我想在 @L_406_0@文件中显示Postgresql@R_301_2489@中的图像.
在尝试了很多不同的可能性后,我不会走得太远.
这是我做的：

上传：

echo "<div>".
    "<table border=\"1\" cellpadding=\"3\">".
       "<form action=\"test2.PHP\" method=\"post\" enctyPE=\"multipart/form-data\">".
         "<tr>".
           "<td>".
             "<input type=\"file\" name=\"file\" id=\"file\">".
             "<input class=\"button\" type=\"submIT\" name=\"submit\" value=\"Submit\">".
           "<tr>".
         "</tr>".
      "</form>".
     "</table>".
   "</div>";

显示：

if(isset($_FILES['file'])) //file uploaded
{
  $file_raw = file_get_contents($_FILES['file']['tmp_name']);
  $file     = pg_escape_bytea($file_raw);
  //Here is Code to insert into Database but just to test,I try it directly:

  header('Content-type: image/png');
  echo pg_unescape_bytea($file);
}

我已将显示部分放在一个额外的文件中,依此类推,但这些是您需要了解的基本信息.

我不会显示任何图像,只是浏览器中的“broken Image”图标.
这有什么不对？我怎么解决这个问题？
谢谢！

<?PHP 
// Connect to the database
$dbconn = pg_connect( 'dbname=foo' );

// Read in a binary file
$data = file_get_contents( 'image1.jpg' );

// Escape the binary data to avoid PRoblems with encoding
$escaped = bin2hex( $data );

// Insert it into the database
pg_query( "INSERT INTO gallery (name,data) VALUES ('Pine trees',decode('{$escaped}','hex'))" );

// Get the bytea data
$res = pg_query("SELECT encode(data,'base64') AS data From gallery WHERE name='Pine trees'");  
$raw = pg_fetch_result($res,'data');

// Convert to binary and send to the browser
header('Content-type: image/jpeg');
echo base64_decode($raw);
?>