How to create Index of json array with only values?

Couchbase Server
1

I have a field called storeImageNameList which stores image names,
Now if i want to get value by image name then how to create index on json array,
Eg: storeImageNameList :[“image1”,“image2”,“image3”…]

Below is the Document values image,

sample1
sample1.PNG554×524 7.33 KB

I want to fetch Bridge001 from document what can be done?

3 
CREATE INDEX ix1 ON default(DISTINCT ARRAY c FOR c IN usedcarimages END) 
WHERE _class = "com.usedCars.ServiceUsedCars.pojo.UsedCarDetails";

SELECT d.* 
FROM default AS d
WHERE  d._class = "com.usedCars.ServiceUsedCars.pojo.UsedCarDetails"
                     AND ANY c IN d.usedcarimages SATISFIES c = "Bridge001" END;

Last Image name in the array

SELECT d.usedcarimages[-1] AS lastimage
FROM default AS d
WHERE  d._class = "com.usedCars.ServiceUsedCars.pojo.UsedCarDetails"
                     AND ANY c IN d.usedcarimages SATISFIES c = "Bridge001" END;

If you want sort and get last image

SELECT ARRAY_SORT(d.usedcarimages)[-1] AS lastimage
FROM default AS d
WHERE  d._class = "com.usedCars.ServiceUsedCars.pojo.UsedCarDetails"
                     AND ANY c IN d.usedcarimages SATISFIES c = "Bridge001" END;
1 Like
4

did not fetch the result…

5

You might not have matching record. Post actual document you expecting

7

thank you very much got the records

8

if image names are stored in following way,

Bridge01-1
Bridge01-2
Bridge01-3

then how can i get all the names by just giving input Bridge01 and not mentioning full name like Bridge01-1?

9

SELECT * FROM UsedCar
WHERE ANY v IN usedcarimages SATISFIES v LIKE ‘%Bridge01%’ END;

10

If you need string started with Bridge01 don’t give leading %. It will impact performance.

SELECT * FROM UsedCar
WHERE ANY v IN usedcarimages SATISFIES v LIKE "Bridge01%" END;