'realm react-native: how to query correctly an array of strings
can someone show me how to query an array of strings with realm in react-native?
assume i have an array like the following:
const preferences = ["automatic","suv","blue",eco]
What I want is to get realm results where ALL strings in the attribute "specifications" of Cars is in "preferences".
E.g.: If an instance of Cars.specifications contains ["automatic","suv"] a result should be returned.
But if an instance of Cars.specifications contained ["automatic,"suv","green"] this instance shouldn't be returned.
The length of preferences can vary.
Thank you very much.
Update:
What i tried is the following:
const query = realm.objects("Cars").filtered('specifications = preferences[0] OR specifications = preferences[1]')
As you see it is an OR operator which is surely wrong and it is hardcoded. Looping with realm really confuses me.
Solution 1:[1]
This code will work!
const collection = realm.objects('Cars');
const preferences = ["automatic","suv","blue","eco"];
let queryString = 'ANY ';
for (let i = 0; i < preferences.length; i++) {
if (i === 0) {
queryString += `specifications CONTAINS '${preferences[i]}'`;
}
if (i !== 0 && i + 1 <= preferences.length) {
queryString += ` OR specifications CONTAINS '${preferences[i]}'`;
}
}
const matchedResult = collection.filtered(queryString);
Solution 2:[2]
example of function to test if a word is inside an array of word
function inArray(word, array) {
var lgth = array.length;
word = word.toLowerCase();
for (var i = 0; i < lgth; i++) {
array[i] = (array[i]).toLowerCase();
if (array[i] == word) return true;
}
return false;
}
const preferences = ["automatic","suv","blue","eco"];
const specifications = ["automatic","suv"] ;
const specifications2 = ["automatic","suv", "boat"] ;
function test(spec,pref){
for (var i in spec){
if(!inArray(spec[i],pref)){
return false ;
}
}
return true;
}
console.log(test(specifications,preferences));
console.log(test(specifications2,preferences));
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | AlainIb |