How to update records in related data models in Google App Maker
I’m creating an app with two data models:
1/ Model “A” with fields: Id, DATE, STATUS, EMAIL_ADDRESS, CASE_TYPE, CASE_DESCRIPTION.
2/ Model “B” with another fields (Id, DATE, ADMIN_EMAIL, CASE_RESOLUTION, and no field called STATUS).
Models”A” and “B” are related ONE-ONE (“A” is the Owner).
Users fill two forms. First form creates record in model “A”, and the second one: in model “B”.
When saving form number one I’m setting record.STATUS = “NEW” automatically, using event onBeforeCreate.
I’d like to update the STATUS to “BOSS_ACCEPT” in model “A” after saving form number two (datasource: “A”: “B” (related)). What should I do?
google-app-maker
asked Nov 26 '18 at 18:37
JolantaJolanta
41
add a comment |
I’m creating an app with two data models:
1/ Model “A” with fields: Id, DATE, STATUS, EMAIL_ADDRESS, CASE_TYPE, CASE_DESCRIPTION.
2/ Model “B” with another fields (Id, DATE, ADMIN_EMAIL, CASE_RESOLUTION, and no field called STATUS).
Models”A” and “B” are related ONE-ONE (“A” is the Owner).
Users fill two forms. First form creates record in model “A”, and the second one: in model “B”.
When saving form number one I’m setting record.STATUS = “NEW” automatically, using event onBeforeCreate.
I’d like to update the STATUS to “BOSS_ACCEPT” in model “A” after saving form number two (datasource: “A”: “B” (related)). What should I do?
google-app-maker
asked Nov 26 '18 at 18:37
JolantaJolanta
41
How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45
add a comment |
I’m creating an app with two data models:
1/ Model “A” with fields: Id, DATE, STATUS, EMAIL_ADDRESS, CASE_TYPE, CASE_DESCRIPTION.
2/ Model “B” with another fields (Id, DATE, ADMIN_EMAIL, CASE_RESOLUTION, and no field called STATUS).
Models”A” and “B” are related ONE-ONE (“A” is the Owner).
Users fill two forms. First form creates record in model “A”, and the second one: in model “B”.
When saving form number one I’m setting record.STATUS = “NEW” automatically, using event onBeforeCreate.
I’d like to update the STATUS to “BOSS_ACCEPT” in model “A” after saving form number two (datasource: “A”: “B” (related)). What should I do?
google-app-maker
asked Nov 26 '18 at 18:37
JolantaJolanta
41
I’m creating an app with two data models:
1/ Model “A” with fields: Id, DATE, STATUS, EMAIL_ADDRESS, CASE_TYPE, CASE_DESCRIPTION.
2/ Model “B” with another fields (Id, DATE, ADMIN_EMAIL, CASE_RESOLUTION, and no field called STATUS).
Models”A” and “B” are related ONE-ONE (“A” is the Owner).
Users fill two forms. First form creates record in model “A”, and the second one: in model “B”.
When saving form number one I’m setting record.STATUS = “NEW” automatically, using event onBeforeCreate.
I’d like to update the STATUS to “BOSS_ACCEPT” in model “A” after saving form number two (datasource: “A”: “B” (related)). What should I do?
google-app-maker
google-app-maker
asked Nov 26 '18 at 18:37
JolantaJolanta
41
asked Nov 26 '18 at 18:37
JolantaJolanta
41
asked Nov 26 '18 at 18:37
JolantaJolanta
41
asked Nov 26 '18 at 18:37
JolantaJolanta
41
asked Nov 26 '18 at 18:37
JolantaJolanta
41
41
How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45
add a comment |
How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45
How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45
add a comment |
1 Answer
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oldest
votes
There are really two ways to handle updating your related Model A record after creating your Model B record.
Server Side, which is more secure.
You would need the following in your OnAfterCreate server event script in Model B:
var relatedrecord = app.models.Model_A.getRecord(record.Model_A._key);
relatedrecord.STATUS = "BOSS_ACCEPT";
app.saveRecords([relatedrecord]);
And in order to reflect this change on the client, you would need to include a datasource load in your 'Submit' button createItem() call back function. This would look something like this:
widget.datasource.createItem(function() {
app.datasources.Model_A.load();
});
Client Side solution.
In your Form 2 'Submit' button include the following code:
widget.datasource.createItem(function(record) {
record.Model_A.STATUS = 'BOSS_ACCEPT';
});
If this answer helps you please mark it as accepted. Thank you.
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
There are really two ways to handle updating your related Model A record after creating your Model B record.
Server Side, which is more secure.
You would need the following in your OnAfterCreate server event script in Model B:
var relatedrecord = app.models.Model_A.getRecord(record.Model_A._key);
relatedrecord.STATUS = "BOSS_ACCEPT";
app.saveRecords([relatedrecord]);
And in order to reflect this change on the client, you would need to include a datasource load in your 'Submit' button createItem() call back function. This would look something like this:
widget.datasource.createItem(function() {
app.datasources.Model_A.load();
});
Client Side solution.
In your Form 2 'Submit' button include the following code:
widget.datasource.createItem(function(record) {
record.Model_A.STATUS = 'BOSS_ACCEPT';
});
If this answer helps you please mark it as accepted. Thank you.
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
add a comment |
There are really two ways to handle updating your related Model A record after creating your Model B record.
Server Side, which is more secure.
You would need the following in your OnAfterCreate server event script in Model B:
var relatedrecord = app.models.Model_A.getRecord(record.Model_A._key);
relatedrecord.STATUS = "BOSS_ACCEPT";
app.saveRecords([relatedrecord]);
And in order to reflect this change on the client, you would need to include a datasource load in your 'Submit' button createItem() call back function. This would look something like this:
widget.datasource.createItem(function() {
app.datasources.Model_A.load();
});
Client Side solution.
In your Form 2 'Submit' button include the following code:
widget.datasource.createItem(function(record) {
record.Model_A.STATUS = 'BOSS_ACCEPT';
});
If this answer helps you please mark it as accepted. Thank you.
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
add a comment |
There are really two ways to handle updating your related Model A record after creating your Model B record.
Server Side, which is more secure.
You would need the following in your OnAfterCreate server event script in Model B:
var relatedrecord = app.models.Model_A.getRecord(record.Model_A._key);
relatedrecord.STATUS = "BOSS_ACCEPT";
app.saveRecords([relatedrecord]);
And in order to reflect this change on the client, you would need to include a datasource load in your 'Submit' button createItem() call back function. This would look something like this:
widget.datasource.createItem(function() {
app.datasources.Model_A.load();
});
Client Side solution.
In your Form 2 'Submit' button include the following code:
widget.datasource.createItem(function(record) {
record.Model_A.STATUS = 'BOSS_ACCEPT';
});
If this answer helps you please mark it as accepted. Thank you.
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
There are really two ways to handle updating your related Model A record after creating your Model B record.
Server Side, which is more secure.
You would need the following in your OnAfterCreate server event script in Model B:
var relatedrecord = app.models.Model_A.getRecord(record.Model_A._key);
relatedrecord.STATUS = "BOSS_ACCEPT";
app.saveRecords([relatedrecord]);
And in order to reflect this change on the client, you would need to include a datasource load in your 'Submit' button createItem() call back function. This would look something like this:
widget.datasource.createItem(function() {
app.datasources.Model_A.load();
});
Client Side solution.
In your Form 2 'Submit' button include the following code:
widget.datasource.createItem(function(record) {
record.Model_A.STATUS = 'BOSS_ACCEPT';
});
If this answer helps you please mark it as accepted. Thank you.
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
answered Nov 27 '18 at 14:57
Markus MalessaMarkus Malessa
74649
74649
add a comment |
add a comment |
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How is your form for model B set up? Is the datasource 'Model_B (create) or is it 'Model_A: Model_B (relation) (create)? If it is the first option how are you setting the relation to Model_A? Because if you set the form for Model_B the second way, it automatically selects your current selected record in Model_A as your relation.
– Markus Malessa
Nov 26 '18 at 20:04
Thank for the answer. Actually, model B i set in the second way. I've already tried add record.modelA.STATUS = "BOSS_ACCEPT" onBeforeCreate and it didn't work. Maybe I'm doing something wrong? Is there any other way to change the STATUS in model A?
– Jolanta
Nov 27 '18 at 9:45