JavaScript: output of a function which returns a string outputs "undefined". Why that?

Question

console.log in the function "getNameByDate" will produce "Ben" and "Tom". But the other console.log outside of the function will produce "undefined". Why not the same output?

  let arr = [
    {
      databaseObject: { geburt: "24.01.2012", name: "Ben"}
    }, 
    {
      databaseObject: { geburt: "29.02.2012", name: "Tom"}
    }, 
  ];

  function getNameByDate(date) {
    jQuery.each( arr, function( i, val ) {         
      if ( val.databaseObject.geburt == date ) {
        console.log (val.databaseObject.name);        
        return val.databaseObject.name;
      }
    });
  }

  let dates = ["24.01.2012", "29.02.2012"];

  jQuery.each( dates, function( i, val ) {
    console.log(getNameByDate(val));
  });

You are forgetting that function (i, val) ... is also a function and the return inside it belongs to function (i, val) ... not getNameByDate(). The function getNameByDate() returns nothing which in javascript is the same as undefined — slebetman
– slebetman, Commented Jun 12, 2019 at 19:13

g4g4 · Accepted Answer · 2019-06-12 01:33:51Z

1

Returning the result in the jQuery.each-loop was not correct. Instead we have to assign the result in that loop to a variable, break the loop by returning false and return the variable after the loop:

 let arr = [
    {
      databaseObject: { geburt: "24.01.2012", name: "Ben"}
    }, 
    {
      databaseObject: { geburt: "29.02.2012", name: "Tom"}
    }, 
  ];

  function getNameByDate(date) {
    var friendsName;
    jQuery.each( arr, function( i, val ) {         
      if ( val.databaseObject.geburt == date ) {
        console.log (val.databaseObject.name);
        friendsName = val.databaseObject.name; 
        return false; // exit loop
      }
    });
    return friendsName;
  }

  let dates = ["24.01.2012", "29.02.2012"];

  jQuery.each( dates, function( i, val ) {
    console.log(getNameByDate(val));
  });

answered Jun 12, 2019 at 1:33

g4g4

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1 Comment

slebetman Over a year ago

Alternatively you can also replace jQuery.each() with a for loop

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