node_save() failing when creating node programatically - drupal-7

Question

I am trying to create a node programatically like so,
$newNode = (object) NULL;
$newNode->type = 'job';
$newNode->title = $data['JobTitle'];
$newNode->uid = $user->uid;
$newNode->created = strtotime("now");
$newNode->changed = strtotime("now");
$newNode->status = 1;
$newNode->comment = 0;
$newNode->promote = 0;
$newNode->moderate = 0;
$newNode->sticky = 0;
$newNode->tid = 0;
$newNode->summary['und'][0]['value'] = $data['JobSummary'];
$newNode->body['und'][0]['value'] = $data['JobDescription'];
$newNode->field_employment_type['und'] = strtolower($data['JobType']);
$newNode->field_job_reference['und'][0]['value'] = $data['JobReference'];
$newNode->field_salary['und'][0]['value'] = "";
$newNode->field_salary_from['und'][0]['value'] = $data['SalaryFrom'];
$newNode->field_salary_to['und'][0]['value'] = $data['SalaryTo'];
$newNode->field_salary_override['und'][0]['value'] = $data['Salary'];
$newNode->field_application_email['und'][0]['value'] = $data['ApplicationEmail'];
$newNode->field_job_category['und'][2] = 2;
$newNode->field_job_category['und'][4] = 4;
//die(print_r($newNode));
// save node
node_save($newNode);
Here I have potentially 4 taxonomies to pick from (their id indicated in brackets) Creative (2), Technical (3), Marketing (4), Client Services (6).
On node_save I am getting the following error,
500 Internal Server Error : An error occurred (23000):
SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'tid'
cannot be null
How do I overcome this this I would have thought setting the field_job_category to the id of the taxonomy would be enough?

Answer 1

Delete $newNode->tid = 0; and prepare object :
$newNode = new stdClass();
$newNode->type = 'job';
node_object_prepare($node);
$newNode->title = $data['JobTitle'];
$newNode->uid = $user->uid;
$newNode->created = strtotime("now");
$newNode->changed = strtotime("now");
$newNode->status = 1;
$newNode->comment = 0;
$newNode->promote = 0;
$newNode->moderate = 0;
$newNode->sticky = 0;
$newNode->summary['und'][0]['value'] = $data['JobSummary'];
$newNode->body['und'][0]['value'] = $data['JobDescription'];
$newNode->field_employment_type['und'] = strtolower($data['JobType']);
$newNode->field_job_reference['und'][0]['value'] = $data['JobReference'];
$newNode->field_salary['und'][0]['value'] = "";
$newNode->field_salary_from['und'][0]['value'] = $data['SalaryFrom'];
$newNode->field_salary_to['und'][0]['value'] = $data['SalaryTo'];
$newNode->field_salary_override['und'][0]['value'] = $data['Salary'];
$newNode->field_application_email['und'][0]['value'] = $data['ApplicationEmail'];
$newNode->field_job_category['und'][]['tid'] = 2; // right syntax
$newNode->field_job_category['und'][]['tid']= 4; // right syntax
//die(print_r($newNode));
// save node
node_save($newNode);

Answer 2

If you want to have only one term id associated:
$newNode->field_category[LANGUAGE_NONE][0]['tid'] = <actual term id>
If you have multiple term ids to associate:
Iterate over your array of term ids, and increment index. Something like:
$newNode->field_category[LANGUAGE_NONE][0]['tid'] = <actual term id>
$newNode->field_category[LANGUAGE_NONE][1]['tid'] = <actual term id>
$newNode->field_category[LANGUAGE_NONE][2]['tid'] = <actual term id>
Note the index values above.
And, I usually set following attributes set for a new node: