route with optional parameters in querystring

Question

I have a scenario where search can be performed only either of one value (id or age or sex).

If i use below route where only id is passed,

this.router.navigate(['details'],
{
     queryParams: 
     {
         id: this.id,
         age: this.age,
         sex: this.sex,
     }
 })

url will look like http://path/details?id=10&age=&sex=

I want my url to be like

path/details?id=10 on id search

path/details?age=25 on age search

So i refactored my code to,

if(this.id) {
    this.router.navigate(['details'],
    {
         queryParams: 
         {
             id: this.id
         }
     })
}else if(this.age){
    this.router.navigate(['details'],
    {
         queryParams: 
         {
             age: this.age
         }
     })
}else if(this.sex){
    this.router.navigate(['details'],
    {
         queryParams: 
         {
             sex: this.sex
         }
     })
}

Is there better way to refactor the code to avoid if statement

IMHO, this question belongs to codereview.stackexchange.com

Igor Soloydenko
– Igor Soloydenko

2017-10-13 22:47:29 +00:00
Commented Oct 13, 2017 at 22:47 — Igor Soloydenko
– Igor Soloydenko, Commented Oct 13, 2017 at 22:47

asmmahmud · Accepted Answer · 2017-10-14 07:39:17Z

3

You can simply do this, :

this.router.navigate(['details'],
{
     queryParams: 
     {
         id: this.id || null,
         age: !this.id && this.age || null,
         sex: !this.id && !this.age && this.sex || null
     }
 });

The above will give you the result.

edited Oct 14, 2017 at 7:39

answered Oct 13, 2017 at 23:02

asmmahmud

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