Unexpected token error in for loop, why?

Question

var ccode = ['EUR', 'BRL', 'RUB', 'KRW', 'RON', 'CHF'];
  var i;
  for (var i = 0; i < ccode.length; i++) {
     var ccode[i] = fx.convert(amount, {to: 'ccode[i]'});
  }

The bit above var ccode[i] is causing an error

SyntaxError: Unexpected token '['. Expected ';' after variable declaration.

I am still new to JS so please bear with me.

I am editing my question here since people asked why I was re-declaring var ccode[i] and its because I need to output this:

var EUR = fx.convert(amount, {to: "EUR"});

for (var i = 0; re declares i again. Remove the var, yes and same for ccode — Liam
– Liam, Commented Nov 24, 2017 at 10:24
I had var ccode[i] because I need to output this "var EUR = fx.convert(amount, {to: "EUR"});" — user3501825
– user3501825, Commented Nov 24, 2017 at 10:33

Machigatta · Accepted Answer · 2017-11-24 10:25:08Z

1

var ccode = ['EUR', 'BRL', 'RUB', 'KRW', 'RON', 'CHF'];
var i;
  for (i = 0; i < ccode.length; i++) {
     ccode[i] = fx.convert(amount, {to: 'ccode[i]'});
  }

ccode is already declared.. no need to use var

edited Nov 24, 2017 at 10:25

answered Nov 24, 2017 at 10:23

Machigatta

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1 Comment

Jamie Eltringham Over a year ago

same for goes for i

Jagdish Idhate · Accepted Answer · 2017-11-24 10:28:56Z

0

Its not about declaring ccode twice, as variable declaration syntax is wrong

var ccode[i]; should not be array, see for reference

You should remove var.

var ccode = ['EUR', 'BRL', 'RUB', 'KRW', 'RON', 'CHF'];
  var i;
  for (var i = 0; i < ccode.length; i++) {
     ccode[i] = fx.convert(amount, {to: 'ccode[i]'});
  }

answered Nov 24, 2017 at 10:28

Jagdish Idhate

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Comments

JackDev · Accepted Answer · 2017-11-24 16:19:43Z

0

I would suggest using an array map:

var ccode = ['EUR', 'BRL', 'RUB', 'KRW', 'RON', 'CHF'];
ccode = ccode.map(function(code) {
  return fx.convert(amount, {to: code})
})

edited Nov 24, 2017 at 16:19

answered Nov 24, 2017 at 10:29

JackDev

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1 Comment

Nina Scholz Over a year ago

map returns a new array, but you do not assign this result.

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