Suppose my text is below:
One of the compelling reasons for buying this phone could be its display. Of course, we would have loved full HD, but then again we shut up because of the punchy price-tag. The screen responsiveness is also good.
I just want to get the index of last occurrence of "the" string in my text.(using RegExp)
var re = new RegExp("\\b"+"the"+"\\b",'g');
var pos = re.lastIndex;
gives only the position of first occurence of the string the..
any suggestions?
Why do you need regex to find last occurence of substring . you can use native .lastIndexOf() method:
re.lastIndexOf("the");
One way;
var pos = -1;
while ((match = re.exec(str)) != null)
pos = match.index;
alert("last match found at " + pos);
The regular expression is /\bthe\b(?!(.|\n)*\bthe\b)/ (no global flag!) ...meaning the word "the" which is not followed by the word "the".
To test:
var re = new RegExp('\\b' + input + '\\b(?!(.|\\n)*\\b' + input + '\\b)');
var pos = re.test(str) ? re.exec(str).index : -1;
Here's a working solution, even with non-global regular expresions:
function regexLastIndexOf(str, regex) {
regex = regex.global
? regex
: new RegExp(regex.source, 'g' + (regex.ignoreCase ? 'i' : '') + (regex.multiLine ? 'm' : ''));
var lastIndexOf = -1;
var nextStop = 0;
var result;
while ((result = regex.exec(str)) != null) {
lastIndexOf = result.index;
regex.lastIndex = ++nextStop;
}
return lastIndexOf;
}
Related
I am trying to replace a specific character ( "," -> this is the specific character in my case) in a string only if it's followed by a uppercase letter, but no matter how I try it doesn't work. Does anyone know a good way of doing this ?
const p = 'The ,quick brown ,Fox jumps over the lazy Dog.';
console.log(p.replace(/,([A-Z])/g, 'xxx$1'));
var str1 = "test,Test";
var newstr ;
var index1 = str1.indexOf(',');
if(str1.charAt(index1+1) == str1.charAt(index1+1).toUpperCase()) {
var newstr = str1.substr(0, index1) + "*" + str1.substr(index1+1 , str1.length);
}else {
newstr = str1;
}
console.log(newstr);
I have some problems with replacing every 6th colon in my array. Have tried something with Regex, but that doesn't seem to work. I have red other questions were people are using nth and then set this variabele to the index you want to replace, but can't figure out why that isn't working. I used the join function to replace the ',' in my array with ':'.
arrayProducts[i] = arrayProducts[i].join(':');
When i use console.log(arrayProducts); this is my result:
F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3
This is what I want:
F200:0:0.0000:1:100:0:1,KPO2:0:0.0000:1:200:0:2,HGB1:0:0.0000:1:300:0:3
Thanks for reading!
Edit: F200, KP02 and HGB1, could also be numbers / digits like: 210, 89, 102 so the :[A-Z] method from regex doesn't work.
You can just count the number of colon occurences and replace every nth of them.
var str = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3', counter = 0;
res = str.replace(/:/g, function(v) {
counter++;
return !(counter % 7) ? ',' : v;
});
console.log(res);
A regex solution is viable. You can use a function as the second parameter of the .replace method to make full use of backreferences.
var str = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3';
str = str.replace(/((?:[^:]*:){6}(?:[^:]*)):/g, function() {
var matches = arguments;
return matches[1] + ',';
});
console.log(str);
What you are looking for is to split over the following expression :[A-Z]
(assuming that your rows always start with this range)
a simple solution could be:
mystring.split(/:[A-Z]/).join(',')
/:[A-Z]/ matches any : followed by a uppercase letter
You could use replace with a look for six parts with colon and replace the seventh.
var string = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3',
result = string.replace(/(([^:]*:){6}[^:]*):/g, '$1,');
console.log(result);
Another solution (based on the number of iteration)
using map method:
str.split(':').map((v, i) => (i % 7 === 0 ? ',' : ':') + v ).join('').slice(1)
using reduce method:
str.split(':').reduce((acc,v, i) => {
return acc + (i % 7 === 0 ? ',' : ':' ) + v ;
}, '').slice(1)
Note: arrow expression does not work on old browsers
maybe you can try this approach,
loop your array and join it manually, something like :
var strarr = "F200:0:00000:1:100:0:1:KPO2:0:00000:1:200:0:2:HGB1:0:00000:1:300:0:3";
var arr = strarr.split(":")
var resStr = "";
for(var i = 0; i < arr.length; i++)
{
if(i > 0 && i%7 == 0)
resStr = resStr + "," + arr[i]
else
resStr = resStr + ( resStr == "" ? "" : ":") + arr[i];
}
console.log(resStr);
Im trying to replace a character at a specific indexOf to uppercase.
My string is a surname plus the first letter in the last name,
looking like this: "lovisa t".
I check the position with this and it gives me the right place in the string. So the second gives me 8(in this case).
first = texten.indexOf(" ");
second = texten.indexOf(" ", first + 1);
And with this I replace the first letter to uppercase.
var name = texten.substring(0, second);
name=name.replace(/^./, name[0].toUpperCase());
But how do I replace the character at "second" to uppercase?
I tested with
name=name.replace(/.$/, name[second].toUpperCase());
But it did´t work, so any input really appreciated, thanks.
Your error is the second letter isn't in position 8, but 7.
Also this second = texten.indexOf(" ", first + 1); gives -1, not 8, because you do not have a two spaces in your string.
If you know that the string is always in the format surname space oneLetter and you want to capitalize the first letter and the last letter you can simply do this:
var name = 'something s';
name = name[0].toUpperCase() + name.substring(1, name.length - 1) + name[name.length -1].toUpperCase();
console.log(name)
Here's a version that does exactly what your question title asks for: It uppercases a specific index in a string.
function upperCaseAt(str, i) {
return str.substr(0, i) + str.charAt(i).toUpperCase() + str.substr(i + 1);
}
var str = 'lovisa t';
var i = str.indexOf(' ');
console.log(upperCaseAt(str, i + 1));
However, if you want to look for specific patterns in the string, you don't need to deal with indices.
var str = 'lovisa t';
console.log(str.replace(/.$/, function (m0) { return m0.toUpperCase(); }));
This version uses a regex to find the last character in a string and a replacement function to uppercase the match.
var str = 'lovisa t';
console.log(str.replace(/ [a-z]/, function (m0) { return m0.toUpperCase(); }));
This version is similar but instead of looking for the last character, it looks for a space followed by a lowercase letter.
var str = 'lovisa t';
console.log(str.replace(/(?:^|\s)\S/g, function (m0) { return m0.toUpperCase(); }));
Finally, here we're looking for (and uppercasing) all non-space characters that are preceded by the beginning of the string or a space character; i.e. we're uppercasing the start of each (space-separated) word.
All can be done by regex replace.
"lovisa t".replace(/(^|\s)\w/g, s=>s.toUpperCase());
Try this one (if it will be helpfull, better move constants to other place, due performance issues(yes, regexp creation is not fast)):
function normalize(str){
var LOW_DASH = /\_/g;
var NORMAL_TEXT_REGEXP = /([a-z])([A-Z])/g;
if(!str)str = '';
if(str.indexOf('_') > -1) {
str = str.replace(LOW_DASH, ' ');
}
if(str.match(NORMAL_TEXT_REGEXP)) {
str = str.replace(NORMAL_TEXT_REGEXP, '$1 $2');
}
if(str.indexOf(' ') > -1) {
var p = str.split(' ');
var out = '';
for (var i = 0; i < p.length; i++) {
if (!p[i])continue;
out += p[i].charAt(0).toUpperCase() + p[i].substring(1) + (i !== p.length - 1 ? ' ' : '');
}
return out;
} else {
return str.charAt(0).toUpperCase() + str.substring(1);
}
}
console.log(normalize('firstLast'));//First Last
console.log(normalize('first last'));//First Last
console.log(normalize('first_last'));//First Last
I need to remove substrings containing the word "page=" followed by number.
Ex.
s= "aaaapage=500";
Should be
s = "page=500"
I tried
s = s.replace(/&page=\d/g,"");
and
s = s.replace(/&page=[\d]+/g,"");
to no avail
You can match text before and after while capturing the page=[digits]:
s= "aaaapage=500";
document.write(s.replace(/.*(page=\d+).*/, '$1') + "<br/>");
// or with multiline input
s= "a\na\naapage=500text\nnewline";
document.write(s.replace(/[\s\S]*(page=\d+)[\s\S]*/, '$1'));
This is good when we only have 1 page=[digits].
When we have more, use exec:
var re = /page=\d+/g;
var str = 'apageaaaapage=500apageaaaapage=210';
var m;
while ((m = re.exec(str)) !== null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
document.write(m[0] + "<br/>");
}
I just realized
s = s.replace(/&page=[\d]+/g,"");
actually works
Just remove all the word chars which exists before page=
s.replace(/\w*(page=\d+)/g,"$1");
How can I get from this string
genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990
this
genre: [Drama, Comedy],
cast: [Leonardo DiCaprio, Cmelo Hotentot],
year: [1986-1990]
with one regular expression?
This could be done using one regex and overload of replace function with replacer as a second argument. But honestly, I have to use one more replace to get rid of pluses (+) - I replaced them by a space () char:
var str = 'genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990';
str = str.replace(/\+/g, ' ');
var result = str.replace(/(\w+:)(\s?)([\w,\s-]+?)(\s?)(?=\w+:|$)/g, function (m, m1, m2, m3, m4, o) {
return m1 + ' [' + m3.split(',').join(', ') + ']' + (o + m.length != str.length ? ',' : '') + '\n';
});
You could find the full example on jsfiddle.
You will not get them into arrays from the start, but it can be parsed if the order stays the same all the time.
var str = "genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990";
str = str.replace(/\+/g," ");
//Get first groupings
var re = /genre:\s?(.+)\scast:\s?(.+)\syear:\s(.+)/
var parts = str.match(re)
//split to get them into an array
var genre = parts[1].split(",");
var cast = parts[2].split(",");
var years = parts[3];
console.log(genre);
You can't do this using only regular expressions cause you're trying to parse a (tiny) grammar.