mainom
gavo reakciją nuo
Plx
Kaip padaryti kad php arba html atvaizduotu visas nuotraukas esancias folderyje viena faile thumb ?
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 ?
mainom gavo reakciją nuo Plx Kaip padaryti kad php arba html atvaizduotu visas nuotraukas esancias folderyje viena faile thumb ?