Question

我是构建上传图片并将其存储到数据库中，我已经可以将很多图片上传到文件夹，但我无法插入上传的所有图片名称，我也不知道如何插入数据库，首先我在发生错误时对下面的代码进行了推荐，其次我不知道如果图像计数不同就把它放在数据库中的查询，例如1-10图像，最后一个问题，如果我查询＆＃ 34; SELECT id ...＆＃34;我想返回它，是否有方法将其返回到字符串或int？如果我使用row（），它将返回stdClass对象。请帮我，下面是我的代码：

控制器：

$this->load->library("myupload", "form_validation");
        $this->load->model("testModel");
        $barangImage = array();

        if($this->input->post("formSubmit")) {
            $this->form_validation->set_rules("nama", "Nama", "required|trim");

            if($this->form_validation->run()) {
                $insertData = array(
                    "nama" => $this->input->post("nama")
                );
                if($id = $this->testModel->add($insertData)) {

                    //print_r($id);

                    if(isset($_FILES) && $image = $this->myupload->uploadFile($_FILES)) {
                        //$image here is already fill with all images name

                        if(isset($image["error"]) && $image["error"]) {
                            echo $image["error"];
                        }else {
                            foreach($image as $img) {
                                $barangImage = array(
                                        "gambar" => $img,
                                        "barangid" => $id
                                );

                            }
                            //but when i put into barangImage,
                            //it only stored last image name
                            print_r($barangImage);
                            //output `Array ( [gambar] => 2.JPG [barangid] => Array ( [id] => 52 ) )`
                        }
                    }

                    if($id = $this->testModel->add_images($barangImage)) {
                        echo "SUCCESS !!!";
                    }else {
                        echo "FAIL INSERT IMAGES!!!";
                    }
                }else {
                    echo "FAIL INSERT DATA NAMA";
                }
            }else {
                echo "FAIL VALIDASI RUN";
            }
        }

模特：

public function add($newData){
        $this->db->insert("cobabarang", $newData);

        $nama = $newData["nama"];
        $id = $this->db->query("SELECT id FROM cobabarang WHERE nama = \"$nama\"");

        return $id->row_array();
    }

    public function add_images($newImage) {

        //$this->db->insert("cobagambar", $newImage);

        $id = $newImage["barangid"]["id"];
        $gambar = $newImage["gambar"];

        $this->db->query("INSERT INTO cobagambar(barangid, gambar1) VALUES($id, \"$gambar\")");
    }

Answer 1

这里有一个错误：

foreach($image as $img) 
{
    $barangImage = array(
         "gambar" => $img,
         "barangid" => $id
    );    
}

将$barangImage更改为$barangImage[]

当您将图像放入数据库时，我建议您使用json_encode($barangImage)，然后使用json_decode($images-json-string)时使用图片。

Answer 2

您的server-side-events循环

有问题

foreach

我的猜测是foreach($image as $img) { $barangImage = array( "gambar" => $img //might be img['img'] I guess $img is again an array...you hvae to check that "barangid" => $id //might be $img['id']check on this too..will be $img['id'] I guess ); }又是一个带有一些键的数组。你真的需要检查一下你可以像这样直接调用$img循环中的插入函数，

foreach

注意：数组foreach($image as $img) { $barangImage = array( "gambar1" => $img['img'], //I guess $img is again an array...you hvae to check that "barangid" => $img['id'] //check on this too..will be $img['id'] I guess ); $id = $this->testModel->add_images($barangImage)); }中的keys必须是表格中的列名。即

barangImage

只需更改gambar1 and barangid will be your column names. so you can directly use codeIgniter's active records.功能

即可

add_images

codeigniter将许多图像名称插入数据库

2 个答案: