CONTROLLER
$date=date("Y-d-m");
$succes=$this->pl->insert_ad($brandID,$model,$price,$year,$picture,$resized,$fuels,$id_usera,$date);
模型
public function insert_ad($brand,$model,$price,$year,$picture,$resized,$fuels,$date)
{
$sql1="INSERT INTO `ads`(`model`, `year`, `price`, `photo`, `date`, `resized_photo`, `id_fuel`, `id_brand`)
VALUES ('".$model."','".$year."','".$price."','".$picture."','".$date."','".$resized."','".$fuels."','".$brand."')";
$this->db->query($sql1);
return $this->db->insert_id();
}
我用print_r($date)
打印它并显示2016-03-16
,我现在真的不知道为什么它不会像其他所有内容一样插入。
答案 0 :(得分:1)
I just tried this minutes ago and it worked! ^_^
$d = date("Y-m-d");
$t = date("h:i:s");
$this->db->set($this->col_prefix.'customer_id', $_POST['customer_id']);
$this->db->set($this->col_prefix.'total_amount', $_POST['total_amount']);
$this->db->set('date', "'".$d."'");
$this->db->set('time_in', "'".$t."'");
$this->db->set('status', 1);
$this->db->insert('sln_transactions');
but haven't tried it yet in array :) hope this could help.
答案 1 :(得分:0)
请检查数据库表中“日期”列的数据类型。可能会有问题。
尝试以下功能:
日期( 'Y / M / d');
日期( 'H-M-S');
日期( 'Y / M / d:H-M-S');
我特别使用这些来获取日期和时间。也可以帮助你。
答案 2 :(得分:0)
从
更改$date=date("Y-d-m");
到
$date=date("Y-m-d");
来自整数类型的remvoe '
public function insert_ad($brand,$model,$price,$year,$picture,$resized,$fuels,$date)
{
$sql1="INSERT INTO `ads` (`id_ad`, `model`, `year`, `price`, `photo`, `date`, `resized_photo`, `id_fuel`, `id_brand`) VALUES (NULL, '$model', '$year', '$price', '$picture', '".date('Y-m-d')."', '$resized', '$fuels', '$brand')";
$this->db->query($sql1);
return $this->db->insert_id();
}
答案 3 :(得分:0)
更改insert_ad
功能,如下所示:
$sql1="INSERT INTO `ads`(`model`, `year`, `price`, `photo`,`resized_photo`, `id_fuel`, `id_brand`,`date`) VALUES ('".$model."','".$year."','".$price."','".$picture."','".$resized."','".$fuels."','".$brand."',NOW())";
$this->db->query($sql1);
return $this->db->insert_id();
同时删除控制器函数调用和模型函数定义中的日期参数。
答案 4 :(得分:0)
<强> CONTROLLER 强>
$date=date("Y-m-d");
数据库日期格式为Y-m-d
而非Y-d-m
答案 5 :(得分:0)
像这样进行查询。使用DATE(".$date.")
$sql1="INSERT INTO `ads`(`model`, `year`, `price`, `photo`, `date`, `resized_photo`, `id_fuel`, `id_brand`)
VALUES ('".$model."','".$year."','".$price."','".$picture."',DATE(".$date."),'".$resized."','".$fuels."','".$brand."')";
答案 6 :(得分:0)
请试试这个:
<强>控制器:强>
$date=date("Y-d-m");
$succes=$this->pl->insert_ad(array(
'model'=>$model,
'id_brand'=>$brandID,
'price'=>$price,
'photo'=>$picture,
'date'=>$date,
'resized_photo'=>$resized,
'id_fuel'=>$id_usera,
'year'=>$year));
<强>型号:强>
public function insert_ad($data) {
$this->db->insert('ads', $data);
return $this->db->insert_id();
}
答案 7 :(得分:0)
**答案**
*正确的日期格式* $date=date("Y-m-d");
*和* $date = date("Y-d-m");
以及
public function insert_ad($brand,$model,$price,$year,$picture,$resized,$fuels,$date)
{
$sql1="INSERT INTO `ads` (`id_ad`, `model`, `year`, `price`, `photo`, `date`, `resized_photo`, `id_fuel`, `id_brand`) VALUES (NULL, '$model', '$year', '$price', '$picture', '".date('Y-m-d')."', '$resized', '$fuels', '$brand')";
$this->db->query($sql1);
return $this->db->insert_id();
}
答案 8 :(得分:0)
尝试在数据库结构中使用type varchar作为日期。
答案 9 :(得分:0)
这项工作对我来说很完美
在控制器功能中
$this->model_invoice->invoice_create();
在模型中
$invoice_date = strtotime($this->input->post('invoice_date_time'));
$date=date("Y-m-d", $invoice_date);
$data = array
(
'invoice_date_time' => $date,
);
$this->db->insert('your_table_name', $data);
答案 10 :(得分:0)
尝试
在您的控制器中删除变量$ data 并在您的模型中删除变量
模型
public function insert_ad($brand,$model,$price,$year,$picture,$resized,$fuels)
{
$sql1="INSERT INTO `ads`(`model`, `year`, `price`, `photo`, `resized_photo`, `id_fuel`, `id_brand`)
VALUES ('".$model."','".$year."','".$price."','".$picture."','".$date."','".$resized."','".$fuels."','".$brand."')";
$this->db->query($sql1);
return $this->db->insert_id();
}
以及您的数据库更改 键入为 DATETIME ,然后 PREDETERMINATED 作为 CURRENT_TIME 并运行它
希望这行得通!