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Posted by Luigi Donatello Asero on 07/12/05 15:52
"Richard Conway" <ric@shoe.com> skrev i meddelandet
news:NOOAe.4096$2I4.2995@fe05.ams...
> Luigi Donatello Asero wrote:
> > "Richard Conway" <ric@shoe.com> skrev i meddelandet
> > news:FaOAe.2$HQ1.0@fe08.ams...
> In that case, there is something wrong with your query. Try changing
> the following line from:
>
> $result = mysql_query("SELECT * FROM name of the table", $db);
>
> to:
>
> $result = mysql_query("SELECT * FROM name of the table", $db) or
> die(mysql_error());
>
> This will trap the error and output it for you so you can get some idea
> of where you've gone wrong.
After some changes:
<html> <body>
<?php $db = mysql_connect("localhost", "user",
"password");
mysql_select_db("databas", $db);
$result = mysql_query("SELECT * FROM table", $db) or
die(mysql_error());
echo "<table border=1>\n"; echo "<tr><th>Region</th>
<th>Ort</th><th>Sovplatser</th><th>Rum</th><th>Avstεnd till havet i
km</th><th>Terrass</th></tr>\n";
while ($mycolumn=mysql_fetch_array ($result))
{ printf("<tr><td>%s
%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td></tr>\n",
$mycolumn[1], $mycolumn[2], $mycolumn[3]
$mycolumn[4],$mycolumn[5],$mycolumn[6]);
}
echo "</table>\n"; ?>
</body>
</html>
I get
Parse error: parse error, unexpected T_VARIABLE in
/home/s/scaiecat/www/sv/test20.php on line 10
I do not understand yet, what is wrong.
--
Luigi Donatello (un italiano che vive in Svezia)
https://www.scaiecat-spa-gigi.com/
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