[vkwanengage Contributions: 0Level 1☆]

Hello NS community,

I was wondering, is there a way to get all users who have “Administrative Role” via SuiteScript? I want to be able to get all these users so i can alert them of some actions via email.

Thanks.
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[dominicb Contributions: 0Level 1☆]

Depending on exactly what you’re trying to alert, it may be easier to just use a saved search?

If not, SS2.0 (straight out of David Smith’s search export Chrome extension):

Code:
var employeeSearchObj = search.create({
type: “employee”,
filters:
[
[“role”,”anyof”,”3″]
],
columns:
[
search.createColumn({
name: “entityid”,
sort: search.Sort.ASC,
label: “Name”
}),
search.createColumn({name: “email”, label: “Email”}),
search.createColumn({name: “title”, label: “Job Title”})
]
});
var searchResultCount = employeeSearchObj.runPaged().count;
log.debug(“employeeSearchObj result count”,searchResultCount);
employeeSearchObj.run().each(function(result){
// .run().each has a limit of 4,000 results
return true;
});

[vkwanengage Contributions: 0Level 1☆]

Thanks dominicb

[TheUsualSuspect Contributions: 0Level 1☆]

Adding to dominicb answer somewhat. You can create a dynamic Group in NetSuite for employees with administrator roles and load the record in script using the ‘entitygroup’ record type. The only real benefit over a direct saved search is some easier management / maintenance for non-technical users in case additional criteria or special cases develop.

Code:
getEmailRecipients: function(entityGroup) {
log.debug({ ‘title’: ‘Entity Error Group Id’, ‘details’: entityGroup });
if (entityGroup) {
var entityGroupRec = record.load({ type: ‘entitygroup’, id: entityGroup });
var memberCount = entityGroupRec.getLineCount(‘groupmembers’);
var emailList = [];
for (var i = 0; i <= memberCount; i++) {
var email = entityGroupRec.getSublistValue({ sublistId: 'groupmembers', fieldId: 'memberemail', line: i });
if (email) { emailList.push(email); }
}
}
log.debug({ 'title': 'List of recipients', 'details': JSON.stringify(emailList) });
return emailList.toString();
}

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