Angular conditional class with pipe












1















Trying to do a conditional class if progress == 100, but my syntax is not correct?



    [class.finalDone]="progress[file.name].progress | async == 100" 









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  • Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

    – SiddAjmera
    Nov 20 '18 at 19:40
















1















Trying to do a conditional class if progress == 100, but my syntax is not correct?



    [class.finalDone]="progress[file.name].progress | async == 100" 









share|improve this question























  • Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

    – SiddAjmera
    Nov 20 '18 at 19:40














1












1








1








Trying to do a conditional class if progress == 100, but my syntax is not correct?



    [class.finalDone]="progress[file.name].progress | async == 100" 









share|improve this question














Trying to do a conditional class if progress == 100, but my syntax is not correct?



    [class.finalDone]="progress[file.name].progress | async == 100" 






angular






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asked Nov 20 '18 at 19:33









MarkMark

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  • Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

    – SiddAjmera
    Nov 20 '18 at 19:40



















  • Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

    – SiddAjmera
    Nov 20 '18 at 19:40

















Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

– SiddAjmera
Nov 20 '18 at 19:40





Are you getting any errors? Also, are you sure progress[file.name].progress will return a Promise or an Observable ?

– SiddAjmera
Nov 20 '18 at 19:40












1 Answer
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Make sure that progress[file.name].progress will return a Promise or an Observable. And then use it like this(wrap the expression till the async part in ()):



[class.finalDone]="(progress[file.name].progress | async) === 100" 


Here's a Sample StackBlitz for your ref.






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    1 Answer
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    oldest

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    1














    Make sure that progress[file.name].progress will return a Promise or an Observable. And then use it like this(wrap the expression till the async part in ()):



    [class.finalDone]="(progress[file.name].progress | async) === 100" 


    Here's a Sample StackBlitz for your ref.






    share|improve this answer




























      1














      Make sure that progress[file.name].progress will return a Promise or an Observable. And then use it like this(wrap the expression till the async part in ()):



      [class.finalDone]="(progress[file.name].progress | async) === 100" 


      Here's a Sample StackBlitz for your ref.






      share|improve this answer


























        1












        1








        1







        Make sure that progress[file.name].progress will return a Promise or an Observable. And then use it like this(wrap the expression till the async part in ()):



        [class.finalDone]="(progress[file.name].progress | async) === 100" 


        Here's a Sample StackBlitz for your ref.






        share|improve this answer













        Make sure that progress[file.name].progress will return a Promise or an Observable. And then use it like this(wrap the expression till the async part in ()):



        [class.finalDone]="(progress[file.name].progress | async) === 100" 


        Here's a Sample StackBlitz for your ref.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 '18 at 19:43









        SiddAjmeraSiddAjmera

        13.7k31137




        13.7k31137






























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