Kaip padaryti kad php arba html atvaizduotu visas nuotraukas esancias folderyje viena faile thumb ?

Kaip padaryti kad php arba html atvaizduotu visas nuotraukas esancias folderyje viename faile thumb ? arba gal zinot koki nebloga upload scripta su tokia funkcija kad ikelus nuotrauka ji iskart atsiranda html faile ? bet nereikia tu gigantisku scriptu ... reikia kazko labai paprasto :rolleyes:

<?php
if ($handle = opendir('.')) {
while (false !== ($file = readdir($handle))) {
	if ($file != "." && $file != "..") {
		echo "<img src='$file' />";
	}
}
closedir($handle);
}
?>

<?php
if ($handle = opendir('.')) {
while (false !== ($file = readdir($handle))) {
	if ($file != "." && $file != "..") {
		echo "<img src='$file' />";
	}
}
closedir($handle);
}
?>

tai cia skaitis failus is to folderio kuriame bus php failas ? ar yra kur nurodyti folderi ?

Arba geriau darom taip cia paprastas upload scriptas

<?php
$allowed = array("jpeg","gif","png","bmp","jpg");  
?>
<form action="" method="post" enctype="multipart/form-data">
Choose file to upload <input name="upload" type="file">
<input name="sb" id="sb" value="go" type="submit">
</form>
<?php
if(isset($_POST['sb'])) {
//check for valid extension
$pathInfo = pathinfo($_FILES["upload"]["name"]);
$extension = $pathInfo['extension'];
//choose directory/foolder to place the file in
$dir = "uploadFolder";
if(!in_array($extension, $allowed)) die("Extension not allowed!");
if(move_uploaded_file($_FILES['upload']['tmp_name'], "$dir/".$_FILES['upload']['name'])) {
print "Your new file can be viewed/download at <input name=\"newfile\" value=\"http://$_SERVER[HTTP_HOST]/$dir/".$_FILES['upload']['name']."\" type=\"text\" style=\"width: 300px\">";
}else{
print "File could not be uploaded";
}

$DatabaseHost = "localhost";
$DatabaseUsername = "xxxxx";
$DatabasePassword = "xxxxx";
$DatabaseName = "xxxxx";

$connection = mysql_connect($DatabaseHost, $DatabaseUsername, $DatabasePassword) 
or die("Cannot connect to MySQL!");
mysql_select_db($DatabaseName, $connection) or die("Cannot find database!");

$sql = "insert into files values (null, '".htmlentities($_FILES['upload']['name'])."', 'now()', '$_SERVER[REMOTE_ADDR]')";
$rs = mysql_query($sql);

print mysql_error();
}
?>
<br/>

o cia my sql

CREATE TABLE `files` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`filename` VARCHAR( 255 ) NOT NULL ,
`timestamp` INT NOT NULL ,
`ip` VARCHAR( 50 ) NOT NULL
) ENGINE = MYISAM;

gal tada tiesiog kad tiesiai is mysql atvaizduotu height="100px" thumbnails puslapyje, kaip tai padaryt ?

Arba geriau darom taip cia paprastas upload scriptas

<?php
$allowed = array("jpeg","gif","png","bmp","jpg");  
?>
<form action="" method="post" enctype="multipart/form-data">
Choose file to upload <input name="upload" type="file">
<input name="sb" id="sb" value="go" type="submit">
</form>
<?php
if(isset($_POST['sb'])) {
//check for valid extension
$pathInfo = pathinfo($_FILES["upload"]["name"]);
$extension = $pathInfo['extension'];
//choose directory/foolder to place the file in
$dir = "uploadFolder";
if(!in_array($extension, $allowed)) die("Extension not allowed!");
if(move_uploaded_file($_FILES['upload']['tmp_name'], "$dir/".$_FILES['upload']['name'])) {
print "Your new file can be viewed/download at <input name=\"newfile\" value=\"http://$_SERVER[HTTP_HOST]/$dir/".$_FILES['upload']['name']."\" type=\"text\" style=\"width: 300px\">";
}else{
print "File could not be uploaded";
}

$DatabaseHost = "localhost";
$DatabaseUsername = "xxxxx";
$DatabasePassword = "xxxxx";
$DatabaseName = "xxxxx";

$connection = mysql_connect($DatabaseHost, $DatabaseUsername, $DatabasePassword) 
or die("Cannot connect to MySQL!");
mysql_select_db($DatabaseName, $connection) or die("Cannot find database!");

$sql = "insert into files values (null, '".htmlentities($_FILES['upload']['name'])."', 'now()', '$_SERVER[REMOTE_ADDR]')";
$rs = mysql_query($sql);

print mysql_error();
}
?>
<br/>

 

o cia my sql

 

CREATE TABLE `files` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`filename` VARCHAR( 255 ) NOT NULL ,
`timestamp` INT NOT NULL ,
`ip` VARCHAR( 50 ) NOT NULL
) ENGINE = MYISAM;

 

gal tada tiesiog kad tiesiai is mysql atvaizduotu height="100px" thumbnails puslapyje, kaip tai padaryt ?

Jei uzdesi 100px man vistiek tures pilnas nuotraukas parsiusti, o ne thumbnail'a iki 100kb, i toki puslapi niekada nebegrysciau