Creating a new record in a table for every result in another.

[Richard P]

Hi Girls and Guys,

What I am attempting to do, is to assign all the 7,000+ items in a specific catalog a single variable set.

I have written a fix script (below)
and while it seems to iterate through all 7,000+ items in the sc_cat_item table, it only creates a single record on the io_set_item table.

advice where I'm going wrong?
I've played around with grNew.newRecord(); and moving the initialize around, in and out of the loop.

Thanks in advance

var gr = new GlideRecord("sc_cat_item");
var grNew = new GlideRecord("io_set_item");

gr.addEncodedQuery('sc_catalogs=67ea87b21b448810ea17dd3fbd4bcb29');
gr.query();
  
while (gr.next()) {
    grNew.initialize();
    grNew.sc_cat_item = gr.sys_id;
    grNew.variable_set = "6c62fc831b584090ea17dd3fbd4bcb6e";
    grNew.order = "100";
    grNew.insert();
}

[Maddalena Mores]

Hi, try to write a function to create records in the io_set_item. 

var gr = new GlideRecord("sc_cat_item");
gr.addEncodedQuery('sc_catalogs=67ea87b21b448810ea17dd3fbd4bcb29');
gr.query();
  
while (gr.next()) {
    
	createRecord(gr.sys_id.toString());
}


function createRecord(id){

    var grNew = new GlideRecord("io_set_item");
    grNew.initialize();
    grNew.sc_cat_item = id;
    grNew.variable_set = "6c62fc831b584090ea17dd3fbd4bcb6e";
    grNew.order = "100";
    grNew.insert();
}

[Richard P · Accepted solution]

Hi All,


Thanks for the input, when I started seeing all the varying ways of achieving this and all producing exactly the same result, I started suspecting the issue did not lie in the script itself.

there's an after insert - business rule on the IO table preventing duplicate input of a variable set on the table...which I find odd since multiple catalog items can utilize the same variable set, thats the point of sets!

I disabled that business rule and off it went perfect, first try.

Thanks for all your help, maybe at least the thread gives options to people trying to find multiple ways of achieving the same insert.

I suspect somehow it tried after the fisrt insert to insert the same record again somehow, although I cannot see how in any of the suggested solutions.