Question

我目前正在开发一个网站，需要上传其用户不同产品的图片。我通过php使用MySql数据库来实现它。

我从用户那里获取输入的基本表单的代码是：

<form enctype="multipart/form-data" action="testimage1.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>

我的数据库表是：

 mysql> CREATE TABLE tbl_images (
 > id tinyint(3) unsigned NOT NULL auto_increment,
 > image blob NOT NULL,
 > PRIMARY KEY (id)
 > );

testimage1.php具有以下代码： -

 $username = "root";
 $password = "";
 $host = "localhost";
 $database = "thinstrokes";


 $link = mysql_connect($host, $username, $password);
 if (!$link) {
 die('Could not connect: ' . mysql_error());
 }

// Select your database
mysql_select_db ($database);

    if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

  // Temporary file name stored on the server
  $tmpName  = $_FILES['image']['tmp_name'];  

  // Read the file 
  $fp      = fopen($tmpName, 'r');
  $data = fread($fp, filesize($tmpName));
  $data = addslashes($data);
  fclose($fp);


  // Create the query and insert
  // into our database.
  $query = "INSERT INTO tbl_images ";
  $query .= "(image) VALUES ('$data')";
  $results = mysql_query($query, $link) or die(mysql_error());

  // Print results
  print "Thank you, your file has been uploaded.";

  }
  else {
  print "No image selected/uploaded";
  }

在提交表单时，我收到错误：No image selected/uploaded

我没有收到错误......我之前已经问过这个问题：

但到目前为止，我还没有成功将图像存储在数据库中。

Answer 1

你的脚本运行得很好，这就是我测试的内容：

<?

if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

  // Temporary file name stored on the server
  $tmpName  = $_FILES['image']['tmp_name'];  

  // Read the file 
  $fp      = fopen($tmpName, 'r');
  $data = fread($fp, filesize($tmpName));
  $data = addslashes($data);
  fclose($fp);


  // Create the query and insert

  // Print results
  print "Thank you, your file has been uploaded.";

  }
  else {
  print "No image selected/uploaded";
  }

?>

<form enctype="multipart/form-data" action="" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>

它工作得很好，如果你想看到它在行动我可以发给你链接。

它必须是破坏你的代码的其他东西（*请注意，我删除了数据库查询以避免出现mysql错误，但脚本在那里工作时也是如此。

Answer 2

//这是您问题的解决方案

<form enctype="multipart/form-data" action="" method="post" name="changer">
    <input name="MAX_FILE_SIZE" value="102400" type="hidden">
    <input name="image" accept="image/jpeg" type="file">
    <input value="Submit" type="submit">
</form>
<?php

    // connection to database
    include 'includes/connection.php';
?>

<?php

    if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 
        // Temporary file name stored on the server
        $tmpName  = $_FILES['image']['tmp_name'];  
        // Read the file 
        $fp      = fopen($tmpName, 'r');
        $data = fread($fp, filesize($tmpName));
        $data = addslashes($data);
        fclose($fp);
        $result = mysql_query("INSERT INTO image (image)VALUES ( '$data')", $connection);
        if(!$result)
        {
            die("Database query failed: ". mysql_error());
        }
        // Print results

        print "Thank you, your file has been uploaded.";
    }
    else 
    {
        print "No image selected/uploaded";
    }
?>

<?php
    //close connection
    include 'includes/close.php';
?>

通过php在MySql数据库中上传图像

2 个答案: