PHP没有正确删除和插入数据

时间:2013-10-04 06:15:14

标签: php jquery

此代码的工作原理是每次用户访问该页面时,我都会从表格中删除所有数据并将其传递给以下帖子。如果用户之前没有访问过该页面,并且数据库因其cookie_id而为空,则会插入内容,如果找到,则会从数据库中删除内容并再次插入。我遇到的问题是,当删除发生时,删除所有内容,似乎从$ data1中插入一个值,而不是从$ data中插入所有数据。关于为什么会发生这种情况的任何想法?

后端:

    $cookie_id = $this->input->cookie("session_id");
    $selected_size = $this->input->post('selected_size');
    $product_id = $this->input->post('product_id');
    $product_name = $this->input->post('product_name');
    $product_color = $this->input->post('product_color');

    $q1 = $this->db->query("SELECT * FROM default_cart_temp
                            WHERE cookie_id = '$cookie_id'
                            AND is_paid = 'No'");
    if ($q1->num_rows() > 0) {
        $this->db->where('cookie_id', $cookie_id);
        $this->db->delete('default_cart_temp');

        foreach($product_id as $key => $value) {
            $data1 = array('selected_size' => $selected_size[$key],
                          'product_id' => $value,
                          'product_name' => $product_name[$key],
                          'product_color' => $product_color[$key],
                          'cookie_id' => $cookie_id,
                          'is_paid' => 'No');
        } $this->db->insert('default_cart_temp', $data1);
        echo json_encode(array('success' => true));
    } else {
        foreach($product_id as $key => $value) {
            $data = array('selected_size' => $selected_size[$key],
                          'product_id' => $value,
                          'product_name' => $product_name[$key],
                          'product_color' => $product_color[$key],
                          'cookie_id' => $cookie_id,
                          'is_paid' => 'No');
        } $this->db->insert('default_cart_temp', $data);
    }

前端:

   selected_size = $('.selected_size1').text();
    arr1 = selected_size.split(".");
    ss1 = arr1.slice(0, -1);

    product_id = $('.product_id1').text();
    arr2 = product_id.split(".");
    ss2 = arr2.slice(0, -1);

    product_name = $('.product_name1').text();
    arr3 = product_name.split(".");
    ss3 = arr3.slice(0, -1);

    product_color = $('.product_color1').text();
    arr4 = product_color.split(".");
    ss4 = arr4.slice(0, -1);

      $.ajax({
        type: "POST",
        dataType: "JSON",
        url: "<?=base_url()?>index.php/products/products/new_instance",
        data: { selected_size: ss1, product_id: ss2, product_name: ss3, product_color: ss4 },
        json: {success: true},
        success: function(data) {
          if(data.success == true) {
            alert("True");
          }
        }
      });

1 个答案:

答案 0 :(得分:1)

在每个foreach循环执行中,您将覆盖$data$data1个变量。您必须将它们更改为数组并添加类似$data[] = $subArray的数据。