors like .NET that support unrestricted lookbehinds. You can hack something up, but it's going to be ugly, even in .NET. Here's a better way:
`[^`\\]*(\\.[^`\\]*)*`
The first part starts from the opening delimiter and gobbles up anything that's not the delimiter or a backslash. If the next character is a backslash, it consumes that and the character following it, whatever it may be. It could be the delimiter character, another backslash, or anything else, it doesn't matter.
It repeats those steps as many times as necessary, and when neither [^`\\] nor \\. can match, the next character must be the closing delimiter. Or the end of the string, but I'm assuming the input is well formed. But if it's not well formed, this regex will fail very quickly. I mention that because of this other approach I see a lot:
`(?:[^`\\]+|\\.)*`
This works fine on well-formed input, but what happens if you remove the last backtick from your sample input?
"hello `how\` are you"
According to RegexBuddy, after encountering the first backtick, this regex performed 9,252 distinct operations (or steps) before it could give up and report failure; mine failed in ten steps.
EDIT To extract just the par inside the delimiters, wrap that part in a capturing group. You'll still have to remove the backslashes manually.
`([^`\\]*(?:\\.[^`\\]*)*)`
I also changed the other group to non-capturing, which I should have done from the start. I don't avoid capturing religiously, but if you are using them to capture stuff, any other groups you use should be non-capturing.
EDIT I think I've been reading too much into the question. On StackOverflow, if you want to include literal backticks in an inline-code segment or a comment, you use three backticks as the the delimiter, not just one. Since there's no need to escape backticks, you can ignore backslashes as well. Your regex could turn out to be as simple as this:
```(.*?)```
Dealing with the possibility of false delimiters, you use the same basic technique:
```([^`]*(?:`(?!``)[^`]*)*)```
Is this what you're after?
By the way, this answer doesn't contradict #nneonneo's comment above. This answer doesn't consider the context in which the match is taking place. Is it in the source code of a program or web page? If it is, did the match occur inside a comment or a string literal? How do I even know the first backtick I found wasn't escaped? Regexes don't know anything about the context in which they operate; that's what parsers are for.
If you don't need nesting, regexes can indeed be a proper tool. Lexers of programming languages, for instance, use regexes to tokenize strings, and strings usually allow their own delimiters as an escaped content. Anything more complicated than that will probably need a full-blown parser though.
The "general formula" is to match an escaped character (\\.) or any character that's valid as content but don't need to be escaped ([^{list of invalid chars}]). A "naïve" solution would be joining them with or (|), but for a more efficient variant see #AlanMoore's answer.
The complete example is shown below, in two variants: the first assumes than backslashes should only be used for escaping inside the string, the second assumes that a backslash anywhere in the text escapes the next character.
`((?:\\.|[^`\\])*)`
(?:\\.|[^`\\])*`((?:\\.|[^`\\])*)`
Working examples here and here. However, as #nneonneo commented (and I endorsed), regexes are not meant to do a complete parse, so you'd better keep things simple if you want them to work out right (do you want to find a token in the text, or do you want to delimit it already knowing where it starts? The answer to that question is important to decide which strategy works best for your case).
I need to find strings with * and / using reg-exes, I am writing in Ruby.The reason for this need to find lots of * and / is that I am building a tokenizer for an language and there are multi-line comments that use the C style of multi-line comments (/* */). I have the single line comments handled already.
Is there a way to use reg-ex without having to use the two foreword slashes to indicate some regular expression because I am finding it impossible to find my mistakes due to the insane amount of escaping. Or can someone give me advise on how to handle the escaping in a sane matter? I already tried writing the sequence first then escaping it.
Thank you for your time and advise.
One trick that might help is the %r literal:
%r{http://www\.google\.com}
I like to use pipes myself, when they're not in the regex.
%r|http://www\.google\.com|
You can also create new instances of Regexp via Regexp.new and pass a string.
Finally, you might also look at Regexp.quote:
Escapes any characters that would have special meaning in a regular expression. Returns a new escaped string, or self if no characters are escaped. For any string, Regexp.new(Regexp.escape(str))=~str will be true.
In my Ruby app, I've used the following method and regular expression to remove all HTML tags from a string:
str.gsub(/<\/?[^>]*>/,"")
This regular expression did just about all I was expecting it to, except it caused all quotation marks to be transformed into “
and all single quotes to be changed to ”
.
What's the obvious thing I'm missing to convert the messy codes back into their proper characters?
Edit: The problem occurs with or without the Regular Expression, so it's clear my problem has nothing to do with it. My question now is how to deal with this formatting error and correct it. Thanks!
Use CGI::unescapeHTML after you perform your regular expression substitution:
CGI::unescapeHTML(str.gsub(/<\/?[^>]*>/,""))
See http://www.ruby-doc.org/core/classes/CGI.html#M000547
In the above code snippet, gsub removes all HTML tags. Then, unescapeHTML() reverts all HTML entities (such as <, “) to their actual characters (<, quotes, etc.)
With respect to another post on this page, note that you will never ever be passed HTML such as
<tag attribute="<value>">2 + 3 < 6</tag>
(which is invalid HTML); what you may receive is, instead:
<tag attribute="<value>">2 + 3 < 6</tag>
The call to gsub will transform the above to:
2 + 3 < 6
And unescapeHTML will finish the job:
2 + 3 < 6
You're going to run into more trouble when you see something like:
<doohickey name="<foobar>">
You'll want to apply something like:
gsub(/<[^<>]*>/, "")
...for as long as the pattern matches.
This regular expression did just about
all I was expecting it to, except it
caused all quotation marks to be
transformed into “ and all
single quotes to be changed to ”
.
This doesn't sound as if the RegExp would be doing this. Are you sure it's different before?
See this question here for information about the problem, it has got an excellent answer:
Get non UTF-8 form fields as UTF-8 in php.
I've run into a similar problem with character changes, this happened when my code ran through another module that enforced UTF-8 encoding and then when it came back, I had a different file (slurped array of lines) on my hands.
You could use a multi-pass system to get the results you are looking for.
After running your regular expression, run an expression to convert &8220; to quotes and another to convert &8221; to single quotes.