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Basic JSON Question

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    Basic JSON Question

    Hi,

    I have a nested JSON that was generated from an SQL Query (using client side data cache editor). I am trying to access one of the child elements (using javascript), but can't seem to get the syntax right.

    The main table queried is the 'jobs' table with a sub query for 'site_details' table and a further sub table called 'locations'. I have hightlighted the field I am trying to access in the attached screen shot (which is a display of the JSON after using 'getfromdatacache' to get the data out of the cache, and then 'JSON.stringify(data)' to convert it to a JSON.

    I have tried different syntax to access the child one data elements (which corresponds to the 'site details table) (eg alert(data[0].child_1[0].site_address1);) but with no luck. Any ideas on the correct syntax ?

    Greg
    json issue.PNG

    #2
    Re: Basic JSON Question

    You'd need to post the sample data for proper testing... can't test using a screenshot... but it looks like your taking a JSON Object... an Array... and converting to a string... then trying to get it's array properties.

    Don't stringify it... leave it as an array and access it's properties using the syntax you've posted.

    Comment


      #3
      Re: Basic JSON Question

      Hi David,

      That was how I was doing it initially. However, I keep getting this error "Uncaught type error: cannot read property '0' of undefined". (see screen shot). I assume its due to one of the array indexes in the alert (alert(data[0].child_1[0].site_address1);).

      I have attached the 'stringified' version of the data. Is that sufficient for you to work with ?

      Cheers,
      Greg
      uncaught.PNG
      Attached Files

      Comment


        #4
        Re: Basic JSON Question

        Sorry... if that's what you're working with... then keep doing what you're doing... you're data is an Array of JSON Objects... all good.

        You're only problem is with case. Case matters in this stuff. You're trying to get .child_1 and it's actually .Child_1 - capital C on the child.

        child_1 doesn't exist... and so the error message... [0] of undefined.

        Comment


          #5
          Re: Basic JSON Question

          !!!!@$%%%&&&&!!!!!! .

          Yes, that was it. Thanks, David.

          Comment


            #6
            Re: Basic JSON Question

            Hi David,

            One other small problem I have with this....
            I have this code to push the json data into an array...however one of the sub arrays has no sub-elements, and I'm getting an undefined error. This is the code:
            for(var i = 0; i < data[0].Child_1.length; i++) {
            for(var j = 0; j < data[0].Child_1[i].Child_1a.length; j++) {
            _d.push( [data[0].Child_1[i].Child_1a[j].name, data[0].Child_1[i].Child_1a[j].id_location]);
            }
            }
            }
            }

            The error is relating to Child_1a.length, as Child_1a doesn't exist for a particular Child_1 array element. I tried modifying the code as follows:

            for(var i = 0; i < data[0].Child_1.length; i++) {
            if (data[0].Child_1[i].id_site_details == siteid) {
            if(typeof (data[0].Child_1[i].Child_1a.length) == 'undefined') {
            _d.push("No locations defined",0);
            } else {
            for(var j = 0; j < data[0].Child_1[i].Child_1a.length; j++) {
            _d.push( [data[0].Child_1[i].Child_1a[j].name, data[0].Child_1[i].Child_1a[j].id_location]);
            }
            }
            }
            }

            but still get the same error. Any suggestions on how I can solve it ?

            Thanks,
            Greg
            undefined.PNG

            Comment


              #7
              Re: Basic JSON Question

              By trying to get the length... of something that doesn't exist... you're creating the error.

              Code:
              if(typeof (data[0].Child_1[i].Child_1a) == 'undefined') {
              Test for the actual structure... that can be undefined.

              Comment


                #8
                Re: Basic JSON Question

                That fixed it. Thanks again....

                Comment

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