Firestore query documents startsWith a string

same as answered by Gil Gilbert. Just an enhancement and some sample code. use String.fromCharCode and String.charCodeAt

var strSearch = "start with text here";
var strlength = strSearch.length;
var strFrontCode = strSearch.slice(0, strlength-1);
var strEndCode = strSearch.slice(strlength-1, strSearch.length);

var startcode = strSearch;
var endcode= strFrontCode + String.fromCharCode(strEndCode.charCodeAt(0) + 1);

then filter code like below.

db.collection(c)
.where('foo', '>=', startcode)
.where('foo', '<', endcode);

Works on any Language and any Unicode.

Warning: all search criteria in firestore is CASE SENSITIVE.

You can but it's tricky. You need to search for documents greater than or equal to the string you want and less than a successor key.

For example, to find documents containing a field 'foo' staring with 'bar' you would query:

db.collection(c)
    .where('foo', '>=', 'bar')
    .where('foo', '<', 'bas');

This is actually a technique we use in the client implementation for scanning collections of documents matching a path. Our successor key computation is called by a scanner which is looking for all keys starting with the current user id.

Extending the previous answers with a shorter version:

  const text = 'start with text here';
  const end = text.replace(/.$/, c => String.fromCharCode(c.charCodeAt(0) + 1));

  query
    .where('stringField', '>=', text)
    .where('stringField', '<', end);

IRL example

async function search(startsWith = '') {
  let query = firestore.collection(COLLECTION.CLIENTS);

  if (startsWith) {
      const end = startsWith.replace(
        /.$/, c => String.fromCharCode(c.charCodeAt(0) + 1),
      );

      query = query
        .where('firstName', '>=', startsWith)
        .where('firstName', '<', end);
  }

  const result = await query
    .orderBy('firstName')
    .get();

  return result;
}