Question

我想要一个代码来更新多行数据库，如下所示：

UPDATE values SET data='{"options":["male","female"],"default":"male"}' where project_id=1 and id=1;
UPDATE values SET data='{"options":["male","female"],"default":"male"}' where project_id=1 and id=2;
UPDATE values SET data='{"options":["male","female"],"default":"male"}' where project_id=1 and id=3;

几个小时之后，我可以在laravel框架中得到类似的结果：

$values = Value::where('project_id', $id)->get();
$sql = "";
foreach($request->fields as $field) {
  if(!empty($field->default)) { //New default value is set
    foreach($values as $value) {
      $data = json_decode($value->data, true); /**data column as json object in mysql database **/
      $data["default"] = $field->default;
      $data = json_encode($data);
      $sql .= "update ".DB::connection()->getDatabaseName().".values set data='".$data."' where id="."$value->id;";
    }
  }
}
DB::unprepared($sql);

但这段代码不是一个好习惯！所以我的问题是

有更好的ORM方式可以做到这一点吗？！

Answer 1

这是一种简单的方法。

$values = Value::where('project_id', $id)->update(['data'=>$data]);

我是从this链接

找到的

希望它有所帮助。

Answer 2

    $values = Value::where('project_id', $id)->get();
    $sql = "";
    foreach($request->fields as $field) {
      if(!empty($field->default)) { //New default value is set
        foreach($values as $value) {      
          $tmp=$value->data;
          $tmp->default = $field->default;
          $value->data = $tmp;
          $value->save();
        }
      }
    }

在Value模型中使用Mutator和Accessor，就像这样

public function getDataAttribute($value)
{
    return json_decode($value);
}
public function setDataAttribute($value)
{
    $this->attributes['data'] = json_encode($value);
}

请参阅文档http://laravel.com/docs/4.2/eloquent#accessors-and-mutators

工作 - https://yadi.sk/i/BnC3HYozehoy2

在Laravel中更新多行数据库

2 个答案: