-1

我正在尝试像这样以编程方式创建一个节点，

$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);

在这里，我可能有 4 个分类法可供选择（它们的 id 在括号中表示）创意（2）、技术（3）、营销（4）、客户服务（6）。

在 node_save 我收到以下错误，

500 内部服务器错误：发生错误 (23000)：SQLSTATE [23000]：完整性约束违规：1048 列 'tid' 不能为空

我该如何克服这个我认为将 field_job_category 设置为分类的 id 就足够了？

4

2 回答 2

0

删除$newNode->tid = 0;并准备对象：

$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);
于 2018-09-21T15:50:00.400 回答
0

如果您只想关联一个术语 ID：

$newNode->field_category[LANGUAGE_NONE][0]['tid'] = <actual term id>

如果您有多个要关联的术语 ID：迭代您的术语 ID 数组，并增加索引。就像是：

$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>

注意上面的索引值。

而且，我通常为新节点设置以下属性集：

于 2018-10-08T09:29:28.210 回答