I am new to JavaScript and jQuery.
I have a variable named as str
in JavaScript and it contains very long text, saying something like
"A quick brown fox jumps over a lazy dog".
I want to wrap it and assign it to the same variable str
by inserting the proper \n
or br/
tags at the correct places.
I don't want to use CSS etc. Could you please tell me how to do it with a proper function in JavaScript which takes the str
and returns the proper formatted text to it?
Something like:
str = somefunction(str, maxchar);
I tried a lot but unfortunately nothing turned up the way I wanted it to be! :(
Any help will be much appreciated...
To wrap text in a canvas element with JavaScript, we have to do the calculation for wrapping the text ourselves. to create the canvas. Then we write: const wrapText = (ctx, text, x, y, maxWidth, lineHeight) => { const words = text.
Select the image you want to wrap text around. The Format tab will appear on the right side of the Ribbon. On the Format tab, click the Wrap Text command in the Arrange group. Then select the desired text wrapping option.
To break on words, a qualifier is needed after the greedy quantifier {1,32} to prevent it from choosing sequences ending in the middle of a word. A word-break char \b can cause spaces at the start of new lines, so a white-space char \s must be used instead.
Although this question is quite old, many solutions provided so far are more complicated and expensive than necessary, as user2257198 pointed out - This is completely solvable with a short one-line regular expression.
However I found some issues with his solution including: wrapping after the max width rather than before, breaking on chars not explicitly included in the character class and not considering existing newline chars causing the start of paragraphs to be chopped mid-line.
Which led me to write my own solution:
// Static Width (Plain Regex) const wrap = (s) => s.replace( /(?![^\n]{1,32}$)([^\n]{1,32})\s/g, '$1\n' ); // Dynamic Width (Build Regex) const wrap = (s, w) => s.replace( new RegExp(`(?![^\\n]{1,${w}}$)([^\\n]{1,${w}})\\s`, 'g'), '$1\n' );
The main concept is simply to find contiguous sequences of chars that do not contain new-lines [^\n]
, up to the desired length, e.g 32 {1,32}
. By using negation ^
in the character class it is far more permissive, avoiding missing things like punctuation that would otherwise have to be explicitly added:
str.replace(/([^\n]{1,32})/g, '[$1]\n'); // Matches wrapped in [] to help visualise "[Lorem ipsum dolor sit amet, cons] [ectetur adipiscing elit, sed do ] [eiusmod tempor incididunt ut lab] [ore et dolore magna aliqua.] "
So far this only slices the string at exactly 32 chars. It works because it's own newline insertions mark the start of each sequence after the first.
To break on words, a qualifier is needed after the greedy quantifier {1,32}
to prevent it from choosing sequences ending in the middle of a word. A word-break char \b
can cause spaces at the start of new lines, so a white-space char \s
must be used instead. It must also be placed outside the group so it's eaten, to prevent increasing the max width by 1 char:
str.replace(/([^\n]{1,32})\s/g, '[$1]\n'); // Matches wrapped in [] to help visualise "[Lorem ipsum dolor sit amet,] [consectetur adipiscing elit, sed] [do eiusmod tempor incididunt ut] [labore et dolore magna] aliqua."
Now it breaks on words before the limit, but the last word and period was not matched in the last sequence because there is no terminating space.
An "or end-of-string" option (\s|$)
could be added to the white-space to extend the match, but it would be even better to prevent matching the last line at all because it causes an unnecessary new-line to be inserted at the end. To achieve this a negative look-ahead of exactly the same sequence can be added before, but using an end-of-string char instead of a white-space char:
str.replace(/(?![^\n]{1,32}$)([^\n]{1,32})\s/g, '[$1]\n'); // Matches wrapped in [] to help visualise "[Lorem ipsum dolor sit amet,] [consectetur adipiscing elit, sed] [do eiusmod tempor incididunt ut] labore et dolore magna aliqua."
This should insert a line break at the nearest whitespace of maxChar:
str = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It w as popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum."; str = wordWrap(str, 40); function wordWrap(str, maxWidth) { var newLineStr = "\n"; done = false; res = ''; while (str.length > maxWidth) { found = false; // Inserts new line at first whitespace of the line for (i = maxWidth - 1; i >= 0; i--) { if (testWhite(str.charAt(i))) { res = res + [str.slice(0, i), newLineStr].join(''); str = str.slice(i + 1); found = true; break; } } // Inserts new line at maxWidth position, the word is too long to wrap if (!found) { res += [str.slice(0, maxWidth), newLineStr].join(''); str = str.slice(maxWidth); } } return res + str; } function testWhite(x) { var white = new RegExp(/^\s$/); return white.test(x.charAt(0)); };
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