How do you make a column in a gridField link to a specific record when the recor

reginaldm over 9 years ago

How do you make a column in a gridField link to a specific record when the recordType is expression backed?

I've tried a few things, but it has only worked once, which was a fluke.

If images are needed for a better understanding, let me know.

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0 Chris over 9 years ago

Posting your code here may help. Also, I realized I do this in a report as well - this is my gridTextColumn (service backed record). In this example, fldID is the unique identifier and fldSIRnumber is displayed to the user as the record link text. local!gridData comes directly from a queryEntity() call:

a!gridTextColumn(
label: "Request #",
data: local!gridData.data.fldSIRnumber,
links:
a!applyComponents(a!recordLink(recordType: cons!SIR_RECORD,label: _, identifier: _),
merge(
index(local!gridData.data,"fldSIRnumber",null),
index(local!gridData.data,"fldID",null)
)
),
field: "fldSIRnumber",
alignment: "CENTER"
)
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0 Chris over 9 years ago

Posting your code here may help. Also, I realized I do this in a report as well - this is my gridTextColumn (service backed record). In this example, fldID is the unique identifier and fldSIRnumber is displayed to the user as the record link text. local!gridData comes directly from a queryEntity() call:

a!gridTextColumn(
label: "Request #",
data: local!gridData.data.fldSIRnumber,
links:
a!applyComponents(a!recordLink(recordType: cons!SIR_RECORD,label: _, identifier: _),
merge(
index(local!gridData.data,"fldSIRnumber",null),
index(local!gridData.data,"fldID",null)
)
),
field: "fldSIRnumber",
alignment: "CENTER"
)
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