2016-02-22 140 views
0

我試圖在頁面中顯示我上傳的圖像。我使用dropzone上傳圖片。使用codeigniter從上傳文件夾中顯示圖像php

控制器:

public function do_uploadlogo($id){ 
$restologo=$_POST['restologo']; 
    $tempFile = $_FILES['file']['tmp_name']; 
    $fileName = $_FILES['file']['name']; 
    $targetPath = getcwd() . '/uploads/restaurants/logos/'; 
    $targetFile = $targetPath . $fileName ; 
    $data = array(
     'restologo' => 'uploads/restaurants/logos/'.$fileName 
    ); 
    $this->load->database('findiningcebu'); 
    $this->db->where('id', $id); 
    $this->db->update('restaurants',$data); 
    move_uploaded_file($tempFile, $targetFile); 
} 
public function addrestologolink($id){ 
    $data = array(

      "action" => base_url('/index.php/AdminController/do_uploadlogo/'.$id), 
      "data" => $this->db->get_where('restaurants',array('id'=>$id)) 
    ); 

    $this->load->view('admin/logo',$data); 
} 

查看

<div class="del" style="padding-left: 15px">Current Logo:</div> 
<div class="del" style="padding-left: 15px"><img src="<?=base_url().$restologo?>" /></div> 

請幫幫忙!

+0

哪裏是restologo變量addrestologolink方法您的數據數組中?我認爲你需要通過restologo變量。 – umefarooq

+0

您是否在控制檯中看到錯誤,或圖片是否顯示? –

+0

我沒有收到任何錯誤。圖像不會顯示。該怎麼辦? – QueenR

回答

1

使用此代碼

<div class="del" style="padding-left: 15px"> 
<?php 
echo'<img src="' . base_url().'uploads/restaurants/logos/' . $restologo . '">'; 
?> 
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