Ne mogu da upload-ujem fotku u folder

marecare

Active member
Joined
May 9, 2014
Messages
1,436
Reaction score
6
Drugari, pravim u PHP-u aplikaciju za vesti i u okviru svake vesti trebao bih da imam fotku.
Na Udemy-ju sam kupio neki kurs u kojem je prikazano kako se upload-uje ubacuje fotka u mysql bazu i upload-uje u folder u okviru projekta. Kada sam pratio taj kurs sve je radilo a sada kada sam pokusao to da implementiram u svoj zadataka, ne radi.
Kada kliknem na dugme “Submit” svi podaci iz forme se upisuju u bazu, cak i fotografija, i ne prijavljuje mi nikakvu gresku:
View attachment 2460

Medjutim fajl se nikako ne pojavljuje u folderu koji sam zadao kao putanju.
Folderu sam dodelio sva prava naredbom: sudo chmod -R 777 uploads

Ne znam sta jos moze da mi pravi problem!!?
Ima li ko ideju?!

Unapred zahvalan.
 
Last edited:

gagiD

Member
Joined
Apr 1, 2014
Messages
267
Reaction score
0
View attachment 2461

😃
btw. svaka vijest ima unique ID tako da se slike mogu spremati kao slika-{ID}, tako da je kolona image viska ako mene pitas.
 
Last edited:

marecare

Active member
Joined
May 9, 2014
Messages
1,436
Reaction score
6
PHP:
<?php
include ‘includes/header.php’;

//Create DB Object
$db = new Database();

if(isset($_POST[‘submit’])){
//Assign Vars
$title = mysqli_real_escape_string($db->link, $_POST[‘title’]);
$body = mysqli_real_escape_string($db->link, $_POST[‘body’]);
$category = mysqli_real_escape_string($db->link, $_POST[‘category’]);
$author = mysqli_real_escape_string($db->link, $_POST[‘author’]);

$image = $_FILES[‘image’][‘name’];
$size = $_FILES[‘image’][‘size’];
$type = $_FILES[‘image’][‘type’];
$temp = $_FILES[‘image’][‘tmp_name’];

//Validation
if($title == ‘’ || $body == ‘’ || $category == ‘’ || $author == ‘’){
$error = ‘Neophodno je popuniniti polja: Title, Body, Category i Author’;
echo $error;
}else{
$query = “INSERT INTO news
(title, body, category, author, image)
VALUES(’$title’, ‘$body’, $category, ‘$author’, ‘$image’)”;

if($insert_row = $db->insert($query)){
move_uploaded_file($temp, “…/uploads/”.$image);
}
}
}

//Create Query
$query = “SELECT * FROM categories”;
//Run Query
$categories = $db->select($query);

?>
Article Title
Article Body
Category <?php while($row = $categories->fetch_assoc()){ if($row['id'] == $article['category']){ $selected = 'selected'; }else{ $selected = ''; } ?> value="<?php echo $row['id']; ?>"> <?php echo $row['name']; ?> <?php } ?>
Author
Image input
Cancel

<?php include 'includes/footer.php'; ?>
 
Last edited:

gagiD

Member
Joined
Apr 1, 2014
Messages
267
Reaction score
0
PHP:
move_uploaded_file($temp, “…/uploads/”.$image);
Destinacija ne valja.
A funkcija vraca TRUE ako se uspjesno izvrsi, tako da uvijek testiraj takve stvari.
 
Last edited:

gagiD

Member
Joined
Apr 1, 2014
Messages
267
Reaction score
0
Da moze i relative, zeznuh se.

Testiras:

PHP:
if (move_uploaded_file($temp, “…/uploads/”.$image))
echo ‘Kopirano.’;
else
echo ‘Nesto ne valja’;
 
Last edited:

gagiD

Member
Joined
Apr 1, 2014
Messages
267
Reaction score
0
Evo npr. ovo ce raditi:

index.html
HTML:
add.php
PHP:
<?php

if (isset($_POST[‘submit’])) {
[CODE]$imageName = $_FILES['image']['name'];
$tmpImage = $_FILES['image']['tmp_name'];


if (move_uploaded_file($tmpImage, "uploads/".$imageName))
    echo 'Kopirano.';
else
    echo 'Nesto ne valja';
[/CODE]
}
?>

A sad kod sebe moras naci gdje “pada”, mozda ovaj $db->insert($query) poziv vrati nulu i sl.
 
Last edited:
Top