napravio sam stranicu gde mozete editovati neki post, medjutim javlja se greska kada ga pokrenem.
Ovako izgleda strana gde se ispisuju svi linkovi i do njih pise edit i delete
ako kliknem edit ono predje u stranicu za unosenje vrednosti i popuni vrednosti onog linka ciji je ID u ovom slucaju row
Code:
$conn=db_connect();
$sql="select * from linkovi order by id desc";
$result=mysql_query($sql);
print '<table>';
while($row=mysql_fetch_array($result)){
print '<tr><td style="border:1px dotted silver">';
print '[<a href="'.$row['adress'].'">'.$row['adress'].'</a>]';
print '</td>';
print '<td>';
print $row['about'];
print '</td><td>';
print '[<a href="dellink.php?row='.$row['id'].'">brisi</a>]';
print '</td><td>';
print '[<a href="unesi.php?row='.$row['id'].'">edit</a>]';
print '</td></tr>';
}
echo '</table>';
$conn=db_connect();
$sql="select * from linkovi order by id desc";
$result=mysql_query($sql);
print '<table>';
while($row=mysql_fetch_array($result)){
print '<tr><td style="border:1px dotted silver">';
print '[<a href="'.$row['adress'].'">'.$row['adress'].'</a>]';
print '</td>';
print '<td>';
print $row['about'];
print '</td><td>';
print '[<a href="dellink.php?row='.$row['id'].'">brisi</a>]';
print '</td><td>';
print '[<a href="unesi.php?row='.$row['id'].'">edit</a>]';
print '</td></tr>';
}
echo '</table>';
kada editujem kako hocu kliknem na unos i ono prebaci na sledecu stranu
Code:
$adresa=addslashes($adresa);
$title=addslashes($title);
$about=addslashes($about);
$category=addslashes($category);
$conn=db_connect();
if(!$conn)
{
echo 'Greskica';
}
if (isset($HTTP_POST_VARS['row']) && $HTTP_POST_VARS['row']!='')
{ // It's an update
$row = $HTTP_POST_VARS['row'];
$sql = "update linkovi
set title = '$title',
adress = '$adresa',
about = '$about',
category = '$category',
where id = '$row'";
}
else { // It's a new story
$sql="insert into linkovi (title, adress, about, category) values
('".$title."','".$adresa."','".$about."','".$category."')";
}
$result = mysql_query($sql, $conn);
if (!$result) {
print "There was a database error when executing <pre>$sql</pre>";
print mysql_error();
exit;
}
$adresa=addslashes($adresa);
$title=addslashes($title);
$about=addslashes($about);
$category=addslashes($category);
$conn=db_connect();
if(!$conn)
{
echo 'Greskica';
}
if (isset($HTTP_POST_VARS['row']) && $HTTP_POST_VARS['row']!='')
{ // It's an update
$row = $HTTP_POST_VARS['row'];
$sql = "update linkovi
set title = '$title',
adress = '$adresa',
about = '$about',
category = '$category',
where id = '$row'";
}
else { // It's a new story
$sql="insert into linkovi (title, adress, about, category) values
('".$title."','".$adresa."','".$about."','".$category."')";
}
$result = mysql_query($sql, $conn);
if (!$result) {
print "There was a database error when executing <pre>$sql</pre>";
print mysql_error();
exit;
}
Ukoliko vidi da je postojeci id onda samo updatuje taj unos, a ako vidi da ga nema onda upise novu.
E sada sve je to teoretski tacno medjutim javlja mi se ovo
Code:
There was a database error when executing
update linkovi
set title = 'Auto Pijac',
adress = 'http://www.autopijac.com',
about = 'Auto Pijacasc',
category = 'Politicki',
where id = '20'
You have an error in your SQL syntax near 'where id = '20'' at line 6
There was a database error when executing
update linkovi
set title = 'Auto Pijac',
adress = 'http://www.autopijac.com',
about = 'Auto Pijacasc',
category = 'Politicki',
where id = '20'
You have an error in your SQL syntax near 'where id = '20'' at line 6
U cemu je problem