Styjun

I have a JSON object that contains an array of objects. I also have an array of desired values, and I want to search for these values inside my JSON. I only care about the first match. If no match is found, throw an error.

There might be a better way to do this but this is what I came up with:

function myFunction() 
 $.getJSON('database.json')
 .done(db => 
 for (let i = 0; i < desiredValues.length; i++) 
 db.arrayOfObjects.forEach(object => 
 if (object.propertyValue === desiredValues[i]) 
 console.log("Match found!");
 return; // break out of myFunction()
 
 );
 
 throw Error("Match not found.");
 )
 .fail(error => 
 throw Error("getJSON request failed.n" + error);
 )

My problem is that the return statement only breaks out of the current iteration of forEach (why?). The remaining objects are still tested for all remaining values of desiredValues, and the error is always thrown. How can I exit myFunction() entirely when a match is found, or how can I restructure this function to achieve what I want?

Edit: I probably should have mentioned I also need to do stuff with the object that was matched, so not just return true if there is a match.

Simply use .find on array.

function myFunction() {
 $.getJSON('database.json')
 .done(db => 
 for (let i = 0; i < desiredValues.length; i++) 
 if (
 db.arrayOfObjects
 .find(object => object.propertyValue === desiredValues[i])
 !== undefined
 ) 
 console.log("Found match! Matched value:", desiredValues[i]);
 return;
 
 throw Error("Match not found.");
 )
 .fail(error => 
 throw Error("getJSON request failed.n" + error);
 )

or map array of objects and mix .find with .includes:

function myFunction() 
 $.getJSON('database.json')
 .done(db => 
 const values = db.arrayOfObjects.map(object => object.propertyValue);

 const match = values.find(value => desiredValues.includes(value)); 

 if (match !== undefined) 
 console.log("Found match! Matched value:", match);
 return;
 

 throw Error("Match not found.");
 )
 .fail(error => 
 throw Error("getJSON request failed.n" + error);
 )

or .filter it to get all matched values:

function myFunction() 
 $.getJSON('database.json')
 .done(db => 
 const values = db.arrayOfObjects.map(object => object.propertyValue);

 const matches = values.filter(value => desiredValues.includes(value)); 

 if (matches.length) 
 console.log("Found", matches.length, "matches! Matched values:", matches);
 return;
 

 throw Error("Match not found.");
 )
 .fail(error => 
 throw Error("getJSON request failed.n" + error);
 )

The problem with return inside forEach is that it terminates the current forEach callback only - that's what return does, it stops the current execution of a function and (possibly) returns a value, without affecting possible future calls of that function.

Use two nested Array.prototype.some instead, which will break out of both immediately and return true once a match is found, and iterate completely through and return false otherwise:

const isFound = desiredValues.some(valueToFind => (
 db.arrayOfObjcts.some(( propertyValue ) => (
 object.propertyValue === valueToFind
 ))
));
if (!isFound) 
 throw new Error('Match not found');

Or, if you're not comfortable with destructuring arguments:

const isFound = desiredValues.some(valueToFind => (
 db.arrayOfObjcts.some(object => (
 propertyValue === valueToFind
 ))
));
if (!isFound) 
 throw new Error('Match not found');

To identify which value was found, use .find instead for the outer loop:

const valueFound = desiredValues.find(valueToFind => (
 db.arrayOfObjcts.some(( propertyValue ) => (
 object.propertyValue === valueToFind
 ))
));
if (!valueFound) 
 throw new Error('Match not found');