Проблема е, че когато са къчени примерно снимки с имена 1.jpg, 2.jpg, 3.jpg, <а hrefa..> се задейства от 2.jpg, т.е. 1.jpg няма линк към по-голямата картина, а пък линка от 2.jpg сочи към картина 1.jpg.
Ето и скриптовете:
index.php
<?php
$target = "images/";
$target = $target . basename( $_FILES['photo']['name']);
$pic=($_FILES['photo']['name']);
mysql_connect("localhost", "root", "pass") or die(mysql_error()) ;
mysql_select_db("test") or die(mysql_error()) ;
mysql_query("INSERT INTO `employees` VALUES ( '$pic')") ;
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " kacheno e ";
}
else {
echo "error";
}
?>
<form enctype="multipart/form-data" action="index.php" method="POST">
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Add">
</form>
Файла, с който се гледат кртинките:
<?php
mysql_connect("localhost", "root", "pass") or die(mysql_error()) ;
mysql_select_db("test") or die(mysql_error()) ;
$data = mysql_query("SELECT * FROM employees") or die(mysql_error());
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
Echo "<ul><li><img src=http://localhost/picture/images/".$info['photo'] ." width=200><a href=http://localhost/picture/images/".$info['photo'] ."></li></ul>";
}
?>
sql-a:
CREATE TABLE employees (photo VARCHAR(30))
Ето и скриптовете:
index.php
<?php
$target = "images/";
$target = $target . basename( $_FILES['photo']['name']);
$pic=($_FILES['photo']['name']);
mysql_connect("localhost", "root", "pass") or die(mysql_error()) ;
mysql_select_db("test") or die(mysql_error()) ;
mysql_query("INSERT INTO `employees` VALUES ( '$pic')") ;
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " kacheno e ";
}
else {
echo "error";
}
?>
<form enctype="multipart/form-data" action="index.php" method="POST">
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Add">
</form>
Файла, с който се гледат кртинките:
<?php
mysql_connect("localhost", "root", "pass") or die(mysql_error()) ;
mysql_select_db("test") or die(mysql_error()) ;
$data = mysql_query("SELECT * FROM employees") or die(mysql_error());
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
Echo "<ul><li><img src=http://localhost/picture/images/".$info['photo'] ." width=200><a href=http://localhost/picture/images/".$info['photo'] ."></li></ul>";
}
?>
sql-a:
CREATE TABLE employees (photo VARCHAR(30))