这必须是一个简单的解决方案,因为我之前已经做了很多次。但就目前而言,我完全被难倒了。我使用以下代码来保存父对象Unknown_Tag及其许多子对象。
方法:
public function saveUnknown(Request $request)
{
$url = $request->url;
$tag = new Unknown_Tag();
$tag->url = $url;
$protocol =
substr($url, 0, strpos($url, ':'));
$tag->protocol = $protocol;
$domain =
parse_url($url, PHP_URL_HOST);
$tag->domain = $domain;
$tag->save();
//get the path
$Path = parse_url($url, PHP_URL_PATH);
if ($Path) {
$splitPath = substr($Path, 1);
$paths = explode('/', $splitPath);
foreach ($paths as $p) {
$path = new Path();
$path->path = $p;
$tag->Paths()->save($path);
}
}
//get Queries
$splitQuery = parse_url($url, PHP_URL_QUERY);
if ($splitQuery) {
$queries = explode('&', $splitQuery);
foreach ($queries as $q) {
$query = new Query();
$q = substr($q, 0, strpos($q, '='));
IF (SUBSTR($q, -1) != ' ') {
$q .= ' ';
}
$query->var = $q;
$value = $q = preg_replace('/^[^=]*:/', '', $q);
$query->value = $value;
$tag->Queries()->save($query);
}
}
}
父对象
class Unknown_Tag extends Model
{
protected $table = 'unknown_tags';
public $timestamps = false;
public function Paths()
{
return $this->hasMany('App\Path', 'tag_id', 'ID');
}
public function Queries()
{
return $this->hasMany('App\Query', 'tag_id', 'ID');
}
}
儿童对象
class Query extends Model
{
protected $table = 'queries';
public $timestamps = false;
public function Tag()
{
return $this->belongsTo('App\Unknown_Tag', 'tag_id', 'ID');
}
}
class Path extends Model
{
protected $table = 'paths';
public $timestamps = false;
public function Tag()
{
return $this->belongsTo('App\Unknown_Tag', 'tag_id', 'ID');
}
}
当我通过post请求运行所有这些时,Parent和所有子节点都被正确保存,但是所有子对象都有一个设置为null的外键。如果我手动将外键更改为它应该是什么,一切正常,所以我很确定这不是我的数据库的问题。任何人都能看到我在这里失踪的显而易见的事情吗?
编辑:要清楚,这不会返回任何错误
答案 0 :(得分:1)
如果有人见过这个,laravel会认为默认的主键是' id'。我已将我的设置为“ID'”,所以我必须通过使用
让laravel知道protected $ primaryKey =' ID';
在我的Unknown_tag定义中