herghost Posted April 13, 2009 Share Posted April 13, 2009 How do you echo a link? I have this: <?php $sql = "SELECT * FROM bandlinks WHERE userid = '$userid'"; $result = mysql_query($sql) or die(mysql_error()); echo "<table>"; if ($result) { while ($r = mysql_fetch_array($result)) { $homepage = $r["homepage"]; $myspace = $r["myspace"]; $facebook = $r["facebook"]; echo "<tr>"; echo "<td>Home Page</td>"; echo "<td><a href="$homepage">Go!</a></td></tr>"; echo "<br>"; echo "<tr>"; echo "<td>Myspace</td>"; echo "<td>$myspace</td></tr>"; echo "<br>"; echo "<tr>"; echo "<td>Facebook</td>"; echo "<td>$facebook</td></tr>"; echo "<br>"; } echo "</table>"; } else { echo "No data."; } ?> And I think that the " are breaking it? The error code is Parse error: parse error, expecting `','' or `';'' in C:\wamp\www\fanjunky\newprofile.php on line 169 which is the last line Quote Link to comment https://forums.phpfreaks.com/topic/153927-solved-echo-help/ Share on other sites More sharing options...
Maq Posted April 13, 2009 Share Posted April 13, 2009 Use single quotes for the attributes: echo "Go!"; Quote Link to comment https://forums.phpfreaks.com/topic/153927-solved-echo-help/#findComment-808974 Share on other sites More sharing options...
herghost Posted April 13, 2009 Author Share Posted April 13, 2009 simple then Thanks Maq Quote Link to comment https://forums.phpfreaks.com/topic/153927-solved-echo-help/#findComment-808981 Share on other sites More sharing options...
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