Sequelize: How to query model by 1:n association, but include all associated objects in a single query...
I have a Ticket model with hasMany relation to Approval model (multi-level approval workflow). Approval belongsTo a User.
For a particular use case, I'll have to filter all the pending approvals to be made by a User and show him his to be approved Tickets.
Have solved like so -
models.Ticket.findAll({
include: [{
model: models.Approval,
where: { userId: options.userId }
}]
}).then(function(tickets){...
Gives me the filtered tickets but I've to also get the list of all approvals for the ticket to show the approval workflow. Can this be done by a single query in Sequelize?
Data (T is Ticket, A is Approval and U is User) -
T1 - A1/U1, A2/U2, A3/U3
T2 - A4/U2
T3 - A5/U1, A6/U4
Output for U1 -
T1 - A1/U1
T3 - A5/U1
Expected for U1 -
T1 - A1/U1, A2/U2, A3/U3
T3 - A5/U1, A6/U4
javascript node.js postgresql sequelize.js
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
add a comment |
I have a Ticket model with hasMany relation to Approval model (multi-level approval workflow). Approval belongsTo a User.
For a particular use case, I'll have to filter all the pending approvals to be made by a User and show him his to be approved Tickets.
Have solved like so -
models.Ticket.findAll({
include: [{
model: models.Approval,
where: { userId: options.userId }
}]
}).then(function(tickets){...
Gives me the filtered tickets but I've to also get the list of all approvals for the ticket to show the approval workflow. Can this be done by a single query in Sequelize?
Data (T is Ticket, A is Approval and U is User) -
T1 - A1/U1, A2/U2, A3/U3
T2 - A4/U2
T3 - A5/U1, A6/U4
Output for U1 -
T1 - A1/U1
T3 - A5/U1
Expected for U1 -
T1 - A1/U1, A2/U2, A3/U3
T3 - A5/U1, A6/U4
javascript node.js postgresql sequelize.js
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
add a comment |
I have a Ticket model with hasMany relation to Approval model (multi-level approval workflow). Approval belongsTo a User.
For a particular use case, I'll have to filter all the pending approvals to be made by a User and show him his to be approved Tickets.
Have solved like so -
models.Ticket.findAll({
include: [{
model: models.Approval,
where: { userId: options.userId }
}]
}).then(function(tickets){...
Gives me the filtered tickets but I've to also get the list of all approvals for the ticket to show the approval workflow. Can this be done by a single query in Sequelize?
Data (T is Ticket, A is Approval and U is User) -
T1 - A1/U1, A2/U2, A3/U3
T2 - A4/U2
T3 - A5/U1, A6/U4
Output for U1 -
T1 - A1/U1
T3 - A5/U1
Expected for U1 -
T1 - A1/U1, A2/U2, A3/U3
T3 - A5/U1, A6/U4
javascript node.js postgresql sequelize.js
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
I have a Ticket model with hasMany relation to Approval model (multi-level approval workflow). Approval belongsTo a User.
For a particular use case, I'll have to filter all the pending approvals to be made by a User and show him his to be approved Tickets.
Have solved like so -
models.Ticket.findAll({
include: [{
model: models.Approval,
where: { userId: options.userId }
}]
}).then(function(tickets){...
Gives me the filtered tickets but I've to also get the list of all approvals for the ticket to show the approval workflow. Can this be done by a single query in Sequelize?
Data (T is Ticket, A is Approval and U is User) -
T1 - A1/U1, A2/U2, A3/U3
T2 - A4/U2
T3 - A5/U1, A6/U4
Output for U1 -
T1 - A1/U1
T3 - A5/U1
Expected for U1 -
T1 - A1/U1, A2/U2, A3/U3
T3 - A5/U1, A6/U4
javascript node.js postgresql sequelize.js
javascript node.js postgresql sequelize.js
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
edited Nov 23 '18 at 21:54
Himavanth
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
asked Nov 22 '18 at 10:35
HimavanthHimavanth
12529
12529
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This is kind of tricky , but this way you can achieve the result in single query ,
models.Approval.findAll({
where: { userId: options.userId } // <---- Get all the approvals for user
include: [{
model: models.Ticket, // <----- Get all tickets for it
include: [{
model: models.Approval // <---- Get all the approvals for that ticket
}]
}]
})
NOTE : I haven't tried this one but as long as I know this should
work.
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
add a comment |
Found a workaround to solve it in one query.
Update Ticket to have two associations on Approval with the same foreign key (one with an alias defined) -
Ticket.hasMany(models.Approval, {
foreignKey: 'ticketId',
};
Ticket.hasMany(models.Approval, {
as: 'ApprovalFilter',
foreignKey: 'ticketId',
};
Then filter using multiple includes -
models.Ticket.findAll({
include: [{
model: models.Approval,
as: 'ApprovalFilter',
where: { userId: options.userId },
}, {
model: models.Approval,
}]
}).then(function(tickets){...
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is kind of tricky , but this way you can achieve the result in single query ,
models.Approval.findAll({
where: { userId: options.userId } // <---- Get all the approvals for user
include: [{
model: models.Ticket, // <----- Get all tickets for it
include: [{
model: models.Approval // <---- Get all the approvals for that ticket
}]
}]
})
NOTE : I haven't tried this one but as long as I know this should
work.
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
add a comment |
This is kind of tricky , but this way you can achieve the result in single query ,
models.Approval.findAll({
where: { userId: options.userId } // <---- Get all the approvals for user
include: [{
model: models.Ticket, // <----- Get all tickets for it
include: [{
model: models.Approval // <---- Get all the approvals for that ticket
}]
}]
})
NOTE : I haven't tried this one but as long as I know this should
work.
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
add a comment |
This is kind of tricky , but this way you can achieve the result in single query ,
models.Approval.findAll({
where: { userId: options.userId } // <---- Get all the approvals for user
include: [{
model: models.Ticket, // <----- Get all tickets for it
include: [{
model: models.Approval // <---- Get all the approvals for that ticket
}]
}]
})
NOTE : I haven't tried this one but as long as I know this should
work.
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
This is kind of tricky , but this way you can achieve the result in single query ,
models.Approval.findAll({
where: { userId: options.userId } // <---- Get all the approvals for user
include: [{
model: models.Ticket, // <----- Get all tickets for it
include: [{
model: models.Approval // <---- Get all the approvals for that ticket
}]
}]
})
NOTE : I haven't tried this one but as long as I know this should
work.
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
answered Nov 22 '18 at 11:21
Vivek DoshiVivek Doshi
21.6k33351
21.6k33351
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
add a comment |
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
Thanks for the reply and it works definitely. But isn't it possible in any other way to do the query on Ticket as base model?User ID here is being used as an optional filter in a bigger use case.
– Himavanth
Nov 22 '18 at 17:26
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
@Himavanth , in that case you need to fetch the ticket's id first , means 2 queries , that way you can do that.
– Vivek Doshi
Nov 22 '18 at 17:29
add a comment |
Found a workaround to solve it in one query.
Update Ticket to have two associations on Approval with the same foreign key (one with an alias defined) -
Ticket.hasMany(models.Approval, {
foreignKey: 'ticketId',
};
Ticket.hasMany(models.Approval, {
as: 'ApprovalFilter',
foreignKey: 'ticketId',
};
Then filter using multiple includes -
models.Ticket.findAll({
include: [{
model: models.Approval,
as: 'ApprovalFilter',
where: { userId: options.userId },
}, {
model: models.Approval,
}]
}).then(function(tickets){...
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
add a comment |
Found a workaround to solve it in one query.
Update Ticket to have two associations on Approval with the same foreign key (one with an alias defined) -
Ticket.hasMany(models.Approval, {
foreignKey: 'ticketId',
};
Ticket.hasMany(models.Approval, {
as: 'ApprovalFilter',
foreignKey: 'ticketId',
};
Then filter using multiple includes -
models.Ticket.findAll({
include: [{
model: models.Approval,
as: 'ApprovalFilter',
where: { userId: options.userId },
}, {
model: models.Approval,
}]
}).then(function(tickets){...
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
add a comment |
Found a workaround to solve it in one query.
Update Ticket to have two associations on Approval with the same foreign key (one with an alias defined) -
Ticket.hasMany(models.Approval, {
foreignKey: 'ticketId',
};
Ticket.hasMany(models.Approval, {
as: 'ApprovalFilter',
foreignKey: 'ticketId',
};
Then filter using multiple includes -
models.Ticket.findAll({
include: [{
model: models.Approval,
as: 'ApprovalFilter',
where: { userId: options.userId },
}, {
model: models.Approval,
}]
}).then(function(tickets){...
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
Found a workaround to solve it in one query.
Update Ticket to have two associations on Approval with the same foreign key (one with an alias defined) -
Ticket.hasMany(models.Approval, {
foreignKey: 'ticketId',
};
Ticket.hasMany(models.Approval, {
as: 'ApprovalFilter',
foreignKey: 'ticketId',
};
Then filter using multiple includes -
models.Ticket.findAll({
include: [{
model: models.Approval,
as: 'ApprovalFilter',
where: { userId: options.userId },
}, {
model: models.Approval,
}]
}).then(function(tickets){...
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
answered Nov 23 '18 at 21:34
HimavanthHimavanth
12529
12529
add a comment |
add a comment |
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