I want to replace all the occurrences of a string that doesn't start with "<pre>" and doesn't end in "</pre>".
So let's say I wanted to find new-line characters and replace them with "<p/>". I can get the "not followed by" part:
var revisedHtml = html.replace(/[\n](?![<][/]pre[>])/g, "<p/>");
But I don't know the "not starting with" part to put at the front.
Any help please? :)
Here's how Steve Levithan's first lookbehind-alternative can be applied to your problem:
var output = s.replace(/(<pre>[\s\S]*?<\/pre>)|\n/g, function($0, $1){
return $1 ? $1 : '<p/>';
});
When it reaches a <pre> element, it captures the whole thing and plugs it right back into the output. It never really sees the newlines inside the element, just gobbles them up along with all other content. Thus, when the \n in the regex does match a newline, you know it's not inside a <pre> element, and should be replaced with a <p/>.
But don't make the mistake of regarding this technique as a hack or a workaround; I would recommend this approach even if lookbehinds were available. With the lookaround approach, the regex has to examine every single newline and apply the lookarounds each time to see if it should be replaced. That's a lot of unnecessary work it has to do, plus the regex is a lot more complicated and less maintainable.
As always when using regexes on HTML, I'm ignoring a lot of factors that can affect the result, like SGML comments, CDATA sections, angle brackets in attribute values, etc. You'll have to determine which among those factors you have to deal with in your case, and which ones you can ignore. When it comes to processing HTML with regexes, there's no such thing as a general solution.
Why not do the reverse. Look for all the substrings enclosed in <pre> tags. Then you know which parts of your string are not enclosed in <pre>.
EDIT: More elegant solution: use split() and use the <pre> HTML as the delimiters. This gives you the HTML outside the <pre> blocks.
var s = "blah blah<pre>formatted</pre>blah blah<pre>another formatted</pre>end";
var rgx = /<pre>.*?<\/pre>/g
var nonPreStrings = s.split(rgx);
for (var idx in nonPreStrings)
alert(nonPreStrings[idx]);
Related
There may be a very simple answer to this, probably because of my familiarity (or possibly lack thereof) of the replace method and how it works with regex.
Let's say I have the following string: abcdefHellowxyz
I just want to strip the first six characters and the last four, to return Hello, using regex... Yes, I know there may be other ways, but I'm trying to explore the boundaries of what these methods are capable of doing...
Anyway, I've tinkered on http://regex101.com and got the following Regex worked out:
/^(.{6}).+(.{4})$/
Which seems to pass the string well and shows that abcdef is captured as group 1, and wxyz captured as group 2. But when I try to run the following:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
to replace those captured groups with "" I receive an empty string as my final output... Am I doing something wrong with this syntax? And if so, how does one correct it, keeping my original stance on wanting to use Regex in this manner...
Thanks so much everyone in advance...
The code below works well as you wish
"abcdefHellowxyz".replace(/^.{6}(.+).{4}$/,"$1")
I think that only use ()to capture the text you want, and in the second parameter of replace(), you can use $1 $2 ... to represent the group1 group2.
Also you can pass a function to the second parameter of replace,and transform the captured text to whatever you want in this function.
For more detail, as #Akxe recommend , you can find document on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace.
You are replacing any substring that matches /^(.{6}).+(.{4})$/, with this line of code:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
The regex matches the whole string "abcdefHellowxyz"; thus, the whole string is replaced. Instead, if you are strictly stripping by the lengths of the extraneous substrings, you could simply use substring or substr.
Edit
The answer you're probably looking for is capturing the middle token, instead of the outer ones:
var str = "abcdefHellowxyz";
var matches = str.match(/^.{6}(.+).{4}$/);
str = matches[1]; // index 0 is entire match
console.log(str);
I have the following string
class=use><em>use</em>
that when searched using us I want to transform into
class=use><em><b>us</b>e</em>
I've tried looking at relating answers but I can't quite get it working the way I want it to. I'm especially interested in this answer's callback approach.
Help appreciated
This is a good exercise for writing regular expressions, and here's a possible solution.
"useclass=use><em>use</em>".replace(/([^=]|^)(us)/g, "$1<b>$2</b>");
// returns "<b>us</b>eclass=use><em><b>us</b>e</em>"
([^=]|^) ensures that the prefix of any matched us is either not an equal sign, or it's the start of the string.
As #jamiec pointed out in the comments, if you are using this to parse/modify HTML, just stop right now. It's mathematically impossible to parse a CFG with a regular grammar (even with enhanced JS regexps you will have a bad time trying to achieve that.)
If you can make any assumptions about the structure of your document, you may be better off using an approach that operates on DOM elements directly rather than parsing the whole document with a regex.
Parsing HTML with a regex has certain problems that can be painful to deal with.
var element = document.querySelector('em');
element.innerHTML = element.innerHTML.replace('us', '<b>us</b>');
<div class=use><em>use</em>
</div>
I would first look for any character other than the equals sign [^=] and separate it by parentheses so that I can use it again in my replacement. Then another set of parentheses around the two characters us ought to do it:
var re = /([^=]|^)(us)/
That will give you two capture groups to work with (inside the parentheses), which you can represent with $1 and $2 in your replacement string.
str.replace( /([^=|^])(us)/, '$1<b>$2</b>' );
I'm trying to improve my understanding of Regex, but this one has me quite mystified.
I started with some text defined as:
var txt = "{\"columns\":[{\"text\":\"A\",\"value\":80},{\"text\":\"B\",\"renderer\":\"gbpFormat\",\"value\":80},{\"text\":\"C\",\"value\":80}]}";
and do a replace as follows:
txt.replace(/\"renderer\"\:(.*)(?:,)/g,"\"renderer\"\:gbpFormat\,");
which results in:
"{"columns":[{"text":"A","value":80},{"text":"B","renderer":gbpFormat,"value":80}]}"
What I expected was for the renderer attribute value to have it's quotes removed; which has happened, but also the C column is completely missing! I'd really love for someone to explain how my Regex has removed column C?
As an extra bonus, if you could explain how to remove the quotes around any value for renderer (i.e. so I don't have to hard-code the value gbpFormat in the regex) that'd be fantastic.
You are using a greedy operator while you need a lazy one. Change this:
"renderer":(.*)(?:,)
^---- add here the '?' to make it lazy
To
"renderer":(.*?)(?:,)
Working demo
Your code should be:
txt.replace(/\"renderer\"\:(.*?)(?:,)/g,"\"renderer\"\:gbpFormat\,");
If you are learning regex, take a look at this documentation to know more about greedyness. A nice extract to understand this is:
Watch Out for The Greediness!
Suppose you want to use a regex to match an HTML tag. You know that
the input will be a valid HTML file, so the regular expression does
not need to exclude any invalid use of sharp brackets. If it sits
between sharp brackets, it is an HTML tag.
Most people new to regular expressions will attempt to use <.+>. They
will be surprised when they test it on a string like This is a
first test. You might expect the regex to match and when
continuing after that match, .
But it does not. The regex will match first. Obviously not
what we wanted. The reason is that the plus is greedy. That is, the
plus causes the regex engine to repeat the preceding token as often as
possible. Only if that causes the entire regex to fail, will the regex
engine backtrack. That is, it will go back to the plus, make it give
up the last iteration, and proceed with the remainder of the regex.
Like the plus, the star and the repetition using curly braces are
greedy.
Try like this:
txt = txt.replace(/"renderer":"(.*?)"/g,'"renderer":$1');
The issue in the expression you were using was this part:
(.*)(?:,)
By default, the * quantifier is greedy by default, which means that it gobbles up as much as it can, so it will run up to the last comma in your string. The easiest solution would be to turn that in to a non-greedy quantifier, by adding a question mark after the asterisk and change that part of your expression to look like this
(.*?)(?:,)
For the solution I proposed at the top of this answer, I also removed the part matching the comma, because I think it's easier just to match everything between quotes. As for your bonus question, to replace the matched value instead of having to hardcode gbpFormat, I used a backreference ($1), which will insert the first matched group into the replacement string.
Don't manipulate JSON with regexp. It's too likely that you will break it, as you have found, and more importantly there's no need to.
In addition, once you have changed
'{"columns": [..."renderer": "gbpFormat", ...]}'
into
'{"columns": [..."renderer": gbpFormat, ...]}' // remove quotes from gbpFormat
then this is no longer valid JSON. (JSON requires that property values be numbers, quoted strings, objects, or arrays.) So you will not be able to parse it, or send it anywhere and have it interpreted correctly.
Therefore you should parse it to start with, then manipulate the resulting actual JS object:
var object = JSON.parse(txt);
object.columns.forEach(function(column) {
column.renderer = ghpFormat;
});
If you want to replace any quoted value of the renderer property with the value itself, then you could try
column.renderer = window[column.renderer];
Assuming that the value is available in the global namespace.
This question falls into the category of "I need a regexp, or I wrote one and it's not working, and I'm not really sure why it has to be a regexp, but I heard they can do all kinds of things, so that's just what I imagined I must need." People use regexps to try to do far too many complex matching, splitting, scanning, replacement, and validation tasks, including on complex languages such as HTML, or in this case JSON. There is almost always a better way.
The only time I can imagine wanting to manipulate JSON with regexps is if the JSON is broken somehow, perhaps due to a bug in server code, and it needs to be fixed up in order to be parseable.
From this q/a, I deduced that matching all instances of a given regex not inside quotes, is impossible. That is, it can't match escaped quotes (ex: "this whole \"match\" should be taken"). If there is a way to do it that I don't know about, that would solve my problem.
If not, however, I'd like to know if there is any efficient alternative that could be used in JavaScript. I've thought about it a bit, but can't come with any elegant solutions that would work in most, if not all, cases.
Specifically, I just need the alternative to work with .split() and .replace() methods, but if it could be more generalized, that would be the best.
For Example:
An input string of: +bar+baz"not+or\"+or+\"this+"foo+bar+
replacing + with #, not inside quotes, would return: #bar#baz"not+or\"+or+\"this+"foo#bar#
Actually, you can match all instances of a regex not inside quotes for any string, where each opening quote is closed again. Say, as in you example above, you want to match \+.
The key observation here is, that a word is outside quotes if there are an even number of quotes following it. This can be modeled as a look-ahead assertion:
\+(?=([^"]*"[^"]*")*[^"]*$)
Now, you'd like to not count escaped quotes. This gets a little more complicated. Instead of [^"]* , which advanced to the next quote, you need to consider backslashes as well and use [^"\\]*. After you arrive at either a backslash or a quote, you need to ignore the next character if you encounter a backslash, or else advance to the next unescaped quote. That looks like (\\.|"([^"\\]*\\.)*[^"\\]*"). Combined, you arrive at
\+(?=([^"\\]*(\\.|"([^"\\]*\\.)*[^"\\]*"))*[^"]*$)
I admit it is a little cryptic. =)
Azmisov, resurrecting this question because you said you were looking for any efficient alternative that could be used in JavaScript and any elegant solutions that would work in most, if not all, cases.
There happens to be a simple, general solution that wasn't mentioned.
Compared with alternatives, the regex for this solution is amazingly simple:
"[^"]+"|(\+)
The idea is that we match but ignore anything within quotes to neutralize that content (on the left side of the alternation). On the right side, we capture all the + that were not neutralized into Group 1, and the replace function examines Group 1. Here is full working code:
<script>
var subject = '+bar+baz"not+these+"foo+bar+';
var regex = /"[^"]+"|(\+)/g;
replaced = subject.replace(regex, function(m, group1) {
if (!group1) return m;
else return "#";
});
document.write(replaced);
Online demo
You can use the same principle to match or split. See the question and article in the reference, which will also point you code samples.
Hope this gives you a different idea of a very general way to do this. :)
What about Empty Strings?
The above is a general answer to showcase the technique. It can be tweaked depending on your exact needs. If you worry that your text might contain empty strings, just change the quantifier inside the string-capture expression from + to *:
"[^"]*"|(\+)
See demo.
What about Escaped Quotes?
Again, the above is a general answer to showcase the technique. Not only can the "ignore this match" regex can be refined to your needs, you can add multiple expressions to ignore. For instance, if you want to make sure escaped quotes are adequately ignored, you can start by adding an alternation \\"| in front of the other two in order to match (and ignore) straggling escaped double quotes.
Next, within the section "[^"]*" that captures the content of double-quoted strings, you can add an alternation to ensure escaped double quotes are matched before their " has a chance to turn into a closing sentinel, turning it into "(?:\\"|[^"])*"
The resulting expression has three branches:
\\" to match and ignore
"(?:\\"|[^"])*" to match and ignore
(\+) to match, capture and handle
Note that in other regex flavors, we could do this job more easily with lookbehind, but JS doesn't support it.
The full regex becomes:
\\"|"(?:\\"|[^"])*"|(\+)
See regex demo and full script.
Reference
How to match pattern except in situations s1, s2, s3
How to match a pattern unless...
You can do it in three steps.
Use a regex global replace to extract all string body contents into a side-table.
Do your comma translation
Use a regex global replace to swap the string bodies back
Code below
// Step 1
var sideTable = [];
myString = myString.replace(
/"(?:[^"\\]|\\.)*"/g,
function (_) {
var index = sideTable.length;
sideTable[index] = _;
return '"' + index + '"';
});
// Step 2, replace commas with newlines
myString = myString.replace(/,/g, "\n");
// Step 3, swap the string bodies back
myString = myString.replace(/"(\d+)"/g,
function (_, index) {
return sideTable[index];
});
If you run that after setting
myString = '{:a "ab,cd, efg", :b "ab,def, egf,", :c "Conjecture"}';
you should get
{:a "ab,cd, efg"
:b "ab,def, egf,"
:c "Conjecture"}
It works, because after step 1,
myString = '{:a "0", :b "1", :c "2"}'
sideTable = ["ab,cd, efg", "ab,def, egf,", "Conjecture"];
so the only commas in myString are outside strings. Step 2, then turns commas into newlines:
myString = '{:a "0"\n :b "1"\n :c "2"}'
Finally we replace the strings that only contain numbers with their original content.
Although the answer by zx81 seems to be the best performing and clean one, it needes these fixes to correctly catch the escaped quotes:
var subject = '+bar+baz"not+or\\"+or+\\"this+"foo+bar+';
and
var regex = /"(?:[^"\\]|\\.)*"|(\+)/g;
Also the already mentioned "group1 === undefined" or "!group1".
Especially 2. seems important to actually take everything asked in the original question into account.
It should be mentioned though that this method implicitly requires the string to not have escaped quotes outside of unescaped quote pairs.
I'm trying to remove a rectangular brackets(bbcode style) using javascript, this is for removing unwanted bbcode.
I try with this.
theString .replace(/\[quote[^\/]+\]*\[\/quote\]/, "")
it works with this string sample:
theString = "[quote=MyName;225]Test 123[/quote]";
it will fail within this sample:
theString = "[quote=MyName;225]Test [quote]inside quotes[/quote]123[/quote]";
if there any solution beside regex no problem
The other 2 solutions simply do not work (see my comments). To solve this problem you first need to craft a regex which matches the innermost matching quote elements (which contain neither [QUOTE..] nor [/QUOTE]). Next, you need to iterate, applying this regex over and over until there are no more QUOTE elements left. This tested function does what you want:
function filterQuotes(text)
{ // Regex matches inner [QUOTE]non-quote-stuff[/quote] tag.
var re = /\[quote[^\[]+(?:(?!\[\/?quote\b)\[[^\[]*)*\[\/quote\]/ig;
while (text.search(re) !== -1)
{ // Need to iterate removing QUOTEs from inside out.
text = text.replace(re, "");
}
return text;
}
Note that this regex employs Jeffrey Friedl's "Unrolling the loop" efficiency technique and is not only accurate, but is quite fast to boot.
See: Mastering Regular Expressions (3rd Edition) (highly recommended).
Try this one:
/\[quote[^\/]+\].*\[\/quote\]$/
The $ sign indicates that only the closing quote element at the end of the string should be used to determine the ending of the quote you're trying to remove.
And i added a "." before the asterisk so that this will match any sign in between. I tested this with your two strings and it worked.
edit: I don't exactly know how you are using that. But just as an addition. If you want the pattern also to match to a string where no attributes are added for example:
[quote]Hello[/quote]
You should change the "+" sign into an asterisk as well like this:
/\[quote[^\/]*\].*\[\/quote\]$/
This answer has flaws, see Ridgerunner's answer for a more correct one.
Here's my crack at it.
function filterQuotes(text)
{
return text.replace(/\[(\/)?quote([^\/]*)?\]/g,"");
}