I need help with a Regex in JavaScript (for a Photoshop script) to match bold tags around words in a string. (not worried about italic or bolditalic at this time).
I don't want to split the string at this stage, I just want to chop it up into certain alternating chunks into using match.
// Be <b>bold!</b> Be fabulous!
Should get match to // ("Be ", "bold!", "Be fabulous!") // line commented for obvious reasons
After that, I'll remove the bold tags - unless Regex can do that in one pass - don't underestimate it's power!
This is what I have so far
(.*?)([<b>]+[\S]+[<\/b>]+[\s]+)+(.*)/g
Only it doesn't match everything as seen here
Just for the record, before anyone suggests a much easier JS solution:
In the Photoshop DOM you can't script regular text mixed with bold. You probably can with Action Manager code, but with generating text that could be a big headache.
To get around this (not an ideal solution) I'll be using regular text & splitting it up at the appropriate places & swapping to bold.
[<b>] is character class, use simply <b> instead.
/(.*?)(<b>+\S+<\/b>+\s+)+(.*)/g
and change \S to [^<]
/(.*?)(<b>+[^<]+<\/b>+\s+)+(.*)/g
You can try:
<b>(.*?)<\/b>
Here is online demo
sample code:
var re = /<b>(.*?)<\/b>/gi;
var str = 'Be <b>bold!</b> Be fabulous! ';
var subst = '$1';
var result = str.replace(re, subst);
output:
Be bold! Be fabulous!
Better try with String.split() function:
var re = /\s*<\/?b>\s*/gi;
var str = 'Be <b>bold!</b> Be fabulous!';
console.log(str.split(re));
output:
["Be", "bold!", "Be fabulous!"]
Related
My goal is to take a markdown text and create the necessary bold/italic/underline html tags.
Looked around for answers, got some inspiration but I'm still stuck.
I have the following typescript code, the regex matches the expression including the double asterisk:
var text = 'My **bold\n\n** text.\n'
var bold = /(?=\*\*)((.|\n)*)(?<=\*\*)/gm
var html = text.replace(bold, '<strong>$1</strong>');
console.log(html)
Now the result of this is : My <\strong>** bold\n\n **<\strong> text.
Everything is great aside from the leftover double asterisk.
I also tried to remove them in a later 'replace' statement, but this creates further issues.
How can I ensure they are removed properly?
With your pattern (?=\*\*)((.|\n)*)(?<=\*\*) you assert (not match) with (?=\*\*) that there is ** directly to the right.
Then directly after that, you capture the ** using ((.|\n)*) so then it becomes part of the match.
Then at the end you assert again with (?<=\*\*) that there is ** directly to the left, but ((.|\n)*) has already matched it.
This way so you will end up with all the ** in the match.
You don't need lookarounds at all, as you are already using a capture group.
In Javascript you could match the ** on the left and right and capture any character in a capture group:
\*\*([^]*?)\*\*
Regex demo
But I would suggest using a dedicated parser to parse markdown instead of using a regex.
Just make another call to replaceAll removing the ** with and empty string.
var text = 'My **bold\n\n** text.\n'
var bold = /(?=\*\*)((.|\n)*)(?<=\*\*)/gm
var html = text.replace(bold, '<strong>$1</strong>');
html = html.replaceAll(/\*\*/gm,'');
console.log(html)
I thought it was very simple to find out. But how many ways I tried still not work properly.
Below is the test snippet.
"100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)".replace(/[\.,\d]*/g, '{n}')
And I want the result like below.
{n}$ and {n}EUR {n}USD {n}$ ({n})
The * is your problem, change the regex to /[.,\d]+/g instead.
"100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)".replace(/[.,\d]+/g, '{n}');
Output
{n}$ and {n}EUR {n}USD {n}$ ({n})
JSFiddle Example Check console screen for the output.
The problem here is that [\.,\d]* can match an empty string. The first step would be to use [.,\d]+ so that at least one of these characters matches.
But a better regex would be \d[.,\d]* because it ensures the replaced characters begin with a digit, so it won't replace periods in sentences.
If you want to go further, you can also use (?=[.,\d]*\d)[.,\d]+ if to handle numbers starting with periods. This one would be the proper answer for your case. The lookahead ensures there's at least one digit anywhere in the replaced text.
Note that you don't need to escape the . inside a character class.
\.?\d[^\s]*\d
Try this.Replace with {n}.See demo.
http://regex101.com/r/kP8uF5/3
var re = /\.?\d[^\s]*\d/gm;
var str = '100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)';
var subst = '{n}';
var result = str.replace(re, subst);
I have made a javascript function to replace some words with other words in a text area, but it doesn't work. I have made this:
function wordCheck() {
var text = document.getElementById("eC").value;
var newText = text.replace(/hello/g, '<b>hello</b>');
document.getElementById("eC").innerText = newText;
}
When I alert the variable newText, the console says that the variable doesn't exist.
Can anyone help me?
Edit:
Now it replace the words, but it replaces it with <b>hello</b>, but I want to have it bold. Is there a solution?
Update:
In response to your edit, about your wanting to see the word "hello" show up in bold. The short answer to that is: it can't be done. Not in a simple textarea, at least. You're probably looking for something more like an online WYSIWYG editor, or at least a RTE (Richt Text Editor). There are a couple of them out there, like tinyMCE, for example, which is a decent WYSIWYG editor. A list of RTE's and HTML editors can be found here.
First off: As others have already pointed out: a textarea element's contents is available through its value property, not the innerText. You get the contents alright, but you're trying to update it through the wrong property: use value in both cases.
If you want to replace all occurrences of a string/word/substring, you'll have to resort to using a regular expression, using the g modifier. I'd also recommend making the matching case-insensitive, to replace "hello", "Hello" and "HELLO" all the same:
var txtArea = document.querySelector('#eC');
txtArea.value = txtArea.value.replace(/(hello)/gi, '<b>$1</b>');
As you can see: I captured the match, and used it in the replacement string, to preserve the caps the user might have used.
But wait, there's more:
What if, for some reason, the input already contains <b>Hello</b>, or contains a word containing the string "hello" like "The company is called hellonearth?" Enter conditional matches (aka lookaround assertions) and word boundaries:
txtArea.value = txtArea.value.replace(x.value.replace(/(?!>)\b(hello)\b(?!<)/gi, '<b>$1</b>');
fiddle
How it works:
(?!>): Only match the rest if it isn't preceded by a > char (be more specific, if you want to and use (?!<b>). This is called a negative look-ahead
\b: a word boundary, to make sure we're not matching part of a word
(hello): match and capture the string literal, provided (as explained above) it is not preceded by a > and there is a word boundary
(?!<): same as above, only now we don't want to find a matching </b>, so you can replace this with the more specific (?!<\/b>)
/gi: modifiers, or flags, that affect the entire pattern: g for global (meaning this pattern will be applied to the entire string, not just a single match). The i tells the regex engine the pattern is case-insensitive, ie: h matches both the upper and lowercase character.
The replacement string <b>$1</b>: when the replacement string contains $n substrings, where n is a number, they are treated as backreferences. A regex can group matches into various parts, each group has a number, starting with 1, depending on how many groups you have. We're only grouping one part of the pattern, but suppose we wrote:
'foobar hello foobar'.replace(/(hel)(lo)/g, '<b>$1-$2</b>');
The output would be "foobar <b>hel-lo</b> foobar", because we've split the match up into 2 parts, and added a dash in the replacement string.
I think I'll leave the introduction to RegExp at that... even though we've only scratched the surface, I think it's quite clear now just how powerful regex's can be. Put some time and effort into learning more about this fantastic tool, it is well worth it.
If <textarea>, then you need to use .value property.
document.getElementById("eC").value = newText;
And, as mentioned Barmar, replace() replaces only first word. To replace all word, you need to use simple regex. Note that I removed quotes. /g means global replace.
var newText = text.replace(/hello/g, '<b>hello</b>');
But if you want to really bold your text, you need to use content editable div, not text area:
<div id="eC" contenteditable></div>
So then you need to access innerHTML:
function wordCheck() {
var text = document.getElementById("eC").innerHTML;
var newText = text.replace(/hello/g, '<b>hello</b>');
newText = newText.replace(/<b><b>/g,"<b>");//These two lines are there to prevent <b><b>hello</b></b>
newText = newText.replace(/<\/b><\/b>/g,"</b>");
document.getElementById("eC").innerHTML = newText;
}
I am pulling content from an RSS feed, before using jquery to format and edit the rss feed (string) that is returned. I am using replace to replace strings and characters like so:
var spanish = $("#wod a").text();
var newspan = spanish.replace("=","-");
$("#wod a").text(newspan);
This works great. I am also trying to remove all text after a certain point. Similar to truncation, I would like to hide all text starting from the word "Example".
In this particular RSS feed, the word example is in every feed. I would like to hide "example" and all text the follows that word. How can I accomplish this?
Though there is not enough jQuery, you even don't need it to remove everything after a certain word in the given string. The first approach is to use substring:
var new_str = str.substring(0, str.indexOf("Example"));
The second is a trick with split:
var new_str = str.split("Example")[0];
If you also want to keep "Example" and just remove everything after that particular word, you can do:
var str = "aaaa1111?bbb&222:Example=123456",
newStr = str.substring(0, str.indexOf('Example') + 'Example'.length);
// will output: aaaa1111?bbb&222:Example
jQuery isn't intended for string manipulation, you should use Vanilla JS for that:
newspan = newspan.replace(/example.*$/i, "");
The .replace() method accepts a regular expression, so in this case I've used /example.*$/i which does a case-insensitive match against the word "example" followed by zero or more of any other characters to the end of the string and replaces them with an empty string.
I would like to hide all text starting from the word "Example"
A solution that uses the simpler replace WITH backreferences so as to "hide" everything starting with the word Example but keeping the stuff before it.
var str = "my house example is bad"
str.replace(/(.*?) example.*/i, "$1") // returns "my house"
// case insensitive. also note the space before example because you
// probably want to throw that out.
I've a string done like this: "http://something.org/dom/My_happy_dog_%28is%29cool!"
How can I remove all the initial domain, the multiple underscore and the percentage stuff?
For now I'm just doing some multiple replace, like
str = str.replace("http://something.org/dom/","");
str = str.replace("_%28"," ");
and go on, but it's really ugly.. any help?
Thanks!
EDIT:
the exact input would be "My happy dog is cool!" so I would like to get rid of the initial address and remove the underscores and percentage and put the spaces in the right place!
The problem is that trying to put a regex on Chrome "something goes wrong". Is it a problem of Chrome or my regex?
I'd suggest:
var str = "http://something.org/dom/My_happy_dog_%28is%29cool!";
str.substring(str.lastIndexOf('/')+1).replace(/(_)|(%\d{2,})/g,' ');
JS Fiddle demo.
The reason I took this approach is that RegEx is fairly expensive, and is often tricky to fine tune to the point where edge-cases become less troublesome; so I opted to use simple string manipulation to reduce the RegEx work.
Effectively the above creates a substring of the given str variable, from the index point of the lastIndexOf('/') (which does exactly what you'd expect) and adding 1 to that so the substring is from the point after the / not before it.
The regex: (_) matches the underscores, the | just serves as an or operator and the (%\d{2,}) serves to match digit characters that occur twice in succession and follow a % sign.
The parentheses surrounding each part of the regex around the |, serve to identify matching groups, which are used to identify what parts should be replaced by the ' ' (single-space) string in the second of the arguments passed to replace().
References:
lastIndexOf().
replace().
substring().
You can use unescape to decode the percentages:
str = unescape("http://something.org/dom/My_happy_dog_%28is%29cool!")
str = str.replace("http://something.org/dom/","");
Maybe you could use a regular expression to pull out what you need, rather than getting rid of what you don't want. What is it you are trying to keep?
You can also chain them together as in:
str.replace("http://something.org/dom/", "").replace("something else", "");
You haven't defined the problem very exactly. To get rid of all stretches of characters ending in %<digit><digit> you'd say
var re = /.*%\d\d/g;
var str = str.replace(re, "");
ok, if you want to replace all that stuff I think that you would need something like this:
/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g
test
var string = "http://something.org/dom/My_happy_dog_%28is%29cool!";
string = string.replace(/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g,"");