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Ayeye Brazo
Ayeye Brazo

Reputation: 3476

Angular filter to replace all underscores to spaces

I need a filter to replace all the underscores to spaces in a string

Upvotes: 26

Views: 87638

Answers (6)

Aliahmad Pasa
Aliahmad Pasa

Reputation: 1

this is i used in angularjs 1.4.7

<li ng-show="filter.degree.length"> 
    <label>Search by Degree :- </label> {{
        filter.degree.toString().split('|').join(', ')
    }} 
</li>

Upvotes: 0

Damien C
Damien C

Reputation: 1137

In some case, you can use split() function.
.replace function is not compliant with regexp syntax (i.e. .replace(/,/g,'\n') syntax)

Full syntax:
{{myVar.toString().split(',').join('\n')}}

.toString() function is in case myVar is not typed as String in typescript.

Upvotes: 7

Avellino
Avellino

Reputation: 89

This simple function can do it:

public getCleanedString(cadena) {
    cadena = cadena.replace(/_/g, ' ');
    return cadena;
  }

Upvotes: 1

BratisLatas
BratisLatas

Reputation: 527

There is a easyer method:

You could replace it inline without a defined filter. This is the way.

This example its for replace just in the view.

{{ value.replace(/_/g, ' ') }}

I hope its could help in a simple change, if you want to change in more places, use the filter.

Upvotes: 1

John Rix
John Rix

Reputation: 6693

Here's a generic replace filter alternative

App.filter('strReplace', function () {
  return function (input, from, to) {
    input = input || '';
    from = from || '';
    to = to || '';
    return input.replace(new RegExp(from, 'g'), to);
  };
});

Use it as follows in your HTML:

{{ addText | strReplace:'_':' ' }}

Minor note: Any HTML tags in the to parameter will cause the expression to fail due to Angular content security rules.

Upvotes: 23

Avinash Raj
Avinash Raj

Reputation: 174756

string.replace not only accepts string as first argument but also it accepts regex as first argument. So put _ within regex delimiters / and aslo add g modifier along with that. g called global modifier which will do the replacement globally.

App.filter('underscoreless', function () {
  return function (input) {
      return input.replace(/_/g, ' ');
  };
});

Upvotes: 44

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