How to pass query string parameters through URL

Question

I am using this:

$state.go("/path/to");

I want to do this:

$state.go("/path/to?param=myParam");

But this is not working.

I have tried:

$location.path("/path/to?param=myParam");
$location.path("/path/to").search({param: "myParam"});

Both of them actually change the url but not updated the angularjs app - since this is not connected to the actual router.

How can I pass query string parameters using AngularJS ui router? Note I know I can pass a full url path like this:

/path/to/param/myParam

But I need in this case that it will be query string params.

you can do this .state('path', { url: '/to/:param', template: '<res-template></res-template>' }) — Owais Alam
– Owais Alam, Commented Jul 23, 2019 at 8:23

NTP · Accepted Answer · 2019-07-23 08:29:08Z

2

According to the documentation $state.go takes another parameter that you can use to pass query string.

$state.go("/path/to",{param:myParam});

answered Jul 23, 2019 at 8:29

NTP

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1 Comment

Raz Buchnik Over a year ago

This doesn't working. I am naming the name of the state the same as the url of it. Does it matters?

georgeawg · Accepted Answer · 2019-07-23 15:17:25Z

1

Url config use like this

  .state('path/:param', { 
                url: '/path/:param',
                templateUrl: 'views/path.html',
      })

you get value in controller use to $stateParams

edited Jul 23, 2019 at 15:17

georgeawg

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answered Jul 23, 2019 at 8:28

Raja Ramachandran

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1 Comment

Theo Over a year ago

While your code may provide the answer to the question, please add context around it so others will have some idea what it does and why it is there.

Raz Buchnik · Accepted Answer · 2019-07-23 11:18:46Z

0

I don't know why, but angularjs ui router is very tedious vs the regular ngroute made by Angular team. The reason I am using ui router is just because that it supports for lazyload of the view.

Anyway, this is the solution I was found - using $state.go and $location.path both:

$location.path(`/path/to`).search({param: "myParam}});
$state.go(`/path/to`);

This can be wrapped for easy use:

module.exports = (path, qsObj) => {
    if (qsObj) {
      $location.path(path).search(qsObj);
    }
    $state.go(path);
}

answered Jul 23, 2019 at 11:18

Raz Buchnik

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