Firebase query results are mistaken
This is my database design.
foodie-ab2b4
Foods
0
FoodName: "Baked Beans In Tomato Sauce"
FoodRecipe:
FoodUri:
Image:
1
FoodName: "Another bean
I'm working on an ios project and this is how my firebase json structured.
let ref = Database.database().reference()
func searchFoodByName(FoodName: String)
let foodsRef = ref.child("Foods")
let input = FoodName
let query = foodsRef.child(key).queryOrdered(byChild: "FoodName").queryEnding(atValue: input)
query.observeSingleEvent(of: .value, with: snapshot in
print(snapshot)
for child in snapshot.children
let snap = child as! DataSnapshot
let dict = snap.value as! [String: Any]
let fName = dict["FoodName"] as! String
let fIngredients = dict["Ingredients"] as! [String]
print(fName)
print(fIngredients)
let key = snapshot.key
print(key)
)
I'm trying to filter my food objects by their names. I have 10 objects in my database. In most cases this search return true objects. But there is one example i can not solve.
When i query the "tomato" word i need to get 1 object. But there is two. Although there is no tomato word in my second object.
The two object is in my json file.
If it is necessary i can upload my whole JSON file to here.
ios swift firebase firebase-realtime-database
add a comment |
This is my database design.
foodie-ab2b4
Foods
0
FoodName: "Baked Beans In Tomato Sauce"
FoodRecipe:
FoodUri:
Image:
1
FoodName: "Another bean
I'm working on an ios project and this is how my firebase json structured.
let ref = Database.database().reference()
func searchFoodByName(FoodName: String)
let foodsRef = ref.child("Foods")
let input = FoodName
let query = foodsRef.child(key).queryOrdered(byChild: "FoodName").queryEnding(atValue: input)
query.observeSingleEvent(of: .value, with: snapshot in
print(snapshot)
for child in snapshot.children
let snap = child as! DataSnapshot
let dict = snap.value as! [String: Any]
let fName = dict["FoodName"] as! String
let fIngredients = dict["Ingredients"] as! [String]
print(fName)
print(fIngredients)
let key = snapshot.key
print(key)
)
I'm trying to filter my food objects by their names. I have 10 objects in my database. In most cases this search return true objects. But there is one example i can not solve.
When i query the "tomato" word i need to get 1 object. But there is two. Although there is no tomato word in my second object.
The two object is in my json file.
If it is necessary i can upload my whole JSON file to here.
ios swift firebase firebase-realtime-database
What is the value ofkey
?
– Frank van Puffelen
Nov 10 '18 at 17:47
add a comment |
This is my database design.
foodie-ab2b4
Foods
0
FoodName: "Baked Beans In Tomato Sauce"
FoodRecipe:
FoodUri:
Image:
1
FoodName: "Another bean
I'm working on an ios project and this is how my firebase json structured.
let ref = Database.database().reference()
func searchFoodByName(FoodName: String)
let foodsRef = ref.child("Foods")
let input = FoodName
let query = foodsRef.child(key).queryOrdered(byChild: "FoodName").queryEnding(atValue: input)
query.observeSingleEvent(of: .value, with: snapshot in
print(snapshot)
for child in snapshot.children
let snap = child as! DataSnapshot
let dict = snap.value as! [String: Any]
let fName = dict["FoodName"] as! String
let fIngredients = dict["Ingredients"] as! [String]
print(fName)
print(fIngredients)
let key = snapshot.key
print(key)
)
I'm trying to filter my food objects by their names. I have 10 objects in my database. In most cases this search return true objects. But there is one example i can not solve.
When i query the "tomato" word i need to get 1 object. But there is two. Although there is no tomato word in my second object.
The two object is in my json file.
If it is necessary i can upload my whole JSON file to here.
ios swift firebase firebase-realtime-database
This is my database design.
foodie-ab2b4
Foods
0
FoodName: "Baked Beans In Tomato Sauce"
FoodRecipe:
FoodUri:
Image:
1
FoodName: "Another bean
I'm working on an ios project and this is how my firebase json structured.
let ref = Database.database().reference()
func searchFoodByName(FoodName: String)
let foodsRef = ref.child("Foods")
let input = FoodName
let query = foodsRef.child(key).queryOrdered(byChild: "FoodName").queryEnding(atValue: input)
query.observeSingleEvent(of: .value, with: snapshot in
print(snapshot)
for child in snapshot.children
let snap = child as! DataSnapshot
let dict = snap.value as! [String: Any]
let fName = dict["FoodName"] as! String
let fIngredients = dict["Ingredients"] as! [String]
print(fName)
print(fIngredients)
let key = snapshot.key
print(key)
)
I'm trying to filter my food objects by their names. I have 10 objects in my database. In most cases this search return true objects. But there is one example i can not solve.
When i query the "tomato" word i need to get 1 object. But there is two. Although there is no tomato word in my second object.
The two object is in my json file.
If it is necessary i can upload my whole JSON file to here.
ios swift firebase firebase-realtime-database
ios swift firebase firebase-realtime-database
edited Nov 10 '18 at 17:46
Frank van Puffelen
228k28374398
228k28374398
asked Nov 10 '18 at 15:46
AsgencerAsgencer
62
62
What is the value ofkey
?
– Frank van Puffelen
Nov 10 '18 at 17:47
add a comment |
What is the value ofkey
?
– Frank van Puffelen
Nov 10 '18 at 17:47
What is the value of
key
?– Frank van Puffelen
Nov 10 '18 at 17:47
What is the value of
key
?– Frank van Puffelen
Nov 10 '18 at 17:47
add a comment |
1 Answer
1
active
oldest
votes
You seem to assume that Firebase can filter based on strings that contain a value, but it can't. See Firebase query - Find item with child that contains string (and the many links from there).
What Firebase can do is search for string values that start with a certain value. You do that by using a combination of queryStarting(atValue:)
and queryEnding(atValue:)
:
let query = foodsRef
.queryOrdered(byChild: "FoodName")
.queryStarting(atValue: input)
.queryEnding(atValue: input+"\uf8ff")
If you use this query, and input
is ``Baked`, it will only match the first food from your JSON.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53240586%2ffirebase-query-results-are-mistaken%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You seem to assume that Firebase can filter based on strings that contain a value, but it can't. See Firebase query - Find item with child that contains string (and the many links from there).
What Firebase can do is search for string values that start with a certain value. You do that by using a combination of queryStarting(atValue:)
and queryEnding(atValue:)
:
let query = foodsRef
.queryOrdered(byChild: "FoodName")
.queryStarting(atValue: input)
.queryEnding(atValue: input+"\uf8ff")
If you use this query, and input
is ``Baked`, it will only match the first food from your JSON.
add a comment |
You seem to assume that Firebase can filter based on strings that contain a value, but it can't. See Firebase query - Find item with child that contains string (and the many links from there).
What Firebase can do is search for string values that start with a certain value. You do that by using a combination of queryStarting(atValue:)
and queryEnding(atValue:)
:
let query = foodsRef
.queryOrdered(byChild: "FoodName")
.queryStarting(atValue: input)
.queryEnding(atValue: input+"\uf8ff")
If you use this query, and input
is ``Baked`, it will only match the first food from your JSON.
add a comment |
You seem to assume that Firebase can filter based on strings that contain a value, but it can't. See Firebase query - Find item with child that contains string (and the many links from there).
What Firebase can do is search for string values that start with a certain value. You do that by using a combination of queryStarting(atValue:)
and queryEnding(atValue:)
:
let query = foodsRef
.queryOrdered(byChild: "FoodName")
.queryStarting(atValue: input)
.queryEnding(atValue: input+"\uf8ff")
If you use this query, and input
is ``Baked`, it will only match the first food from your JSON.
You seem to assume that Firebase can filter based on strings that contain a value, but it can't. See Firebase query - Find item with child that contains string (and the many links from there).
What Firebase can do is search for string values that start with a certain value. You do that by using a combination of queryStarting(atValue:)
and queryEnding(atValue:)
:
let query = foodsRef
.queryOrdered(byChild: "FoodName")
.queryStarting(atValue: input)
.queryEnding(atValue: input+"\uf8ff")
If you use this query, and input
is ``Baked`, it will only match the first food from your JSON.
edited Nov 25 '18 at 21:44
marc_s
572k12811051252
572k12811051252
answered Nov 10 '18 at 17:54
Frank van PuffelenFrank van Puffelen
228k28374398
228k28374398
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53240586%2ffirebase-query-results-are-mistaken%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What is the value of
key
?– Frank van Puffelen
Nov 10 '18 at 17:47