Adding numbers that are not in array format? Or how to filter to array so I can sum up
up vote
1
down vote
favorite
In previous versions of jq I was able to run the following:
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount | add'
On this sample data:
{
"data": {
"organization": {
"repositories": {
"nodes": [{
"pullRequests": {
"totalCount": 2
}
},
{
"pullRequests": {
"totalCount": 8
}
},
{
"pullRequests": {
"totalCount": 23
}
}
]
}
}
}
}
And I would get the correct result.
But currently on jq-1.6 I am getting the following error:
jq: error (at <stdin>:24): Cannot iterate over number (2)
What I noticed from the output without the add
filter is that is not an array:
➤ cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount'
2
8
23
So my question is how to add these numbers up?
I also tried casting it to array by using [.pullRequests.totalCount]
but I was unable to merge, meld, join the arrays to get the final count.
stream add jq
add a comment |
up vote
1
down vote
favorite
In previous versions of jq I was able to run the following:
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount | add'
On this sample data:
{
"data": {
"organization": {
"repositories": {
"nodes": [{
"pullRequests": {
"totalCount": 2
}
},
{
"pullRequests": {
"totalCount": 8
}
},
{
"pullRequests": {
"totalCount": 23
}
}
]
}
}
}
}
And I would get the correct result.
But currently on jq-1.6 I am getting the following error:
jq: error (at <stdin>:24): Cannot iterate over number (2)
What I noticed from the output without the add
filter is that is not an array:
➤ cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount'
2
8
23
So my question is how to add these numbers up?
I also tried casting it to array by using [.pullRequests.totalCount]
but I was unable to merge, meld, join the arrays to get the final count.
stream add jq
For now I am doingcat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.
– Danilo Cabello
Nov 22 at 15:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In previous versions of jq I was able to run the following:
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount | add'
On this sample data:
{
"data": {
"organization": {
"repositories": {
"nodes": [{
"pullRequests": {
"totalCount": 2
}
},
{
"pullRequests": {
"totalCount": 8
}
},
{
"pullRequests": {
"totalCount": 23
}
}
]
}
}
}
}
And I would get the correct result.
But currently on jq-1.6 I am getting the following error:
jq: error (at <stdin>:24): Cannot iterate over number (2)
What I noticed from the output without the add
filter is that is not an array:
➤ cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount'
2
8
23
So my question is how to add these numbers up?
I also tried casting it to array by using [.pullRequests.totalCount]
but I was unable to merge, meld, join the arrays to get the final count.
stream add jq
In previous versions of jq I was able to run the following:
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount | add'
On this sample data:
{
"data": {
"organization": {
"repositories": {
"nodes": [{
"pullRequests": {
"totalCount": 2
}
},
{
"pullRequests": {
"totalCount": 8
}
},
{
"pullRequests": {
"totalCount": 23
}
}
]
}
}
}
}
And I would get the correct result.
But currently on jq-1.6 I am getting the following error:
jq: error (at <stdin>:24): Cannot iterate over number (2)
What I noticed from the output without the add
filter is that is not an array:
➤ cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount'
2
8
23
So my question is how to add these numbers up?
I also tried casting it to array by using [.pullRequests.totalCount]
but I was unable to merge, meld, join the arrays to get the final count.
stream add jq
stream add jq
edited Nov 22 at 15:49
peak
29.6k83955
29.6k83955
asked Nov 22 at 15:35
Danilo Cabello
1,102815
1,102815
For now I am doingcat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.
– Danilo Cabello
Nov 22 at 15:36
add a comment |
For now I am doingcat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.
– Danilo Cabello
Nov 22 at 15:36
For now I am doing
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.– Danilo Cabello
Nov 22 at 15:36
For now I am doing
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.– Danilo Cabello
Nov 22 at 15:36
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
You are mistaken in thinking that the jq filter as shown used to work on the JSON as shown.
There are fortunately two simple fixes:
[ .data.organization.repositories.nodes
| .pullRequests.totalCount ]
| add
or:
.data.organization.repositories.nodes
| map(.pullRequests.totalCount)
| add
Using sigma/1
Another option is to use a stream-oriented summation function:
def sigma(s): reduce s as $s (null; .+$s);
.data.organization.repositories.nodes
| sigma(..pullRequests.totalCount)
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You are mistaken in thinking that the jq filter as shown used to work on the JSON as shown.
There are fortunately two simple fixes:
[ .data.organization.repositories.nodes
| .pullRequests.totalCount ]
| add
or:
.data.organization.repositories.nodes
| map(.pullRequests.totalCount)
| add
Using sigma/1
Another option is to use a stream-oriented summation function:
def sigma(s): reduce s as $s (null; .+$s);
.data.organization.repositories.nodes
| sigma(..pullRequests.totalCount)
add a comment |
up vote
2
down vote
You are mistaken in thinking that the jq filter as shown used to work on the JSON as shown.
There are fortunately two simple fixes:
[ .data.organization.repositories.nodes
| .pullRequests.totalCount ]
| add
or:
.data.organization.repositories.nodes
| map(.pullRequests.totalCount)
| add
Using sigma/1
Another option is to use a stream-oriented summation function:
def sigma(s): reduce s as $s (null; .+$s);
.data.organization.repositories.nodes
| sigma(..pullRequests.totalCount)
add a comment |
up vote
2
down vote
up vote
2
down vote
You are mistaken in thinking that the jq filter as shown used to work on the JSON as shown.
There are fortunately two simple fixes:
[ .data.organization.repositories.nodes
| .pullRequests.totalCount ]
| add
or:
.data.organization.repositories.nodes
| map(.pullRequests.totalCount)
| add
Using sigma/1
Another option is to use a stream-oriented summation function:
def sigma(s): reduce s as $s (null; .+$s);
.data.organization.repositories.nodes
| sigma(..pullRequests.totalCount)
You are mistaken in thinking that the jq filter as shown used to work on the JSON as shown.
There are fortunately two simple fixes:
[ .data.organization.repositories.nodes
| .pullRequests.totalCount ]
| add
or:
.data.organization.repositories.nodes
| map(.pullRequests.totalCount)
| add
Using sigma/1
Another option is to use a stream-oriented summation function:
def sigma(s): reduce s as $s (null; .+$s);
.data.organization.repositories.nodes
| sigma(..pullRequests.totalCount)
answered Nov 22 at 15:46
peak
29.6k83955
29.6k83955
add a comment |
add a comment |
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For now I am doing
cat pull_requests.json | jq '.data.organization.repositories.nodes | .pullRequests.totalCount' | awk '{s+=$1}END{print s}'
but I would like to use only jq in this case because it used to work.– Danilo Cabello
Nov 22 at 15:36