How to determine if case owner is queue with specific developer name
I have a problem with selecting specific custom queue in apex code.
For example I have:
Case c = (Case) controller.getRecord();
String queueName = c.Owner.Name;
if (queueName != 'ATM_queue') {
//DO SOMETHING
}
Thanks for any advice.
apex queue
asked Nov 19 '18 at 14:03
David
548
add a comment |
I have a problem with selecting specific custom queue in apex code.
For example I have:
Case c = (Case) controller.getRecord();
String queueName = c.Owner.Name;
if (queueName != 'ATM_queue') {
//DO SOMETHING
}
Thanks for any advice.
apex queue
asked Nov 19 '18 at 14:03
David
548
add a comment |
I have a problem with selecting specific custom queue in apex code.
For example I have:
Case c = (Case) controller.getRecord();
String queueName = c.Owner.Name;
if (queueName != 'ATM_queue') {
//DO SOMETHING
}
Thanks for any advice.
apex queue
asked Nov 19 '18 at 14:03
David
548
I have a problem with selecting specific custom queue in apex code.
For example I have:
Case c = (Case) controller.getRecord();
String queueName = c.Owner.Name;
if (queueName != 'ATM_queue') {
//DO SOMETHING
}
Thanks for any advice.
apex queue
apex queue
asked Nov 19 '18 at 14:03
David
548
asked Nov 19 '18 at 14:03
David
548
asked Nov 19 '18 at 14:03
David
548
asked Nov 19 '18 at 14:03
David
548
asked Nov 19 '18 at 14:03
David
548
548
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There's no direct relationship between Owner and DeveloperName, because it is a Name object. If you really want to check by developer name, you'll have to query for it:
if(c.ownerid.getsobjecttype() == Group.SobjectType) {
Group queue = [select developername from group where id = :c.ownerid];
if(Queue.DeveloperName == 'atm_queue') {
As an alternative, you could also create a formula field to return the value, something like owner:queue.developername, and then you can check that value instead.
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
add a comment |
I'm not sure if you specifically need Developer Name or if you can use friendly name instead, but the Owner.Name field is populated when the Owner is a queue - but it uses the display name. So if your queue is called "ATM Queue" then the Owner.Name will be ATM Queue and your if statement would be:
if(queueName != 'ATM Queue'){
// Do Something
}
Additionally, to make sure that the owner IS a queue before you assign the queueName string, you should do a Owner Check to see if the Owner is a User or a queue, like this:
String queueName;
String ownerId;
Case c = (Case) controller.getRecord();
ownerId = c.OwnerId;
if(String.valueOf(ownerId).startsWith('00G')){
queueName = c.Owner.Name;
}
if (queueName != 'ATM Queue') {
//DO SOMETHING
}
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
4
if(c.OwnerId.getSobjectType() == Group.SobjectType)is more self-documenting instead of relying on the key prefix.
– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
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
There's no direct relationship between Owner and DeveloperName, because it is a Name object. If you really want to check by developer name, you'll have to query for it:
if(c.ownerid.getsobjecttype() == Group.SobjectType) {
Group queue = [select developername from group where id = :c.ownerid];
if(Queue.DeveloperName == 'atm_queue') {
As an alternative, you could also create a formula field to return the value, something like owner:queue.developername, and then you can check that value instead.
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
add a comment |
There's no direct relationship between Owner and DeveloperName, because it is a Name object. If you really want to check by developer name, you'll have to query for it:
if(c.ownerid.getsobjecttype() == Group.SobjectType) {
Group queue = [select developername from group where id = :c.ownerid];
if(Queue.DeveloperName == 'atm_queue') {
As an alternative, you could also create a formula field to return the value, something like owner:queue.developername, and then you can check that value instead.
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
add a comment |
There's no direct relationship between Owner and DeveloperName, because it is a Name object. If you really want to check by developer name, you'll have to query for it:
if(c.ownerid.getsobjecttype() == Group.SobjectType) {
Group queue = [select developername from group where id = :c.ownerid];
if(Queue.DeveloperName == 'atm_queue') {
As an alternative, you could also create a formula field to return the value, something like owner:queue.developername, and then you can check that value instead.
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
There's no direct relationship between Owner and DeveloperName, because it is a Name object. If you really want to check by developer name, you'll have to query for it:
if(c.ownerid.getsobjecttype() == Group.SobjectType) {
Group queue = [select developername from group where id = :c.ownerid];
if(Queue.DeveloperName == 'atm_queue') {
As an alternative, you could also create a formula field to return the value, something like owner:queue.developername, and then you can check that value instead.
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
answered Nov 19 '18 at 14:53
sfdcfox
248k11189424
248k11189424
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
add a comment |
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
+1 for formula field as it saves queries, Downside(Only 20 object references on a case and that can be bit tricky)
– Pranay Jaiswal
Nov 19 '18 at 15:05
add a comment |
I'm not sure if you specifically need Developer Name or if you can use friendly name instead, but the Owner.Name field is populated when the Owner is a queue - but it uses the display name. So if your queue is called "ATM Queue" then the Owner.Name will be ATM Queue and your if statement would be:
if(queueName != 'ATM Queue'){
// Do Something
}
Additionally, to make sure that the owner IS a queue before you assign the queueName string, you should do a Owner Check to see if the Owner is a User or a queue, like this:
String queueName;
String ownerId;
Case c = (Case) controller.getRecord();
ownerId = c.OwnerId;
if(String.valueOf(ownerId).startsWith('00G')){
queueName = c.Owner.Name;
}
if (queueName != 'ATM Queue') {
//DO SOMETHING
}
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
4
if(c.OwnerId.getSobjectType() == Group.SobjectType)is more self-documenting instead of relying on the key prefix.
– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
add a comment |
I'm not sure if you specifically need Developer Name or if you can use friendly name instead, but the Owner.Name field is populated when the Owner is a queue - but it uses the display name. So if your queue is called "ATM Queue" then the Owner.Name will be ATM Queue and your if statement would be:
if(queueName != 'ATM Queue'){
// Do Something
}
Additionally, to make sure that the owner IS a queue before you assign the queueName string, you should do a Owner Check to see if the Owner is a User or a queue, like this:
String queueName;
String ownerId;
Case c = (Case) controller.getRecord();
ownerId = c.OwnerId;
if(String.valueOf(ownerId).startsWith('00G')){
queueName = c.Owner.Name;
}
if (queueName != 'ATM Queue') {
//DO SOMETHING
}
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
4
if(c.OwnerId.getSobjectType() == Group.SobjectType)is more self-documenting instead of relying on the key prefix.
– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
add a comment |
I'm not sure if you specifically need Developer Name or if you can use friendly name instead, but the Owner.Name field is populated when the Owner is a queue - but it uses the display name. So if your queue is called "ATM Queue" then the Owner.Name will be ATM Queue and your if statement would be:
if(queueName != 'ATM Queue'){
// Do Something
}
Additionally, to make sure that the owner IS a queue before you assign the queueName string, you should do a Owner Check to see if the Owner is a User or a queue, like this:
String queueName;
String ownerId;
Case c = (Case) controller.getRecord();
ownerId = c.OwnerId;
if(String.valueOf(ownerId).startsWith('00G')){
queueName = c.Owner.Name;
}
if (queueName != 'ATM Queue') {
//DO SOMETHING
}
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
I'm not sure if you specifically need Developer Name or if you can use friendly name instead, but the Owner.Name field is populated when the Owner is a queue - but it uses the display name. So if your queue is called "ATM Queue" then the Owner.Name will be ATM Queue and your if statement would be:
if(queueName != 'ATM Queue'){
// Do Something
}
Additionally, to make sure that the owner IS a queue before you assign the queueName string, you should do a Owner Check to see if the Owner is a User or a queue, like this:
String queueName;
String ownerId;
Case c = (Case) controller.getRecord();
ownerId = c.OwnerId;
if(String.valueOf(ownerId).startsWith('00G')){
queueName = c.Owner.Name;
}
if (queueName != 'ATM Queue') {
//DO SOMETHING
}
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
answered Nov 19 '18 at 14:52
Morgan Marchese
1,455426
1,455426
4
if(c.OwnerId.getSobjectType() == Group.SobjectType)is more self-documenting instead of relying on the key prefix.
– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
add a comment |
4
if(c.OwnerId.getSobjectType() == Group.SobjectType)is more self-documenting instead of relying on the key prefix.
– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
4
4
if(c.OwnerId.getSobjectType() == Group.SobjectType) is more self-documenting instead of relying on the key prefix.– sfdcfox
Nov 19 '18 at 14:54
if(c.OwnerId.getSobjectType() == Group.SobjectType) is more self-documenting instead of relying on the key prefix.– sfdcfox
Nov 19 '18 at 14:54
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
I actually didn't even know that you could use .getSObjectType() in that context, that's very good information, thanks!
– Morgan Marchese
Nov 19 '18 at 14:56
add a comment |
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