React ternary operator with array.map return

Question

In my react render function, I am using array.map to return some JSX code in an array and then rendering it. This doesn't seem to work, I read a few questions here that suggested using return statement inside if/else block but that won't work in my case. I want to check whether round and duration are set on each array element and only then pass it in the JSX code.Could someone tell me a different approach please.

render() {
 var interviewProcessMapped = interviewProcess.map((item, index) => {

  return
    {item.round ?
        <div className="row title">
          <div className="__section--left"><span className="section-title">Round {index + 1}</span></div>
          <div className="__section--right">
            <h3>{item.round}</h3>
          </div>
        </div>
        : null
    }

    {
      item.durationHours > 0 || item.durationMinutes > 0 ?
        <div className="row">
          <div className="__section--left">Duration</div>
          <div className="__section--right border">
            {item.durationHours > 0 ? <span>{item.durationHours} Hours</span> : null} {item.durationMinutes > 0 ? <span>{item.durationMinutes} Minutes</span> : null}
          </div>
        </div>
        : null
    }
  });

 return <div>{interviewProcessMapped}</div>
}

Answer 1

{ not required here:

return {item.round ?

If you use it that means you are returning an object.

Another issue is alone return means return; ( automatic semicolon insertion ) so either put the condition in same line or use () .

Write it like this:

render() {
    let a, b;
    var interviewProcessMapped = interviewProcess.map((item, index) => {
        a = item.round ?
                <div className="row title">
                    ....
                </div>
            : null;

        b =  (item.durationHours > 0 || item.durationMinutes > 0) ?
                  <div className="row">
                      ....
                  </div>
              : null;

        if(!a || !b)
            return a || b;
        return [a, b];
    });

    return (....)
}

Answer 2

You should probably use a combination of Array.prototype.map() and Array.prototype.filter()

No need for if at all

Here the pseudo code:

interviewProcess.filter(() => {
    // return only if round and duration set
}).map(() => {
    // transform the filtered list
});