I have this product feed from website, and in imageUrlStr column there are three images URL, I have added three more columns like Image1, Image2 and Image3.
imageUrlStr value
https://rukminim1.flixcart.com/image/200/200/l48s9zk0/water-geyser/g/o/i/2022-finesse-2000-ao-smith-10-original-imagf5zqzevs48p5.jpeg?q=90;https://rukminim1.flixcart.com/image/400/400/l48s9zk0/water-geyser/g/o/i/2022-finesse-2000-ao-smith-10-original-imagf5zqzevs48p5.jpeg?q=90;https://rukminim1.flixcart.com/image/800/800/l48s9zk0/water-geyser/g/o/i/2022-finesse-2000-ao-smith-10-original-imagf5zqzevs48p5.jpeg?q=90
I have tried this =LEFTB(D2,100) but as the length of the url is different this solution does not works.
https://docs.google.com/spreadsheets/d/1kLzW8_pg_wUiLg3rzhUX5fZfXlwIANm2r1mi9cvcrCk/edit?usp=sharing
You can edit and view product feed sheet from above url.
How can I retrieve image url as in different columns?
>Solution :
try:
=ARRAYFORMULA(IMAGE(IFERROR(SPLIT(D2:D; ";"))))
or just url would be:
=ARRAYFORMULA(IFERROR(SPLIT(D2:D; ";")))
or to get only first link:
=INDEX(IFERROR(SPLIT(D2:D; ";"));;1)