图像不使用路径从数据库中获取

问题描述

我正在尝试使用路径从数据库上传和获取图像.上传过程完美.但是，我无法从数据库中获取图像.我试过 print_r($row['image']);.我得到这样的路径 C:/xampp/htdocs/xampp/htdocs/www/images/0d13808ad672c2713d306efbb0e42918.我不知道为什么这段代码没有从 db 中获取图像?

I'm trying to upload and fetch images from database using path. Upload process working perfectly. But, I cannot able to fetch image from db. I've tried print_r($row['image']);. I'm getting the path like this C:/xampp/htdocs/xampp/htdocs/www/images/0d13808ad672c2713d306efbb0e42918. I don't know why this code doesn't fetch image from db?

fetch.php

        <?php
            include('config.php');
            ini_set('display_startup_errors',1); a
            ini_set('display_errors',1);
            error_reporting(-1);

            try
            {
                  $stmt = $conn->prepare("SELECT * FROM imgdb WHERE id = 3");
                  $conn->errorInfo();
                  // $stmt->bindParam('1', $imgid, PDO::PARAM_INT);
                  $stmt->execute();

                  // $path = "/xampp/htdocs/www/images/";
                  // $imgpath = $_SERVER['DOCUMENT_ROOT'].$path;
                   while($row = $stmt->fetch(PDO::FETCH_ASSOC))
                    {
                        echo "<img src=".$row['image']." height='100' width='100'/>";
                        print_r($row['image']);
                    }
            }
            catch (PDOException $e)
            {
                 echo 'Database Error'.$e->getMessage();
            }
        ?>

上传.php

<?php

    ini_set('display_startup_errors',1);
    ini_set('display_errors',1);
    error_reporting(-1);

      include('config.php');

      if ($_FILES["file"]["error"] > 0)
      {
          echo "Error: " . $_FILES["file"]["error"] . "<br>";
      }
      else
      {
//          echo "Upload: " . $_FILES["file"]["name"] . "<br>";
//          echo "Type: " . $_FILES["file"]["type"] . "<br>";
//          echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
//          echo "Stored in: " . $_FILES["file"]["tmp_name"];
      }


      $filename  = basename($_FILES['file']['tmp_name']);
      $newname = md5($filename);

      $final_save_dir = '/xampp/htdocs/www/images/'.$newname ;
        if(move_uploaded_file($_FILES['file']['tmp_name'], $final_save_dir . $_FILES['file']['name']))
        {
            echo "Uploaded";
        }
        else
        {
           echo "File was not uploaded";
        }

      $imgid = $_SERVER['DOCUMENT_ROOT'].$final_save_dir;

      try
      {
          $stmt = $conn->prepare("INSERT INTO imgdb ( image ) VALUES ( ? ) ");
          $stmt->bindParam('1', $imgid, PDO::PARAM_STR);
          $conn->errorInfo();
          $stmt->execute();
      }
      catch (PDOException $e)
      {
          echo 'Database Error'.$e->getMessage();
      }
?>

推荐答案

试试这个并和你的比较一下，我不知道它是否有效，因为我没有测试过，但它应该.

Try this and compare with yours, I don't know if it works because I didn't tested, but it should.

上传.php

include('config.php');

if ($_FILES["file"]["error"] > 0 )
{
    echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
  else
{
    $filename  = basename($_FILES['file']['tmp_name']);
    $ext = pathinfo($_FILES["file"]["name"], PATHINFO_EXTENSION);

    $new_file_name = md5($filename).'.'.$ext;

    $final_save_dir = $_SERVER['DOCUMENT_ROOT'].DS.'www'.DS.'images'.DS;
    if(move_uploaded_file($_FILES['file']['tmp_name'], $final_save_dir . $new_file_name))
    {
        echo "Uploaded";
    }
    else
    {
       echo "File was not uploaded";
    }

    try
    {
        $stmt = $conn->prepare("INSERT INTO imgdb ( image ) VALUES ( ? ) ");
        $stmt->bindParam('1', $new_file_name, PDO::PARAM_STR);
        $conn->errorInfo();
        $stmt->execute();
    }
    catch (PDOException $e)
    {
      echo 'Database Error'.$e->getMessage();
    }
}

fetch.php

<?php
include('config.php');
ini_set('display_startup_errors',1);
ini_set('display_errors',1);
error_reporting(-1);

try
{
      $stmt = $conn->prepare("SELECT * FROM imgdb WHERE id = 3");
      $conn->errorInfo();
      $stmt->execute();

       while($row = $stmt->fetch(PDO::FETCH_ASSOC))
        {
            echo "<img src='images/".$row['image']."' height='100' width='100'/>";
        }
}
catch (PDOException $e)
{
     echo 'Database Error'.$e->getMessage();
}

这篇关于图像不使用路径从数据库中获取的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！