Google Firestore: Query on substring of a property value (text search)

Question

I am looking to add a simple search field, would like to use something like

collectionRef.where('name', 'contains', 'searchTerm')

I tried using where('name', '==', '%searchTerm%'), but it didn't return anything.

Answer 1

There's no such operator, allowed ones are ==, <, <=, >, >=.

You can filter by prefixes only, for example for everything that starts between bar and foo you can use

collectionRef.where('name', '>=', 'bar').where('name', '<=', 'foo')

You can use external service like Algolia or ElasticSearch for that.

Answer 2

I agree with @Kuba's answer, But still, it needs to add a small change to work perfectly for search by prefix. here what worked for me

For searching records starting with name queryText

collectionRef.where('name', '>=', queryText).where('name', '<=', queryText+ '\uf8ff').

The character \uf8ff used in the query is a very high code point in the Unicode range (it is a Private Usage Area [PUA] code). Because it is after most regular characters in Unicode, the query matches all values that start with queryText.

Answer 3

While Kuba's answer is true as far as restrictions go, you can partially emulate this with a set-like structure:

{
  'terms': {
    'reebok': true,
    'mens': true,
    'tennis': true,
    'racket': true
  }
}

Now you can query with

collectionRef.where('terms.tennis', '==', true)

This works because Firestore will automatically create an index for every field. Unfortunately this doesn't work directly for compound queries because Firestore doesn't automatically create composite indexes.

You can still work around this by storing combinations of words but this gets ugly fast.

You're still probably better off with an outboard full text search.

Answer 4

While Firebase does not explicitly support searching for a term within a string,

Firebase does (now) support the following which will solve for your case and many others:

As of August 2018 they support array-contains query. See: https://firebase.googleblog.com/2018/08/better-arrays-in-cloud-firestore.html

You can now set all of your key terms into an array as a field then query for all documents that have an array that contains 'X'. You can use logical AND to make further comparisons for additional queries. (This is because firebase does not currently natively support compound queries for multiple array-contains queries so 'AND' sorting queries will have to be done on client end)

Using arrays in this style will allow them to be optimized for concurrent writes which is nice! Haven't tested that it supports batch requests (docs don't say) but I'd wager it does since its an official solution.

---

Usage:

collection("collectionPath").
    where("searchTermsArray", "array-contains", "term").get()

Answer 5

Full-Text Search, Relevant Search, and Trigram Search!

UPDATE - 2/17/21 - I created several new Full Text Search Options.

See Fireblog.io for details.

---

Also, side note, dgraph now has websockets for realtime... wow, never saw that coming, what a treat! Slash Dgraph - Amazing!

---

--Original Post--

A few notes here:

1.) \uf8ff works the same way as ~

2.) You can use a where clause or start end clauses:

ref.orderBy('title').startAt(term).endAt(term + '~');

is exactly the same as

ref.where('title', '>=', term).where('title', '<=', term + '~');

3.) No, it does not work if you reverse startAt() and endAt() in every combination, however, you can achieve the same result by creating a second search field that is reversed, and combining the results.

Example: First you have to save a reversed version of the field when the field is created. Something like this:

// collection
const postRef = db.collection('posts')

async function searchTitle(term) {

  // reverse term
  const termR = term.split("").reverse().join("");

  // define queries
  const titles = postRef.orderBy('title').startAt(term).endAt(term + '~').get();
  const titlesR = postRef.orderBy('titleRev').startAt(termR).endAt(termR + '~').get();

  // get queries
  const [titleSnap, titlesRSnap] = await Promise.all([
    titles,
    titlesR
  ]);
  return (titleSnap.docs).concat(titlesRSnap.docs);
}

With this, you can search the last letters of a string field and the first, just not random middle letters or groups of letters. This is closer to the desired result. However, this won't really help us when we want random middle letters or words. Also, remember to save everything lowercase, or a lowercase copy for searching, so case won't be an issue.

4.) If you have only a few words, Ken Tan's Method will do everything you want, or at least after you modify it slightly. However, with only a paragraph of text, you will exponentially create more than 1MB of data, which is bigger than firestore's document size limit (I know, I tested it).

5.) If you could combine array-contains (or some form of arrays) with the \uf8ff trick, you might could have a viable search that does not reach the limits. I tried every combination, even with maps, and a no go. Anyone figures this out, post it here.

6.) If you must get away from ALGOLIA and ELASTIC SEARCH, and I don't blame you at all, you could always use mySQL, postSQL, or neo4Js on Google Cloud. They are all 3 easy to set up, and they have free tiers. You would have one cloud function to save the data onCreate() and another onCall() function to search the data. Simple...ish. Why not just switch to mySQL then? The real-time data of course! When someone writes DGraph with websocks for real-time data, count me in!

Algolia and ElasticSearch were built to be search-only dbs, so there is nothing as quick... but you pay for it. Google, why do you lead us away from Google, and don't you follow MongoDB noSQL and allow searches?

Answer 6

Per the Firestore docs, Cloud Firestore doesn't support native indexing or search for text fields in documents. Additionally, downloading an entire collection to search for fields client-side isn't practical.

Third-party search solutions like Algolia and Elastic Search are recommended.

Answer 7

Late answer but for anyone who's still looking for an answer, Let's say we have a collection of users and in each document of the collection we have a "username" field, so if want to find a document where the username starts with "al" we can do something like

 FirebaseFirestore.getInstance().collection("users").whereGreaterThanOrEqualTo("username", "al")

Answer 8

I'm sure Firebase will come out with "string-contains" soon to capture any index[i] startAt in the string... But I’ve researched the webs and found this solution thought of by someone else set up your data like this

state = {title:"Knitting"}
...
const c = this.state.title.toLowerCase()

var array = [];
for (let i = 1; i < c.length + 1; i++) {
 array.push(c.substring(0, i));
}

firebase
.firestore()
.collection("clubs")
.doc(documentId)
.update({
 title: this.state.title,
 titleAsArray: array
})

query like this

firebase
.firestore()
.collection("clubs")
.where(
 "titleAsArray",
 "array-contains",
 this.state.userQuery.toLowerCase()
)

Answer 9

EDIT 05/2021:

Google Firebase now has an extension to implement Search with Algolia. Algolia is a full text search platform that has an extensive list of features. You are required to have a "Blaze" plan on Firebase and there are fees associated with Algolia queries, but this would be my recommended approach for production applications. If you prefer a free basic search, see my original answer below.

https://firebase.google.com/products/extensions/firestore-algolia-search https://www.algolia.com

ORIGINAL ANSWER:

The selected answer only works for exact searches and is not natural user search behavior (searching for "apple" in "Joe ate an apple today" would not work).

I think Dan Fein's answer above should be ranked higher. If the String data you're searching through is short, you can save all substrings of the string in an array in your Document and then search through the array with Firebase's array_contains query. Firebase Documents are limited to 1 MiB (1,048,576 bytes) (Firebase Quotas and Limits) , which is about 1 million characters saved in a document (I think 1 character ~= 1 byte). Storing the substrings is fine as long as your document isn't close to 1 million mark.

Example to search user names:

Step 1: Add the following String extension to your project. This lets you easily break up a string into substrings. (I found this here).

extension String {

var length: Int {
    return count
}

subscript (i: Int) -> String {
    return self[i ..< i + 1]
}

func substring(fromIndex: Int) -> String {
    return self[min(fromIndex, length) ..< length]
}

func substring(toIndex: Int) -> String {
    return self[0 ..< max(0, toIndex)]
}

subscript (r: Range<Int>) -> String {
    let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
                                        upper: min(length, max(0, r.upperBound))))
    let start = index(startIndex, offsetBy: range.lowerBound)
    let end = index(start, offsetBy: range.upperBound - range.lowerBound)
    return String(self[start ..< end])
}

Step 2: When you store a user's name, also store the result of this function as an array in the same Document. This creates all variations of the original text and stores them in an array. For example, the text input "Apple" would creates the following array: ["a", "p", "p", "l", "e", "ap", "pp", "pl", "le", "app", "ppl", "ple", "appl", "pple", "apple"], which should encompass all search criteria a user might enter. You can leave maximumStringSize as nil if you want all results, however, if there is long text, I would recommend capping it before the document size gets too big - somewhere around 15 works fine for me (most people don't search long phrases anyway).

func createSubstringArray(forText text: String, maximumStringSize: Int?) -> [String] {
    
    var substringArray = [String]()
    var characterCounter = 1
    let textLowercased = text.lowercased()
    
    let characterCount = text.count
    for _ in 0...characterCount {
        for x in 0...characterCount {
            let lastCharacter = x + characterCounter
            if lastCharacter <= characterCount {
                let substring = textLowercased[x..<lastCharacter]
                substringArray.append(substring)
            }
        }
        characterCounter += 1
        
        if let max = maximumStringSize, characterCounter > max {
            break
        }
    }
    
    print(substringArray)
    return substringArray
}

Step 3: You can use Firebase's array_contains function!

[yourDatabasePath].whereField([savedSubstringArray], arrayContains: searchText).getDocuments....

Answer 10

I just had this problem and came up with a pretty simple solution.

String search = "ca";
Firestore.instance.collection("categories").orderBy("name").where("name",isGreaterThanOrEqualTo: search).where("name",isLessThanOrEqualTo: search+"z")

The isGreaterThanOrEqualTo lets us filter out the beginning of our search and by adding a "z" to the end of the isLessThanOrEqualTo we cap our search to not roll over to the next documents.

Answer 11

If you don't want to use a third-party service like Algolia, Firebase Cloud Functions are a great alternative. You can create a function that can receive an input parameter, process through the records server-side and then return the ones that match your criteria.

Answer 12

As of today, there are basically 3 different workarounds, which were suggested by the experts, as answers to the question.

I have tried them all. I thought it might be useful to document my experience with each one of them.

Method-A: Using: (dbField ">=" searchString) & (dbField "<=" searchString + "\uf8ff")

Suggested by @Kuba & @Ankit Prajapati

.where("dbField1", ">=", searchString)
.where("dbField1", "<=", searchString + "\uf8ff");

A.1 Firestore queries can only perform range filters (>, <, >=, <=) on a single field. Queries with range filters on multiple fields are not supported. By using this method, you can't have a range operator in any other field on the db, e.g. a date field.

A.2. This method does NOT work for searching in multiple fields at the same time. For example, you can't check if a search string is in any of the fileds (name, notes & address).

Method-B: Using a MAP of search strings with "true" for each entry in the map, & using the "==" operator in the queries

Suggested by @Gil Gilbert

document1 = {
  'searchKeywordsMap': {
    'Jam': true,
    'Butter': true,
    'Muhamed': true,
    'Green District': true,
    'Muhamed, Green District': true,
  }
}

.where(`searchKeywordsMap.${searchString}`, "==", true);

B.1 Obviously, this method requires extra processing every time data is saved to the db, and more importantly, requires extra space to store the map of search strings.

B.2 If a Firestore query has a single condition like the one above, no index needs to be created beforehand. This solution would work just fine in this case.

B.3 However, if the query has another condition, e.g. (status === "active",) it seems that an index is required for each "search string" the user enters. In other words, if a user searches for "Jam" and another user searches for "Butter", an index should be created beforehand for the string "Jam", and another one for "Butter", etc. Unless you can predict all possible users' search strings, this does NOT work - in case of the query has other conditions!

.where(searchKeywordsMap["Jam"], "==", true); // requires an index on searchKeywordsMap["Jam"]
.where("status", "==", "active");

**Method-C: Using an ARRAY of search strings, & the "array-contains" operator

Suggested by @Albert Renshaw & demonstrated by @Nick Carducci

document1 = {
  'searchKeywordsArray': [
    'Jam',
    'Butter',
    'Muhamed',
    'Green District',
    'Muhamed, Green District',
  ]
}

.where("searchKeywordsArray", "array-contains", searchString); 

C.1 Similar to Method-B, this method requires extra processing every time data is saved to the db, and more importantly, requires extra space to store the array of search strings.

C.2 Firestore queries can include at most one "array-contains" or "array-contains-any" clause in a compound query.

General Limitations:

1. None of these solutions seem to support searching for partial strings. For example, if a db field contains "1 Peter St, Green District", you can't search for the string "strict."
2. It is almost impossible to cover all possible combinations of expected search strings. For example, if a db field contains "1 Mohamed St, Green District", you may NOT be able to search for the string "Green Mohamed", which is a string having the words in a different order than the order used in the db field.

There is no one solution that fits all. Each workaround has its limitations. I hope the information above can help you during the selection process between these workarounds.

For a list of Firestore query conditions, please check out the documentation https://firebase.google.com/docs/firestore/query-data/queries.

I have not tried https://fireblog.io/blog/post/firestore-full-text-search, which is suggested by @Jonathan.

Answer 13

I actually think the best solution to do this within Firestore is to put all substrings in an array, and just do an array_contains query. This allows you to do substring matching. A bit overkill to store all substrings but if your search terms are short it's very very reasonable.

Answer 14

I used trigram just like Jonathan said it.

trigrams are groups of 3 letters stored in a database to help with searching. so if I have data of users and I let' say I want to query 'trum' for donald trump I have to store it this way

and I just to recall this way

 onPressed: () {
      //LET SAY YOU TYPE FOR 'tru' for trump
      List<String> search = ['tru', 'rum'];
      Future<QuerySnapshot> inst = FirebaseFirestore.instance
          .collection("users")
          .where('trigram', arrayContainsAny: search)
          .get();
      print('result=');
      inst.then((value) {
        for (var i in value.docs) {
          print(i.data()['name']);
        }
      });

that will get correct result no matter what

Answer 15

This worked for me perfectly but might cause performance issues.

Do this when querying firestore:

   Future<QuerySnapshot> searchResults = collectionRef
        .where('property', isGreaterThanOrEqualTo: searchQuery.toUpperCase())
        .getDocuments();

Do this in your FutureBuilder:

    return FutureBuilder(
          future: searchResults,
          builder: (context, snapshot) {           
            List<Model> searchResults = [];
            snapshot.data.documents.forEach((doc) {
              Model model = Model.fromDocumet(doc);
              if (searchQuery.isNotEmpty &&
                  !model.property.toLowerCase().contains(searchQuery.toLowerCase())) {
                return;
              }

              searchResults.add(model);
            })
   };

Answer 16

Following code snippet takes input from user and acquires data starting with the typed one.

Sample Data:

Under Firebase Collection 'Users'

user1: {name: 'Ali', age: 28},

user2: {name: 'Khan', age: 30},

user3: {name: 'Hassan', age: 26},

user4: {name: 'Adil', age: 32}

TextInput: A

Result:

{name: 'Ali', age: 28},

{name: 'Adil', age: 32}

let timer;

// method called onChangeText from TextInput

const textInputSearch = (text) => {

const inputStart = text.trim();
let lastLetterCode = inputStart.charCodeAt(inputStart.length-1);
lastLetterCode++;
const newLastLetter = String.fromCharCode(lastLetterCode);
const inputEnd = inputStart.slice(0,inputStart.length-1) + lastLetterCode;

clearTimeout(timer);

timer = setTimeout(() => {
    firestore().collection('Users')
        .where('name', '>=', inputStart)
        .where('name', '<', inputEnd)
        .limit(10)
        .get()
        .then(querySnapshot => {
            const users = [];
                querySnapshot.forEach(doc => {
                    users.push(doc.data());
                })
            setUsers(users); //  Setting Respective State
        });
    }, 1000);

};

Answer 17

2021 Update

Took a few things from other answers. This one includes:

• Multi word search using split (acts as OR)
• Multi key search using flat

A bit limited on case-sensitivity, you can solve this by storing duplicate properties in uppercase. Ex: query.toUpperCase() user.last_name_upper

// query: searchable terms as string

let users = await searchResults("Bob Dylan", 'users');

async function searchResults(query = null, collection = 'users', keys = ['last_name', 'first_name', 'email']) {

    let querySnapshot = { docs : [] };

    try {
        if (query) {
            let search = async (query)=> {
                let queryWords = query.trim().split(' ');
                return queryWords.map((queryWord) => keys.map(async (key) =>
                    await firebase
                        .firestore()
                        .collection(collection)
                        .where(key, '>=', queryWord)
                        .where(key, '<=', queryWord +  '\uf8ff')
                        .get())).flat();
            }

            let results = await search(query);

            await (await Promise.all(results)).forEach((search) => {
                querySnapshot.docs = querySnapshot.docs.concat(search.docs);
            });
        } else {
            // No query
            querySnapshot = await firebase
                .firestore()
                .collection(collection)
                // Pagination (optional)
                // .orderBy(sortField, sortOrder)
                // .startAfter(startAfter)
                // .limit(perPage)
                .get();
        }
    } catch(err) {
        console.log(err)
    }

    // Appends id and creates clean Array
    const items = [];
    querySnapshot.docs.forEach(doc => {
        let item = doc.data();
        item.id = doc.id;
        items.push(item);
    });

    // Filters duplicates
    return items.filter((v, i, a) => a.findIndex(t => (t.id === v.id)) === i);
}

Note: the number of Firebase calls is equivalent to the number of words in the query string * the number of keys you're searching on.

Answer 18

With Firestore you can implement a full text search but it will still cost more reads than it would have otherwise, and also you'll need to enter and index the data in a particular way, So in this approach you can use firebase cloud functions to tokenise and then hash your input text while choosing a linear hash function h(x) that satisfies the following - if x < y < z then h(x) < h (y) < h(z). For tokenisation you can choose some lightweight NLP Libraries in order to keep the cold start time of your function low that can strip unnecessary words from your sentence. Then you can run a query with less than and greater than operator in Firestore. While storing your data also, you'll have to make sure that you hash the text before storing it, and store the plain text also as if you change the plain text the hashed value will also change.

Answer 19

Same as @nicksarno but with a more polished code that doesn't need any extension:

Step 1

func getSubstrings(from string: String, maximumSubstringLenght: Int = .max) -> [Substring] {
    let string = string.lowercased()
    let stringLength = string.count
    let stringStartIndex = string.startIndex
    var substrings: [Substring] = []
    for lowBound in 0..<stringLength {
        for upBound in lowBound..<min(stringLength, lowBound+maximumSubstringLenght) {
            let lowIndex = string.index(stringStartIndex, offsetBy: lowBound)
            let upIndex = string.index(stringStartIndex, offsetBy: upBound)
            substrings.append(string[lowIndex...upIndex])
        }
    }
    return substrings
}

Step 2

let name = "Lorenzo"
ref.setData(["name": name, "nameSubstrings": getSubstrings(from: name)])

Step 3

Firestore.firestore().collection("Users")
  .whereField("nameSubstrings", arrayContains: searchText)
  .getDocuments...

Answer 20

Firebase suggests Algolia or ElasticSearch for Full-Text search, but a cheaper alternative might be MongoDB. The cheapest cluster (approx US$10/mth) allows you to index for full-text.

Answer 21

We can use the back-tick to print out the value of a string. This should work:

where('name', '==', `${searchTerm}`)