Question

我设法将上传的照片插入到数据库以及指定的路径中，问题是我上传的照片没有插入到日志记录用户列而是创建了另一列，我错过了什么？下面是我的代码和截图，以防你们几乎不明白我的意思

MODEL

function create($data){
    $query = $this->db->insert('dosen', $data);
    return $query;
}

CONTROLLER

public function Upload(){
    $upload = $this->input->post('fotoDsn');
    //Foto Set
    $photoName = gmdate("d-m-y-H-i-s", time()+3600*7).".jpg";
    $config['upload_path'] = './assets/';
    $config['allowed_types'] = 'gif||jpg||png';
    $config['max_size'] = '2048000';
    $config['file_name'] = $photoName;
    $this->load->library('upload',$config);
    if($this->upload->do_upload('userfile')){           
        $upload = 1;
    }
    else{
        $upload = 2;
    }
    if($upload==1){
        $data   = array(
                'foto_dosen'=>$photoName);
        $insert = $this->MDosen->create($data);
        if($insert){
            echo 1;
        }else{
            echo 2;
        }
    }//else kalo gagal
}

查看

<a href="#modalUpload" class="upload waves-effect waves-light modal-trigger" id="<?php echo $result->id ?>"><i class="material-icons tiny">assignment_ind</i>&nbsp;Upload</a>    

<div id="modalUpload" class="modal" style="width: 40%; height: auto">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
             <i class="medium material-icons prefix">assignment_ind</i>
                <h4 class="modal-title"> Upload Foto</h4>
            </div>
            <div class="modal-body">
                <form action="upload" id="tambahFormUpload" method="post" enctype="multipart/form-data">
                    <div class="file-field input-field">
                        <div class="btn">
                            <span>File</span>
                            <input type="file" name="userfile">
                        </div>
                        <div class="file-path-wrapper">
                            <input class="file-path validate" type="text">
                        </div>
                    </div>
                    <div class="modal-footer">
                        <button type="submit" class="btn btn-primary" >Simpan</button>
                    </div>  
                </form>
            </div>
        </div>  
    </div>
</div>

JS

$('.upload').click(function(){
        id1 = $(this).attr('id');
    });
    $('#tambahFormUpload').submit(function(e){
        alert(id1);
        e.preventDefault();
        var formData = new FormData($(this)[0]);
        $.ajax({
            url:'Profil/Upload/'+id1,
            data:formData,
            type:'POST',
            contentType: false,
            processData: false,
            success:function(data){
                $("#modalUpload").hide();
                window.location.reload(true);
            }
        });
    });

Answer 1

基于@Sean注释，确实使用public JSONObject makeHttpRequest2(String url2 , String method, String name, String password) throws IOException { // Making HTTP request try { // check for request method if (method == "POST") { // request method is POST url = new URL(url2); conn = (HttpURLConnection) url.openConnection(); conn.setDoOutput(true); conn.setRequestMethod("POST"); conn.setUseCaches(false); conn.setConnectTimeout(10000); conn.setReadTimeout(10000); conn.setRequestProperty("Content-Type", "application/json"); conn.setRequestProperty("Host", "android.schoolportal.gr"); conn.connect(); JSONObject jsonParam = new JSONObject(); jsonParam.put("name", name); //jsonParam.put("email", email); jsonParam.put("password", password); Log.d("json",String.valueOf(jsonParam)); OutputStreamWriter out=new OutputStreamWriter( conn.getOutputStream()); out.write(jsonParam.toString()); out.close(); } } catch (UnsupportedEncodingException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } catch (JSONException e) { e.printStackTrace(); } try { InputStream is = conn.getInputStream(); BufferedReader reader = new BufferedReader(new InputStreamReader( is, "UTF-8"), 8); StringBuilder sb = new StringBuilder(); String line = null; while ((line = reader.readLine()) != null) { sb.append(line + "\n"); } is.close(); json = sb.toString(); Log.e("JSON", json); } catch (Exception e) { Log.e("Buffer Error", "Error converting result " + e.toString()); } // try parse the string to a JSON object try { jObj = new JSONObject(json); } catch (JSONException e) { Log.e("JSON Parser", "Error parsing data " + e.toString()); } Log.d("json3",String.valueOf(json)); // return JSON String return jObj; }，我们应该使用insert，所以正确的控制器将使用update模型，而不是{{1然后把参数放在我的函数上，第二个虽然没有插入正确的列，因为无法找到用户的id（update子句），所以create我们可以设置where子句并将其指向update where，希望此帮助

<强> CONTROLLER

user

<强> MODEL

id

上传照片codeigniter不插入右列

1 个答案: