I want to check that next Element in my array is same
const p = ['12', '13', '13', '14', '15', '15', '16'] var len = p.length for (i = 0; i <= len - 1; i++) { var d1 = p[i] // document.write(d1) for (j = i + 1; j <= i + 1; j++) { var d2 = p[j] // document.write(d2) if (d1 == d2) { console.log(d1) console.log(d2) } else { console.log(d1) console.log('oops!') } break; } }
Here I have 7 Elements in my array in which same element is same as their before element but some are not.
What I want : If the next Element of my Element is not the same with the first one so automatically it print opps in place of next and then check for next. as I write in my code but my code is correct
Output I want = 12 oops 13 13 14 oops 15 15 16 oops
Output I’m geeting= 12 oops! 13 13 13 oops! 14 oops! 15 15 15 oops! 16 oops!
Anyone help me with this? I don’t know what to do.
Thank You
Advertisement
Answer
I’m not very sure that I understand logic and how to handle all edge cases, but here is variant which fits provided expected output:
const p = ['12', '13', '13', '14', '15', '15', '16'] let len = p.length; for (let i = 0; i < len; i++) { console.log(p[i]); if (i+1 >= len || p[i] != p[i+1]) console.log('oops!'); // print oops only if we're at the end of the array OR elements are different else console.log(p[i++]); // this will run only if we're before the end of array AND numbers are the same }
note: what about multiple (more than 2) consecutive equal numbers?