值所在的 Firebase 过滤器

Question

The following call:

fDB.ref(`organization/${uid}/attendance/${date}`)
.once('value')
.then(d => console.log(d.val()))

returns an object like so:

{ '-Lz8YxH_n7a5C6wv5vrX': 'absent',
  '-Lz8YxHbn5iOCxaGUBZt': 'absent',
  '-Lz8YxHfQPKH5BaIJBK_': 'absent',
  '-Lz8YxHiwp8QZW3TqAFn': 'present',
  '-Lz8YxHjtMUkroeyT5bv': 'absent',
  '-Lz8YxHmHqDblmBY4jyx': 'absent',
  '-Lz8YxHo-e4AiOwwex0s': 'absent',
  '-Lz8YxHqQXWoaGOFRLrO': 'present',
}

how can I query ONLY the values that are absent? I tried:

fDB.ref(`organization/${uid}/attendance/`)
    .orderByChild(date)
    .equalTo('absent')
    .once('value')
    .then(d => console.log(d.val()))

The above returned null I also tried

fDB.ref(`organization/${uid}/attendance/${date}`)
    .orderByKey()
    .equalTo('absent')
    .once('value')
    .then(d => console.log(d.val()))

Same thing null. How can I achieve it?

Answer 1

It will filter out using orderByValue on the date node with equalTo , allowing you to run filter in values:

 fDB.ref(`organization/${uid}/attendance/${date}`)
    .orderByValue()
    .equalTo('absent')
    .once('value')
    .then(d => console.log(d.val()));