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PHP实现上传文件并存进数据库的方法

时间:2021-07-01 10:21:17 帮助过:32人阅读

本文实例讲述了PHP实现上传文件并存进数据库的方法。分享给大家供大家参考。具体如下:

show_add.php文件如下:

<?php     
  if(!isset($_REQUEST['id']) or $_REQUEST['id']=="") die("error: id none");
  $id = $_REQUEST['id'];
  //定位记录,读出
  $conn=mysql_connect("localhost","root","admin");
  if(!$conn) die("error: mysql connect failed");
  mysql_select_db("nokiapaymentplat",$conn);
  $sql = "select * from receive where id=$id";
  $result = mysql_query($sql,$conn);
  if(!$result) die("error: mysql query");
  $num=mysql_num_rows($result);
  if($num<1) die("error: no this recorder");
  $data = mysql_result($result,0,"file_data");
  $type = mysql_result($result,0,"file_type");
  $name = mysql_result($result,0,"file_name");
  mysql_close($conn);
  //先
输出相应的文件头,并且恢复原来的文件名 header("Content-type:$type"); header("Content-Disposition: attachment; filename=$name"); echo $data; ?>

show_info.php文件如下:

<?php 
   if(!isset($_REQUEST['id']) or $_REQUEST['id']=="") die("error: id none");
  $id = $_REQUEST['id'];
  //定位记录,读出
  $conn=mysql_connect("localhost","root","admin");
  if(!$conn) die("error: mysql connect failed");
  mysql_select_db("nokiapaymentplat",$conn);
  $sql = "select file_name ,file_size from receive where id=$id";
  $result = mysql_query($sql,$conn);
  if(!$result) die(" error: mysql query");
  //如果没有指定的记录,则报错
  $num=mysql_num_rows($result);
  if($num<1) die("error: no this recorder");
  //下面两句程序也可以这么写
  //$row=mysql_fetch_object($result);
  //$name=$row->name;
  //$size=$row->size;
  $name = mysql_result($result,0,"file_name");
  $size = mysql_result($result,0,"file_size");
  mysql_close($conn);
  echo "<hr>上传的文件的信息:";
  echo "<br>The file's name - $name";  
  echo "<br>The file's size - $size"; 
  echo "<br><a href=show_add.php?id=$id>附件</a>";
?>

submit.php文件如下:

<?php  
  if(is_uploaded_file($_FILES['myfile']['tmp_name'])) {
  //有了上传文件了 
  $myfile=$_FILES["myfile"];
    //设置超时限制时间,缺省时间为 30秒,设置为0时为不限时
    $time_limit=60;     
    set_time_limit($time_limit); //
    //把文件内容读到字符串中
    $fp=fopen($myfile['tmp_name'], "rb");
    if(!$fp) die("file open error");
    $file_data = addslashes(fread($fp, filesize($myfile['tmp_name'])));
    fclose($fp);
    unlink($myfile['tmp_name']); 
    //文件格式,名字,大小
    $file_type=$myfile["type"];
    $file_name=$myfile["name"];
    $file_size=$myfile["size"];
    die($file_type);
    //连接数据库,把文件存到数据库中
    $conn=mysql_connect("localhost","root","admin");
    if(!$conn) die("error : mysql connect failed");
    mysql_select_db("nokiapaymentplat",$conn);
    $sql="insert into receive 
    (file_data,file_type,file_name,file_size) 
    values ('$file_data','$file_type','$file_name',$file_size)";
    $result=mysql_query($sql,$conn);
    //下面这句取出了刚才的insert语句的id
    $id=mysql_insert_id();
    mysql_close($conn);
    set_time_limit(30); //恢复缺省超时设置 
    echo "上传成功--- ";
    echo "<a href='show_info.php?id=$id'>显示上传文件信息</a>";
  }  
  else {  
    echo "你没有上传任何文件";  
  }  
?>

upload.php文件如下:



<html>  
<head>  
<title>文件上传表单</title>  
</head>  
<body>  
<table>  
<form enctype='multipart/form-data' name='myform' action='submit.php' 
method='post'>  
<INPUT TYPE = "hidden" NAME = "MAX_FILE_SIZE" VALUE ="1000000">
<tr><td>选择上传文件</td><td>
<input name='myfile' type='file'></td></tr> 
<tr><td colspan='2'><input name='submit' value='上传'  type='submit'></td></tr>  
</table>  
</body>  
</html>

希望本文所述对大家的PHP程序设计有所帮助。

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