SELECT DB
mojete skazat pojalusta tovarishi specialisti kak mojno ispravit etot kod.ne rabotaet ne kak.ya novichok
spasibo za rane.
<?php
$username="5543";
$password="123456";
$database="5543";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$sql = mysql_query("SELECT * FROM links" ,$db);
$result=mysql_query($sql);
echo ("<table border ='1'>");
echo ("<tr><td>Category</td><td>E-mail</td><td>URL</td><td>Description</td></tr>");
while ($tablerows = mysql_fetch_row($sql))
{
echo("<tr><td><a href='$tablerows[0]'>$tablerows[1]</a></td><td>$tablerows[2]</td><td>$tablerows[3]</td><td>$tablerows[4]</td></tr> ");
}
echo "</table>";
mysql_close($db);
?>
while ($tablerows = mysql_fetch_row([COLOR="Red"]$sql[/COLOR]))
Должно быть так:
Код:
while ($tablerows = mysql_fetch_row($result))
while ($tablerows = mysql_fetch_row([COLOR="Red"]$sql[/COLOR]))
Должно быть так:
Код:
while ($tablerows = mysql_fetch_row($result))
spasibo drug no vse takje ne pokazivaet vot address
http://khaarsen.ueuo.com/list.php
Сделай так:
Код:
<?php
$username="5543";
$password="123456";
$database="5543";
$db = mysql_connect(localhost,$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$sql = "SELECT * FROM links";
$result=mysql_query($sql);
echo "<table border ='1'>";
echo "<tr><td>Category</td><td>E-mail</td><td>URL</td><td>Description</td></tr>";
while ($tablerows = mysql_fetch_row($result))
{
echo "<tr><td><a href='".$tablerows[0]."'>".$tablerows[1]."</a></td><td>".$tablerows[2]."</td><td>".$tablerows[3]."</td><td>".$tablerows[4]."</td></tr> ";
}
echo "</table>";
mysql_close($db);
?>
$username="5543";
$password="123456";
$database="5543";
$db = mysql_connect(localhost,$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$sql = "SELECT * FROM links";
$result=mysql_query($sql);
echo "<table border ='1'>";
echo "<tr><td>Category</td><td>E-mail</td><td>URL</td><td>Description</td></tr>";
while ($tablerows = mysql_fetch_row($result))
{
echo "<tr><td><a href='".$tablerows[0]."'>".$tablerows[1]."</a></td><td>".$tablerows[2]."</td><td>".$tablerows[3]."</td><td>".$tablerows[4]."</td></tr> ";
}
echo "</table>";
mysql_close($db);
?>
Сделай так:
Код:
<?php
$username="5543";
$password="123456";
$database="5543";
$db = mysql_connect(localhost,$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$sql = "SELECT * FROM links";
$result=mysql_query($sql);
echo "<table border ='1'>";
echo "<tr><td>Category</td><td>E-mail</td><td>URL</td><td>Description</td></tr>";
while ($tablerows = mysql_fetch_row($result))
{
echo "<tr><td><a href='".$tablerows[0]."'>".$tablerows[1]."</a></td><td>".$tablerows[2]."</td><td>".$tablerows[3]."</td><td>".$tablerows[4]."</td></tr> ";
}
echo "</table>";
mysql_close($db);
?>
$username="5543";
$password="123456";
$database="5543";
$db = mysql_connect(localhost,$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$sql = "SELECT * FROM links";
$result=mysql_query($sql);
echo "<table border ='1'>";
echo "<tr><td>Category</td><td>E-mail</td><td>URL</td><td>Description</td></tr>";
while ($tablerows = mysql_fetch_row($result))
{
echo "<tr><td><a href='".$tablerows[0]."'>".$tablerows[1]."</a></td><td>".$tablerows[2]."</td><td>".$tablerows[3]."</td><td>".$tablerows[4]."</td></tr> ";
}
echo "</table>";
mysql_close($db);
?>
brat spasibo bolshoe ti geni rabotaet uje vse ok.spasibo esho raz.prixadi v ispaniu tebe preglashayu
:D :D Оставляй отзыв =)