我想像
.$quiz_option1,$quiz_option2,$quiz_option3
一样将$quiz_option
增加 1
在这里,我提供了添加更多功能以将答案插入数据库。通过使用上面的选项字段,我不知道怎么可能。我正在使用最后一个插入 id 来执行以下任务。
$Quiz_ID = $quiz->insert();
for($i=0;$i<count($_REQUEST['quiz_options']);$i++)
{
$data=array
(
"quiz_Id"=>$Quiz_ID,
"quiz_Options1"=>$_REQUEST['quiz_options'][$i]
//To change in above line of code like quiz_Options1, quiz_Options2
);
$quiz->insertOptions($data,'quizoptions');
}
我认为这就是你需要的:
$Quiz_ID = $quiz->insert();
$data = array(
"quiz_Id" => $Quiz_ID;
)
for($i=0; $i<count($_REQUEST['quiz_options']); $i++) {
$key = "quiz_Options" . ($i+1);
$data[$key] = $_REQUEST['quiz_options'][$i];
}
$quiz->insertOptions($data,'quizoptions');
$data (print_r($data)) 的示例如下所示:
Array
(
[quiz_Id] => 111
[quiz_Options1] => aaa
[quiz_Options2] => bbb
[quiz_Options3] => ccc
)
它应该是这样的:
for($i=0;$i<count($_REQUEST['quiz_options']);$i++)
{
$data=array
(
"quiz_Id"=>$Quiz_ID,
"quiz_Options".$i =>$_REQUEST['quiz_options'][$i]
//To change in above line of code like quiz_Options1, quiz_Options2
);
$quiz->insertOptions($data,'quizoptions');
}
您需要
在数组启动之前预先增加 1 到 $i
。
for($i=0;$i<count($_REQUEST['quiz_options']);$i++)
{
$key = $i;
$data=array
(
"quiz_Id"=>$Quiz_ID,
"quiz_Options".++$key =>$_REQUEST['quiz_options'][$i]
//To change in above line of code like quiz_Options1, quiz_Options2
);
$quiz->insertOptions($data,'quizoptions');
}