Autor Zpráva
designed
Profil *
#1 · Zasláno: 10. 8. 2006, 10:38:09
Odpovědět Citovat
Ahoj,

chtel bych poprosit o radu se skriptem php. Resim takovou situaci: potrebuji ukladat fotky do databaze MySQL. Ukladani mi funguje jak ma, ukladam do BLOB. Problem mam pri zobrazovani fotek z DB. Mam skript, kterym umim zobrazit fotky z pole BLOB, ale problem je v tom, ze se vechny nahledy zobrazuji se stejnym obsahem. Tzn. v DB mam napriklad tri fotky 1. hory, 2. jezero, 3. auto. Nyni mi skript zobrazi u nahledu vsech tri snimku napr. 3x hory nebo 3x jezero nebo 3x auto. Pokud kliknu na nahled tak se ale zobrazi velka fotka spravne. Takze problem je jen s nahledy. Spravne se ma zobrazit nahled 1x hory, 1x jezero, 1x auto.

Diky za radu.

Skripty:
- strukrura tabulky
- index.php
- image.php


CREATE TABLE image (
image_id int(10) unsigned NOT NULL auto_increment,
image_type varchar(50) NOT NULL default '',
image longblob NOT NULL,
image_size bigint(20) NOT NULL default '0',
image_name varchar(255) NOT NULL default '',
image_date datetime NOT NULL default '0000-00-00 00:00:00',
UNIQUE KEY image_id (image_id)
);


index.php
-------------

<?php
$conn = mysql_connect("mysql", "jmeno", "heslo") OR DIE (mysql_error());
@mysql_select_db ("nazev_databaze", $conn) OR DIE (mysql_error());

if ($_FILES) {
$image_types = Array ("image/bmp",
"image/jpeg",
"image/pjpeg",
"image/gif",
"image/x-png");

$userfile = addslashes (fread (fopen ($_FILES["userfile"]["tmp_name"], "r"), filesize ($_FILES["userfile"]["tmp_name"])));
$file_name = $_FILES["userfile"]["name"];
$file_size = $_FILES["userfile"]["size"];
$file_type = $_FILES["userfile"]["type"];

if (in_array (strtolower ($file_type), $image_types)) {
$sql = "INSERT INTO image (image_type, image, image_size, image_name, image_date) ";
$sql.= "VALUES (";
$sql.= "'{$file_type}', '{$userfile}', '{$file_size}', '{$file_name}', NOW())";
@mysql_query ($sql, $conn);
Header("Location:".$_SERVER["PHP_SELF"]);
exit();
}
}

if ($_GET) {
$iid = $_GET["iid"];
$act = $_GET["act"];
switch ($act) {
case rem:
$sql = "DELETE FROM image WHERE image_id=$iid";
@mysql_query ($sql, $conn);
Header("Location:./index.php");
exit();
break;
default:
print "<img src=\"image.php?iid=$iid\">";
break;
}
}

?>
<html>
<head>
<title>Foto</title>
</head>
<body>
<form method="post" enctype="multipart/form-data">
Select Image File: <input type="file" name="userfile" size="40"><input type="submit" value="submit">
</form>
<?php
$sql = "SELECT * FROM image ORDER BY image_date DESC";
$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";
$str .= "<img src=\"image.php?iid=$iid\" width=\"100\" height=\"72\"> ";
$str .= "<a href=\"index.php?iid=".$row["image_id"]."\">".$row["image_name"]."</a> ";
$str .= "[".$row["image_date"]."] ";
$str .= "[".$row["image_size"]."] ";
$str .= "[<a href=\"index.php?act=rem&iid=".$row["image_id"]."\">Remove</a>]<br>";
}
print $str;
}
?>
</body>
</html>


image.php
-------------

<?php
$conn = mysql_connect("mysql", "jmeno", "heslo") OR DIE (mysql_error());
@mysql_select_db ("nazev_databaze", $conn) OR DIE (mysql_error());

$sql = "SELECT * FROM image WHERE image_id=".$_GET["iid"];
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
$row = @mysql_fetch_array ($result);
$image_type = $row["image_type"];
$image = $row["image"];
Header ("Content-type: $image_type");
print $image;
}
?>
WertriK
Profil
#2 · Zasláno: 10. 8. 2006, 12:25:31
Odpovědět Citovat
Můj horký tip

$str .= "<img src=\"image.php?iid=$iid\" width=\"100\" height=\"72\"> ";

nahradit tímto :

$str .= "<img src=\"image.php?iid=".$row["image_id"]."\" width=\"100\" height=\"72\"> ";
Časová prodleva: 20 let
Toto téma je uzamčeno. Odpověď nelze zaslat.

0