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イメージをアップロードしてmysql dbにパスを保存するコードの作成方法? 私は試みましたが、何も働いていません。イメージコードをアップロードしてパスをmysql dbに保存する方法
イメージをアップロードしてmysql dbにパスを保存するコードの作成方法? 私は試みましたが、何も働いていません。イメージコードをアップロードしてパスをmysql dbに保存する方法
1つの方法は、画像をアップロードしてサーバー上のフォルダに保存し、名前をmysqlデータベースに保存することです。ここでは一例です::
まず、アップロードするフォームを作成します::
//file.html
はUpload your file to the database...
<form action="upload.php" method="post" enctype="multipart/form-data" name="uploadform">
<input type="hidden" name="MAX_FILE_SIZE" value="350000">
<input name="picture" type="file" id="picture" size="50">
<input name="upload" type="submit" id="upload" value="Upload Picture!">
</form>
その後、我々は
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title> Upload Image </title>
<?php
// if something was posted, start the process...
if(isset($_POST['upload']))
{
// define the posted file into variables
$name = $_FILES['picture']['name'];
$tmp_name = $_FILES['picture']['tmp_name'];
$type = $_FILES['picture']['type'];
$size = $_FILES['picture']['size'];
// if your server has magic quotes turned off, add slashes manually
if(!get_magic_quotes_gpc()){
$name = addslashes($name);
}
// open up the file and extract the data/content from it
$extract = fopen($tmp_name, 'r');
$content = fread($extract, $size);
$content = addslashes($content);
fclose($extract);
// connect to the database
include "connect.php";
// the query that will add this to the database
$addfile = "INSERT INTO files (name, size, type, content) VALUES ('$name', '$size', '$type', '$content')";
mysql_query($addfile) or die(mysql_error());
if(!empty($_FILES))
{
$target = "upload/";
$target = $target . basename($_FILES['picture']['name']) ;
$ok=1;
$picture_size = $_FILES['picture']['size'];
$picture_type=$_FILES['picture']['type'];
//This is our size condition
if ($picture_size > 5000000)
{
echo "Your file is too large.<br>";
$ok=0;
}
//This is our limit file type condition
if ($picture_type =="text/php")
{
echo "No PHP files<br>";
$ok=0;
}
//Here we check that $ok was not set to 0 by an error
if ($ok==0)
{
Echo "Sorry your file was not uploaded";
}
//If everything is ok we try to upload it
else
{
if(move_uploaded_file($_FILES['picture']['tmp_name'], $target))
{
echo "The file ". basename($_FILES['picture']['name']). " has been uploaded <br/>";
}
else
{
echo "Sorry, there was a problem uploading your file.";
}
}
}
mysql_close();
echo "Successfully uploaded your picture!";
}else{die("No uploaded file present");
}
?>
</head>
<body>
<div align="center">
<img src="upload/<?php echo $name; ?>"
<br />
<a href="form.html">upload more images</a>
</div>
</body>
</html>
upload.php
を作成します//getpicture.php
** //最後にconnect.php **
google - > phpファイルのアップロード - > http://php.net/upload –