Forum Discussion
Equiavlent js Map
I have a set variable activity that returns a json object and I need to perform JS Array.prototype.map equiavlent in Data Pipeline.
Can you please help. I don't want to manually hardcode by array index inside a concat
const obj = {
"value": {
"colNames": [
{
"tbl_id": 2,
"tbl_name": "fact_Activity",
"src_colname": "col1",
"sample_datetime": "2024-08-02T04:23:18Z",
"last_run_at": "2024-08-02T00:23:40.071123Z"
},
{
"tbl_id": 2,
"tbl_name": "fact_Activity",
"src_colname": "col2",
"sample_datetime": "2024-08-02T04:23:18Z",
"last_run_at": "2024-08-02T00:23:40.071123Z"
},
{
"tbl_id": 2,
"tbl_name": "fact_Activity",
"src_colname": "col3",
"sample_datetime": "2024-08-02T04:23:18Z",
"last_run_at": "2024-08-02T00:23:40.071123Z"
}
]
}
const result = obj.value.colNames.map(a=>a.src_colname).join(",")
//'col1,col2,col3'
}
@join(activity('Script1').output.resultSets[0].rows,',')
// this joins all the key-value pairs- Anonymous2 years ago
Hi smpa01 ,
Yes, the regular js array methods are supported as of now. I'll let you know as soon as possible if subsequent tests reveal unsupported methods.
Best Regards,
Adamk KongIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 Replies
- AnonymousNot applicable
Hi smpa01 ,
You can try to modify your code like below, this might potentially give you the result 'col1,col2,col3' without hard-coding the array index.
@join(activity('Script1').output.resultSets.rows.map(a => a.src_colname), ',')Best Regards,
Adamk KongIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
- smpa01Community Champion
Anonymous are all js Array methods valid in pipeline expression?
- AnonymousNot applicable
Hi smpa01 ,
Yes, the regular js array methods are supported as of now. I'll let you know as soon as possible if subsequent tests reveal unsupported methods.
Best Regards,
Adamk KongIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.