JavaScript: replace string inside array

Question

Maybe I'm making a very stupid mistake but here it goes. I want to take [ 'hr' ] and turn it into [ '* * *' ] So I did this:

var hr = jsonml[i]

console.log(hr)
// outputs: [ 'hr' ]
hr.replace(/hr/g, '* * *')

But I get this error:

TypeError: Object hr has no method 'replace'

What am I doing wrong?

hr should be of type String

Rakesh_Kumar
– Rakesh_Kumar

2015-03-02 14:14:03 +00:00
Commented Mar 2, 2015 at 14:14 — Rakesh_Kumar
– Rakesh_Kumar, Commented Mar 2, 2015 at 14:14
hr[0].replace(/hr/g, '* * *') ?

Millard
– Millard

2015-03-02 14:14:14 +00:00
Commented Mar 2, 2015 at 14:14 — Millard
– Millard, Commented Mar 2, 2015 at 14:14

Oleksandr T. · Accepted Answer · 2015-03-02 14:14:32Z

3

Because hr is Array, try this

hr[0] = hr[0].replace(/hr/g, '* * *');

or

hr = hr[0].replace(/hr/g, '* * *');

answered Mar 2, 2015 at 14:14

Oleksandr T.

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1 Comment

Peter Herdenborg Over a year ago

replace() returns the result of the operation, it doesn't change the variable on which you invoke it. So try hr = [hr[0].replace(...)]. (Assuming you want to continue having hr being an array)

curlybracket · Accepted Answer · 2015-03-02 14:22:57Z

1

hr is an array containing one string element. I would do this way:

if (hr.length > 0)
    hr[0] = hr[0].replace(/hr/g, '* * *');

EDIT: or maybe

for (var i = 0; i < hr.length; i++)
    hr[i] = hr[i].replace(/hr/g, '* * *');

if hr may contain more than one element

edited Mar 2, 2015 at 14:22

answered Mar 2, 2015 at 14:17

curlybracket

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Comments

Pippi · Accepted Answer · 2015-03-02 14:20:41Z

0

You can see the typeof of object :

alert(typeof hr);

you will see thath this object is an array!

use this :

for (i = 0; i < hr.length; i++) {
  var result = hr[i].replace(/hr/g, '* * *');
}

answered Mar 2, 2015 at 14:20

Pippi

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Comments

Peter Herdenborg · Accepted Answer · 2015-03-02 14:22:53Z

0

Just for the sake of providing an answer that actually does what OP asks for (no more, no less):

hr = [hr[0].replace('hr', '* * *')];

No need to use a regular expression when replacing in this case.

answered Mar 2, 2015 at 14:22

Peter Herdenborg

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