I have a javascript string that contains \\n. When this is displayed out to a webpage, it shows literally as \n (as expected). I used text.replace(/\\n/g, '\n') to get it to act as a newline (which is the desired format).
I'm trying to determine the best way to catch all such instances (including similar instances like tabs \\t -> \t).
Is there a way to use regex (can't determine how to copy the matched wildcard letter to use in the replacement string) or anything else?
As mentioned by dandavis in the comments in original post, JSON.parse() ended up working for me.
i.e. text = JSON.parse(text);
Second answer, the first was wrong.
JavaScript works on a special way in this case. Read this for more details.
In your case it should be one of this ...
var JSCodeNewLine = "\u000A";
text.replace(/\\n/g, JSCodeNewLine);
var JSCodeCarriageReturnNewLine = "\u000D\u000A";
text.replace(/\\n/g, JSCodeCarriageReturnNewLine);
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 ran into the below monster of a regex in the wild today. The regex is meant to validate a url.
function superUrlValidation(url) {
return new RegExp(/^/.source + "((.+):\/\/)?" + /(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/.source, "i")
.test(url);
}
I've never seen .source used in a regex like this so I looked it up.
The MDN docs for RegExp.prototype.source states:
The source property returns a String containing the source text of the regexp object, and it doesn't contain the two forward slashes on both sides and any flags.
... and gives this example:
var regex = /fooBar/ig;
console.log(regex.source); // "fooBar", doesn't contain /.../ and "ig".
I understand the MDN example (you're getting the source text of the regex object after it is created, makes sense), but I dont understand how this is being used in the superUrlValidation regex above.
How is the source being used before the regex object is completed and what does this accomplish? I cant find any documentation showing .source being used in this way.
Note that .source is used twice in the regex, at the beginning and the end
Use of .source everywhere in your regex seems totally unnecessary, may be just a trick to avoid double escaping. In fact even use of new RegExp is not needed and you can get away with just the regex literal as this:
var re = /^((.+):\/\/)?(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i;
/^/ is a regex literal, meaning it's a valid regex object in it's own right. This means that /^/.source === "^".
This seems like an arbitrary example of using the source property as this means the author could have just placed a "^" in it's place, or even just put a ^ at the beginning of the next string, and it would have the same effect.
The .source property returns the content of the regex between the forward slashes as you say. so the result of the above is equivalent to this string:
/^((.+):\/\/)?(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i
In JavaScript you can write regexes like this: /matchsomething/ or using the RegExp function/constructor above. It looks like the code you found is the result of someone not know what they were doing. They seem to have taken a few regexes using the literal syntax (i.e /match_here/) and plugged it into the constructor version and stuck them all together.
I can't see any benefit in using the source property this way. I would just use the string version or the constructor version. Or better, find out what the original author intended and write it again or find a respected regex library with the criteria you need.
And, yeah, wow. It's massive.
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.
Thanks for looking!
BACKGROUND
I am writing some front-end code that consumes a JSON service which is returning malformed JSON. Specifically, the keys are not surrounded with quotes:
{foo: "bar"}
I have NO CONTROL over the service, so I am correcting this like so:
var scrubbedJson = dirtyJson.replace(/(['"])?([a-zA-Z0-9_]+)(['"])?:/g, '"$2": ');
This gives me well formed JSON:
{"foo": "bar"}
Problem
However, when I call JSON.parse(scrubbedJson), I still get an error. I suspect it may be because the entire JSON string is surrounded in double quotes but I am not sure.
UPDATE
This has been solved--the above code works fine. I had a rogue single quote in the body of the JSON that was returned. I got that out of there and everything now parses. Thanks.
Any help would be appreciated.
You can avoid using a regexp altogether and still output a JavaScript object from a malformed JSON string (keys without quotes, single quotes, etc), using this simple trick:
var jsonify = (function(div){
return function(json){
div.setAttribute('onclick', 'this.__json__ = ' + json);
div.click();
return div.__json__;
}
})(document.createElement('div'));
// Let's say you had a string like '{ one: 1 }' (malformed, a key without quotes)
// jsonify('{ one: 1 }') will output a good ol' JS object ;)
Here's a demo: http://codepen.io/csuwldcat/pen/dfzsu (open your console)
something like this may help to repair the json ..
$str='{foo:"bar"}';
echo preg_replace('/({)([a-zA-Z0-9]+)(:)/','$1"$2"${3}',$str);
Output:
{"foo":"bar"}
EDIT:
var str='{foo:"bar"}';
str.replace(/({)([a-zA-Z0-9]+)(:)/,'$1"$2"$3')
There is a project that takes care of all kinds of invalid cases in JSON https://github.com/freethenation/durable-json-lint
I was trying to solve the same problem using a regEx in Javascript. I have an app written for Node.js to parse incoming JSON, but wanted a "relaxed" version of the parser (see following comments), since it is inconvenient to put quotes around every key (name). Here is my solution:
var objKeysRegex = /({|,)(?:\s*)(?:')?([A-Za-z_$\.][A-Za-z0-9_ \-\.$]*)(?:')?(?:\s*):/g;// look for object names
var newQuotedKeysString = originalString.replace(objKeysRegex, "$1\"$2\":");// all object names should be double quoted
var newObject = JSON.parse(newQuotedKeysString);
Here's a breakdown of the regEx:
({|,) looks for the beginning of the object, a { for flat objects or , for embedded objects.
(?:\s*) finds but does not remember white space
(?:')? finds but does not remember a single quote (to be replaced by a double quote later). There will be either zero or one of these.
([A-Za-z_$\.][A-Za-z0-9_ \-\.$]*) is the name (or key). Starts with any letter, underscore, $, or dot, followed by zero or more alpha-numeric characters or underscores or dashes or dots or $.
the last character : is what delimits the name of the object from the value.
Now we can use replace() with some dressing to get our newly quoted keys:
originalString.replace(objKeysRegex, "$1\"$2\":")
where the $1 is either { or , depending on whether the object was embedded in another object. \" adds a double quote. $2 is the name. \" another double quote. and finally : finishes it off.
Test it out with
{keyOne: "value1", $keyTwo: "value 2", key-3:{key4:18.34}}
output:
{"keyOne": "value1","$keyTwo": "value 2","key-3":{"key4":18.34}}
Some comments:
I have not tested this method for speed, but from what I gather by reading some of these entries is that using a regex is faster than eval()
For my application, I'm limiting the characters that names are allowed to have with ([A-Za-z_$\.][A-Za-z0-9_ \-\.$]*) for my 'relaxed' version JSON parser. If you wanted to allow more characters in names (you can do that and still be valid), you could instead use ([^'":]+) to mean anything other than double or single quotes or a colon. You can have all sorts of stuff in here with this expression, so be careful.
One shortcoming is that this method actually changes the original incoming data (but I think that's what you wanted?). You could program around that to mitigate this issue - depends on your needs and resources available.
Hope this helps.
-John L.
How about?
function fixJson(json) {
var tempString, tempJson, output;
tempString = JSON.stringify(json);
tempJson = JSON.parse(tempString);
output = JSON.stringify(tempJson);
return output;
}
I'm using Fancy Upload 3 and onSelect of a file I need to run a check to make sure the user doesn't have any bad characters in the filename. I'm currently getting people uploading files with hieroglyphics and such in the names.
What I need is to check if the filename only contains:
A-Z
a-z
0-9
_ (underscore)
- (minus)
SPACE
ÀÈÌÒÙàèìòùÁÉÍÓÚÝáéíóúýÂÊÎÔÛâêîôûÃÑÕãñõÄËÏÖÜäëïöü (as single and double byte)
Obviously you can see the difficult thing there. The non-english single and double byte chars.
I've seen this:
[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF]
And this:
[\x80-\xA5]
But neither of them fully cover the situation right.
Examples that should work:
fást.zip
abc.zip
ABC.zip
Über.zip
Examples that should NOT work:
∑∑ø∆.zip
¡wow!.zip
•§ªº¶.zip
The following is close, but I'm NO RegEx'pert, not even close.
var filenameReg = /^[A-Za-z0-9-_]|[\x00A0-\xD7FF\xF900-\xFDCF\xFDF0-\xFFEF]+$/;
Thanks in advance.
Solution from Zafer mostly works, but it does not catch all of the other symbols, see below.
Uncaught:
¡£¢§¶ª«ø¨¥®´åß©¬æ÷µç
Caught:
™∞•–≠'"πˆ†∑œ∂ƒ˙∆˚…≥≤˜∫√≈Ω
Regex:
var filenameReg = /^([A-Za-z0-9\-_. ]|[\x00A0-\xD7FF\xF900-\xFDCF\xFDF0-\xFFEF])+$/;
Alternation between two character classes (ie. [abc]|[def]) can be simplified to a single character class ([abcdef]) -- the first can be read as "(a or b or c) OR (d or e or f)"; the second as "(a or b or c or d or e or f)". What probably tripped up your regular expression is the unescaped dash in the first class -- if you want a literal dash, it should be the last character in the class.
So we'll modify your expression to get it working:
var filenameReg = /^[A-Za-z0-9_\x00A0-\xD7FF\xF900-\xFDCF\xFDF0-\xFFEF-]+$/;
The problem now is that you're not accounting for the file extension, but that is an easy modification (assuming you're always getting .zip files):
var filenameReg = /^[A-Za-z0-9_\x00A0-\xD7FF\xF900-\xFDCF\xFDF0-\xFFEF-]+\.zip$/;
Replace zip with another pattern if the extension differs.
It looks like it is the character ranges that are causing the problem, because they include some unallowable characters in between. Since you already have the list of allowable characters, the best thing would be to just use that directly:
var filenameReg = /^[A-Za-z0-9_\-\ ÀÈÌÒÙàèìòùÁÉÍÓÚÝáéíóúýÂÊÎÔÛâêîôûÃÑÕãñõÄËÏÖÜäëïöü]+$/;
The following should work:
var filenameReg = /^([A-Za-z0-9\-_. ]|[\x00A0-\xD7FF\xF900-\xFDCF\xFDF0-\xFFEF])+$/;
I've put \ next to - and grouped two expressions otherwise + sign doesn't affect the first expression.
EDIT 1 :I've also put . in the expression.
We have diffrent rules for diffrent platforms. But I think you mean long file names in windows. For that you can use following RegEx:
var longFilenames = #"^[^\./:*\?\""<>\|]{1}[^\/:*\?\""<>\|]{0,254}$";
NOTE: Instead of saying which Character is allowed, you need to say which ones are not allowed!
But keep in mind that this is not 100% complete RegEx. If you really want to make it complete you have to add exceptions for reserved names as well.
You can find more information about filename rules here:
http://msdn.microsoft.com/en-us/library/aa365247%28VS.85%29.aspx