I'm pretty new to this scripting and all, I want to find the length of a sentence in a cell in google sheet. Example, "I want to check the length" is a sentence in cell A1 in google sheet, I want to find the length of this string.
And one more, if I want to replace text like space(" ") with any thing like plus (+), I have used script as shown below
var newdata = olddata.toString().replace(" ","+");
// here old data refers to the string "I want to check length"
// output that displayed in log is like "I+want to check length"
// Output, that I want i like "I+want+to+check+length"
when I used this, in log I can see only, first space is getting replaced in the whole string/sentence. so how can modify/alter my code in order to replace all the spaces with any desired character.
The length of the string is given by newString.length property.
let oldString = "I want to check the length";
console.log(oldString.length);
The replace() function uses regular expressions to replace the strings. To replaces all matched patters, you will need to use /g global match. To match all spaces, you need to provide /\ /g to the replace() function. See example:
let oldString = "I want to check the length";
let newString = oldString.replace(/\ /g, "+");
console.log(newString);
console.log(newString.length);
Related
I'm trying to delete the numbers after a certain word, until a symbol.
So say I have the following string: jessicaamount=9487&doe. I want the final string to look like this: jessica&doe. (There may not always be anything after the number.)
I'm thinking regex is the solution, but not too sure how I would go about it.
An option without regex would be to use indexOf() and .slice().
function removeNumber (string) {
let part1 = string.slice(0, string.indexOf("amount="));
let part2 = string.slice(string.indexOf("&"));
return part1 + part2;
}
console.log(removeNumber("jessicaamount=9487&doe")) // jessica&doe
It will remove what is between strings in variables part1 and part2.
It includes the first character of part2.
I have a following string:
Text
I want to extract from this string, with the use of JavaScript 'pl' or 'pl_company_com'
There are a few variables:
jan_kowalski is a name and surname it can change, and sometimes even have 3 elements
the country code (in this example 'pl') will change to other en / de / fr (this is that part of the string i want to get)
the rest of the string remains the same for every case (beginning + everything after starting with _company_com ...
Ps. I tried to do it with split, but my knowledge of JS is very basic and I cant get what i want, plase help
An alternative to Randy Casburn's solution using regex
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_(.*_company_com)')[1];
console.log(out);
Or if you want to just get that string with those country codes you specified
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_((en|de|fr|pl)_company_com)')[1];
console.log(out);
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_((en|de|fr|pl)_company_com)')[1];
console.log(out);
A proof of concept that this solution also works for other combinations
let urls = [
new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx'),
new URL('https://my.domain.com/personal/firstname_middlename_lastname_pl_company_com/Documents/Forms/All.aspx')
]
urls.forEach(url => console.log(url.href.match('.*_(en|de|fr|pl).*')[1]))
I have been very successful before with this kind of problems with regular expressions:
var string = 'Text';
var regExp = /([\w]{2})_company_com/;
find = string.match(regExp);
console.log(find); // array with found matches
console.log(find[1]); // first group of regexp = country code
First you got your given string. Second you have a regular expression, which is marked with two slashes at the beginning and at the end. A regular expression is mostly used for string searches (you can even replace complicated text in all major editors with it, which can be VERY useful).
In this case here it matches exactly two word characters [\w]{2} followed directly by _company_com (\w indicates a word character, the [] group all wanted character types, here only word characters, and the {}indicate the number of characters to be found). Now to find the wanted part string.match(regExp) has to be called to get all captured findings. It returns an array with the whole captured string followed by all capture groups within the regExp (which are denoted by ()). So in this case you get the country code with find[1], which is the first and only capture group of the regular expression.
I want to test if a sentence like type var1,var2,var3 is matching a text declaration or not.
So, I used the following code :
var text = "int a1,a2,a3",
reg = /int ((([a-z_A-Z]+[0-9]*),)+)$/g;
if (reg.test(text)) console.log(true);
else console.log(false)
The problem is that this regular expression returns false on text that is supposed to be true.
Could someone help me find a good regular expression matching expressions as in the example above?
You have a couple of mistekes.
As you wrote, the last coma is required at the end of the line.
I suppose you also want to match int abc123 as correct string, so you need to include letter to other characters
Avoid using capturing groups for just testing strings.
const str = 'int a1,a2,a3';
const regex = /int (?:[a-zA-Z_](?:[a-zA-Z0-9_])*(?:\,|$))+/g
console.log(regex.test(str));
You will need to add ? after the comma ,.
This token ? matches between zero and one.
Notice that the last number in your text a3 does not have , afterward.
int ((([a-z_A-Z]+[0-9]*),?)+)$
What is the best way to capture everything except when faced with two or more new lines?
ex:
name1
address1
zipcode
name2
address2
zipcode
name3
address3
zipcode
One regex I considered was /[^\n\n]*\s*/g. But this stops when it is faced with a single \n character.
Another way I considered was /((?:.*(?=\n\n)))\s*/g. But this seems to only capture the last line ignoring the previous lines.
What is the best way to handle similar situation?
UPDATE
You can consider replacing the variable length separator with some known fixed length string not appearing in your processed text and then split. For instance:
> var s = "Hi\n\n\nBye\nCiao";
> var x = s.replace(/\n{2,}/, "#");
> x.split("#");
["Hi", "Bye
Ciao"]
I think it is an elegant solution. You could also use the following somewhat contrived regex
> s.match(/((?!\n{2,})[\s\S])+/g);
["Hi", "
Bye
Ciao"]
and then process the resulting array by applying the trim() string method to its members in order to get rid of any \n at the beginning/end of every string in the array.
((.+)\n?)*(you probably want to make the groups non-capturing, left it as is for readability)
The inner part (.+)\n? means "non-empty line" (at least one non-newline character as . does not match newlines unless the appropriate flag is set, followed by an optional newline)
Then, that is repeated an arbitrary number of times (matching an entire block of non-blank lines).
However, depending on what you are doing, regexp probably is not the answer you are looking for. Are you sure just splitting the string by \n\n won't do what you want?
Do you have to use regex? The solution is simple without it.
var data = 'name1...';
var matches = data.split('\n\n');
To access an individual sub section split it by \n again.
//the first section's name
var name = matches[0].split('\n')[0];
I am trying to extract the first character after the last underscore in a string with an unknown number of '_' in the string but in my case there will always be one, because I added it in another step of the process.
What I tried is this. I also tried the regex by itself to extract from the name, but my result was empty.
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var string = match(/[^_]*$/)[1]
string.charAt(0)
So the final desired result is 'D'. If the RegEx can only get me what is behind the last '_' that is fine because I know I can use the charAt like currently shown. However, if the regex can do the whole thing, even better.
If you know there will always be at least one underscore you can do this:
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var firstCharAfterUnderscore = s.charAt(s.lastIndexOf("_") + 1);
// OR, with regex
var firstCharAfterUnderscore = s.match(/_([^_])[^_]*$/)[1]
With the regex, you can extract just the one letter by using parentheses to capture that part of the match. But I think the .lastIndexOf() version is easier to read.
Either way if there's a possibility of no underscores in the input you'd need to add some additional logic.