Ето кода
Като добавя коментар в бд не записва ид на новината Защо?
А при извеждане дава това
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/u873363251/public_html/mem.php on line 180
PHP:
<?php
$per_page = 1;
$start = $_GET['start'];
$max_pages = ceil($foundnum / $per_page);
if(!$start)
$start=0;
ob_start();
include("config.php");
include_once('bbcode.php');
if (!$_GET[ime])
{
$getime = mysql_query("SELECT * from news order by id desc LIMIT $start, $per_page");
while ($ime = mysql_fetch_array($getime))
{
echo (" <a href=\"mem.php?ime=$ime[ime]\">$ime[ime]</a>\n");
}
}
ELSE
{
$getime = mysql_query("SELECT * from news where ime = '$_GET[ime]'");
$imenum = mysql_num_rows($getime);
if ($imenum == ime)
{
echo "no new";
}
else
{
$profile = mysql_fetch_array($getime);
}
}
echo('<div id="container">
<div id="header">
<header>uuu</header>
</div>
<div class="style1" id="bar">
<div id="r"> <input type="text" id="search" size="15" maxlength="25" name=""><input type="submit" name="s" id="btnsearch" value=" " /></div>
<div id="apDiv2"><ul id="MenuBar1" class="MenuBarHorizontal"> <li><a class="MenuBarItemSubmenu" href="#">Начало</a> </li> <li><a class="MenuBarItemSubmenu" href="#">Item2</a> <ul> <li><a class="MenuBarItemSubmenu" href="#">рингови инв колички</a></li> <li><a class="MenuBarItemSubmenu" href="#">Активни инв колички</a></li> <li><a class="MenuBarItemSubmenu" href="#">Item 1.3</a></li> </ul> </li> <li><a class="MenuBarItemSubmenu" href="#">Item 3</a> <ul> <li><a class="MenuBarItemSubmenu" href="#">Item 3.1</a> <ul> <li><a href="#">Item 3.1.1</a></li> <li><a href="#">Item 3.1.2</a></li> </ul> </li> <li><a class="MenuBarItemSubmenu" href="#">Item 3.2</a></li> <li><a class="MenuBarItemSubmenu" href="#">Item 3.3</a></li> </ul> </li> <li><a href="#">Item 4</a></li> </ul>
</div></div>
<div id="right">
<div id="right1"> </div>
<div id="right2"> </div>
</div> <div id="sdrjanie">');
echo("<br><center><b>$profile[ime]</b><br>");
echo bbcode($profile['nowina']);
echo("avtor ".$profile['awtor']." <br/>");
echo("
<center><form method=\"POST\">
<table width=\"100%\">
<input type=\"hidden\" name=\"id\" value=\"".$row['id']."\" />
<tr>
<td align=\"right\" width=\"25%\">
ime
</td>
<td align=\"left\">
<input size=\"25\" name=\"ime\"></td>
</tr>
<tr>
<td align=\"right\" width=\"25%\">
komentar</td>
<td align=\"left\">
<input size=\"25\" name=\"comentar\"\<textarea/></td>
</tr>
<td align=\"center\">
</td>
<td align=\"left\">
<input type=\"submit\" name=\"izprati\" value=\"Обнови\"></td>
</tr>
</table>
</form>
</center>");
$our = mysql_query("SELECT * FROM comments WHERE id=$id ORDER BY id_com DESC");
while($row=mysql_fetch_array($our)){
echo "<table width='300' align='center'>
<tr><td><b>Автор: <font color='red'>$row[ime]</font></b></td></tr>
<tr><td align='center'><b>$row[comentar]</b></td></tr>
<tr><td align='right'><b>Дата: <font color='red'>$row[data]</font><hr/></b></td></tr>
</table>";
}
echo('</div> <div id="footer"> <br/><footer>
uuuuuuu
</footer> </div> </div>');
?>
<?php
if(isset($_POST['izprati'])){
$ime = $_POST['ime'];
$ime = htmlspecialchars($ime);
$ime = strip_tags($ime);
$id = $_POST['id'];
$comentar = $_POST['comentar'];
$comentar = htmlspecialchars($comentar);
$comentar = strip_tags($comentar);
if($ime== NULL){
echo "Попълнете име";
exit;
}
elseif($comentar==NULL){
echo "Попълнете коментара ";
exit;
}
else {
$sql = "INSERT INTO comments (ime, comentar, data, ip, id) VALUES ('$ime', '$comentar', '$data', '$ip', '$id')";
mysql_query($sql) or die (mysql_error());
header("LOCATION: mem.php?id=$id");
}
}
?>
<script type="text/javascript">
</script>
</body>
</html>Като добавя коментар в бд не записва ид на новината Защо?
А при извеждане дава това
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/u873363251/public_html/mem.php on line 180
