如果Javascript中为空或空，则不要传递变量

Question

我需要在service_ids中将if条件设置为空或null。 如果为空或为null，则不应传递service_ids变量。我该怎么办？

const formData = {
      car: form.value.car,
      remarks: form.value.remarks,
      if(form.value.service_ids !== null || form.value.service_ids !== []) {
        service_ids: form.value.service_ids.map((a: Service) => a.service_id ? a.service_id : []),
      }
}

Answer 1

怎么样

const formData = {
    car: form.value.car,
    remarks: form.value.remarks
}

if(form.value.service_ids !== null || form.value.service_ids !== []) {
    formData.service_ids = form.value.service_ids.map(
        (a: Service) => a.service_id ? a.service_id : []
    );
}

Answer 2

var {car,remarks,service_ids} = form.value;
service_ids = service_ids && service_ids.length > 0
     ? service_ids.map((a:Service) => a.service_id ? a.service_id : undefined).filter(val=>!(!val || val === ""))
    : []
var formData = {
    car,
    remarks,
    service_ids
}