继续PHP代码之前的用户操作
我使用下面的php代码创建和存储cookie继续PHP代码之前的用户操作,php,codeigniter,cookies,Php,Codeigniter,Cookies,我使用下面的php代码创建和存储cookie if ($logged_in_user['slots'] > $this->devices_model->count_devices($logged_in_user['id'])) { //HERE SHOULD BE PLACED WHAT I NEED $cookie = array( 'name' => 'kolass_app', 'value' =>
if ($logged_in_user['slots'] > $this->devices_model->count_devices($logged_in_user['id'])) {
//HERE SHOULD BE PLACED WHAT I NEED
$cookie = array(
'name' => 'kolass_app',
'value' => md5(uniqid($logged_in_user['id'], true)),
'device' => $this->agent->browser() . ' ' . $this->agent->version(),
'expire' => time() + (10 * 365 * 24 * 60 * 60),
);
$this->input->set_cookie($cookie);
$data['user_id'] = $logged_in_user['id'];
$data['device'] = $cookie['device'];
$data['value'] = $cookie['value'];
$this->devices_model->add_device($data);
redirect(base_url());
}
在继续创建和保存cookie之前,如何获取用户操作是或否
我使用的是Codeigniter您不能在PHP脚本中与用户交互。您需要做的是首先执行第一部分(如果),然后向用户显示一个页面,用户可以在其中选择是或否。如果用户选择是,则将其发送到执行其余代码的页面。不过要注意,如果有人打开多个选项卡,然后在所有选项卡上选择“是”,那么您还需要在第二页上输入
if
,以获得实际无法获得的“插槽”。使用
<?php
if ($logged_in_user['slots'] > $this->devices_model->count_devices($logged_in_user['id'])) {
?>
<script>
function myFunction() {
var r = confirm("Are You sure??");
if (r == true) {
<?php
$cookie = array(
'name' => 'kolass_app',
'value' => md5(uniqid($logged_in_user['id'], true)),
'device' => $this->agent->browser() . ' ' . $this->agent->version(),
'expire' => time() + (10 * 365 * 24 * 60 * 60),
);
$this->input->set_cookie($cookie);
$data['user_id'] = $logged_in_user['id'];
$data['device'] = $cookie['device'];
$data['value'] = $cookie['value'];
$this->devices_model->add_device($data);
redirect(base_url())
?>
} else {
alert('Not Served');
}
}
</script>
<?php
}
else
{
echo 'Not Loged In';
}
函数myFunction(){
var r=确认(“你确定吗?”);
如果(r==true){
}否则{
警报(“未送达”);
}
}