Get function names from specific string

Question

I have string with slash separated contains function names.

e.g.
my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()

Within above string I want only function name i.e. getCustomer(), getControllerName() & getCsvFileName()

I searched some regex like:

let res = myString.match(/(?<=(function\s))(\w+)/g);

but its returning result as null.

Update: Now I want to get function names without parentheses () i.e. getCustomer, getControllerName & getCsvFileName

Please help me in this

human bean · Accepted Answer · 2022-11-19 19:16:21Z

1

const str = "my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()"

let tokens = [];

for (element of str.split("/"))
  if (element.endsWith("()"))
    tokens.push(element.split("=")[1] ?? element.split("=")[0])

console.log(tokens);

General idea: split the string along slashes, and for each of these tokens, if the token ends with () (as per Nick's suggestion), split the token along =. Append the second index of the token split along = if it exists, otherwise append the first.

A "smaller" version (using purely array methods) could be:

const str = "my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()"

let tokens = str.split("/")
                .filter(element => element.endsWith("()"))
                .map(element => element.split("=")[1] ?? element.split("=")[0]);

console.log(tokens);

edited Nov 19, 2022 at 19:16

answered Nov 19, 2022 at 13:39

human bean

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3 Comments

Nick Parsons Over a year ago

.endsWith() might make more sense than taking a substring.

Garry S Over a year ago

@human-bean your code is working perfectly fine. Can I get function names without brackets ? e.g. getCustomer, getControllerName, getCsvFileName

human bean Over a year ago

Yes. Simply change the expressions element.split("=")[1] and element.split("=")[0] to element.split("=")[1].slice(0, -2) and element.split("=")[0].slice(0, -2), respectively, to remove the trailing brackets for each element added to the array.

Maniraj Murugan · Accepted Answer · 2022-11-19 13:27:03Z

1

You can split the string that has parentheses () first like /.*?\([^)]*\)/g.

This will give array of results, and after that you can iterate the array data and for each item, you can split the = and / before function name with the help of item.split(/=|\//).

Then push the filtered function name into empty array functionNames.

Working Example:

const string = `my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()`;

const functionNames = [];

string.match(/.*?\([^)]*\)/g).forEach(item => {
  const splitString = item.split(/=|\//);
  const functionName = splitString[splitString.length - 1];
  functionNames.push(functionName);
});

console.log(functionNames);

answered Nov 19, 2022 at 13:27

Maniraj Murugan

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Comments

Soorya J · Accepted Answer · 2022-11-19 13:42:08Z

As per, MDN docs the match() method returns null if it does not find a match for the provided regex in the provided search string.

The regular expression which you have provided,/(?<=(function\s))(\w+)/g matches any word that has 'function ' before it. (NOTE: a space after the word function)

Your search string my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName() does not include 'function ' before any characters. That is why you got null as result of match() method.

let yourString = 'my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()';

let myReferenceString = 'SAMPLETEXTfunction sayHi()/function sayHello()';


let res = yourString.match(/(?<=(function\s))(\w+)/g);
let res2 = myReferenceString.match(/(?<=(function\s))(\w+)/g);

console.log("Result of your string", res);
console.log("Result of my string", res2);

My solution here,

let myreferenceString = 'my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()'

let res = myreferenceString.match(/((?<==)(\w+\(\)))|((?<=\/)(\w+\(\)))/g);

console.log("Result", res);

NOTE: I have used the 'Positive Look Behind regex operator', This is not supported in browsers like Safari and IE. Please do reasearch about this before considering this approach.

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