我有一个表单来提交作业，在表单中有一个部分，您可以添加额外信息以及克隆字段的功能。
我在将记录保存到我的任务表时遇到问题，该表链接到带有ID的作业表。
当用户克隆字段时，我无法循环数据并将其保存在数据库中。
我正在削减所有验证以节省空间。这是我在JobController中的store方法：
public function store(Request $request)
{
/*JOB*/
$job = new Job();
$job->title=$request->title;
$job->client_id=$request->client_id;
$job->user_id=$request->user_id;
$job-> number =$request->number;
$job->ponumber =$request->ponumber;
$job->status =$request->status;
$job->invoicestatus =$request->invoicestatus;
$job->save();
$lastInsertedId = $job->id;
/*TASK*/
$task = new Task();
$task->job_id = $lastInsertedId;
$task->description = $request->description;
$task->supplier_hs= $request->supplier_hs;
$task->quote_hs= $request->quote_hs;
$task->quote_hs_date= $request->quote_hs_date;
$task->quote_client= $request->quote_client;
$task->quote_client_date= $request->quote_client_date;
$task->readytoinvoice= $request->readytoinvoice;
$task->invoiced= $request->invoiced;
$result = $task->getAttributes();
foreach($result as$key=>$t){
$i = 0;
foreach($result[$key] as $value){
echo"<pre>";
print_r($value);
echo"</pre>";
$i++;
echo "counter: " .$i;
}
}
//$task->save();
//return redirect('jobs')->with(array('feedback'=>'New Job created successfully','last_insert_id' => $job));
}
这是我从$ result获得的输出：
Array
(
[description] => Array
(
[0] => job 1
[1] => Job 2
[2] => job 3
)
[supplier_hs] => Array
(
[0] => 1000
[1] => 200
[2] => 3000
)
[quote_hs] => Array
(
[0] => 1000
[1] => 2000
[2] => 300
)
[quote_hs_date] => Array
(
[0] => 2017-02-15 11:50:07
[1] => 2017-02-01 00:00:00
[2] => 2017-02-15 11:50:07
)
[quote_client] => Array
(
[0] => 1000
[1] => 2000
[2] => 3000
)
[quote_client_date] => Array
(
[0] => 2017-02-15 11:50:07
[1] => 2017-02-01 00:00:00
[2] => 2017-02-15 11:50:07
)
[readytoinvoice] => Array
(
[0] => 1
[1] => 1
[2] => 1
)
[invoiced] => Array
(
[0] => 1
[1] => 1
[2] => 1
)
)
这就是我想得到的结果：
Array
(
[0] => Array
(
[description] => job 1
[supplier_hs]=> Job 2
[quote_hs] => job 3
[quote_hs_date] => 1000
[quote_client] => 2000
[quote_client_date] => 2017-02-15 11:50:07
[readytoinvoice] => 1
[invoiced] => 1
),
[1] => Array
(
[description] => job 1
[supplier_hs]=> Job 2
[quote_hs] => job 3
[quote_hs_date] => 1000
[quote_client] => 2000
[quote_client_date] => 2017-02-15 11:50:07
[readytoinvoice] => 1
[invoiced] => 1
),
[2] => Array
(
[description] => job 1
[supplier_hs]=> Job 2
[quote_hs] => job 3
[quote_hs_date] => 1000
[quote_client] => 2000
[quote_client_date] => 2017-02-15 11:50:07
[readytoinvoice] => 1
[invoiced] => 1
)
)
我用这种方式工作了。我不确定这是实现这一目标的最佳方式，但这是我的代码：
$tasks['tasks'] = array() ;
foreach($result as $key=>$value){
$i = 0;
if(!is_null($key)){
foreach($result as $value){
if(!is_null($value)){
$index = array_keys($value);
foreach($index as $ind){
if($ind == $i){
$tasks['tasks'][$i]['job_id'] = $lastInsertedId;
$tasks['tasks'][$i][$key] = $result[$key][$i];
}
}
}
$i++;
}
}
}
答案 0 :(得分：0)
如果我理解$ result最初是一个数组数组（每个'标题'，如quote_client等），那么使用它可能会有所帮助：
$finally = [];
$width = 0;
$keys = array_keys($result);
foreach ($keys as $kkk){
foreach($result as $res_key => $res_value){
foreach($res_value as $key => $value)
{
$finally[$width][$res_key] = $value;
}
}
$width ++;
}
PS：当然这个方法有一些嵌套循环，所以不保证效率。但是我的想法是laravel的集合中可能有一个方法可能会有所帮助。
希望有所帮助：）