Result of Mysql Query in a PHP table

[Felix]

Hi,

I've a problem:

I want to have the result of my Mysql Query in a Table in my php file.
Now I've this:
<?

mysql_connect(" localhost","roo t")
or die ("Keine Verbindung moeglich");

mysql_select_db ("datenbank" )
or die ("Die Datenbank existiert nicht");

$abfrage = "SELECT * FROM tabelle";
$ergebnis = mysql_query($ab frage);
while($row = mysql_fetch_obj ect($ergebnis))

{
echo $row->Referenznummer ;
}

?>
But I want that the result is in a Table. With the heading "Referenznummer ".
It should be like:
---------------
Referenznummer:
---------------
Result 1
---------------
Result 2
......
....
...
I searched a long time in google without success, please help me

Thanks
Felix

Btw: sorry for my bad english

[Kevin Thorpe]

Felix wrote:

Hi,

I've a problem:

I want to have the result of my Mysql Query in a Table in my php file.
Now I've this:
<?

mysql_connect(" localhost","roo t")
or die ("Keine Verbindung moeglich");

mysql_select_db ("datenbank" )
or die ("Die Datenbank existiert nicht");
echo "<table>";
echo "<tr><th>Refere nznummer</th></tr>"; $abfrage = "SELECT * FROM tabelle";
$ergebnis = mysql_query($ab frage);
while($row = mysql_fetch_obj ect($ergebnis))

{ echo "<tr><td>"; echo $row->Referenznummer ; echo "</td></tr>"; } echo "</table>";
?>
But I want that the result is in a Table. With the heading "Referenznummer ".
It should be like:

[Marius Mathiesen]

Felix wrote:

mysql_connect(" localhost","roo t")
or die ("Keine Verbindung moeglich");

mysql_select_db ("datenbank" )
or die ("Die Datenbank existiert nicht");

$abfrage = "SELECT * FROM tabelle";
$ergebnis = mysql_query($ab frage); But I want that the result is in a Table. With the heading "Referenznummer ".
It should be like:
---------------
Referenznummer:
---------------
Result 1
---------------
Result 2

Hi, Felix
I've written a small utility function that I use for this. It takes a
MySQL result set as an argument, and returns an HTML table (string)
containing all the column headers and rows, formatted.

Here's the function:

function _mysql_result_a ll($result, $tableFeatures= "") {
$table .= "<!--Debugging output for SQL query-->\n\n";
$table .= "<table $tableFeatures> \n\n";
$noFields = mysql_num_field s($result);
$table .= "<tr>\n";
for ($i = 0; $i < $noFields; $i++) {
$field = mysql_field_nam e($result, $i);
$table .= "\t<th>$fie ld</th>\n";
}
while ($r = mysql_fetch_row ($result)) {
$table .= "<tr>\n";
foreach ($r as $column) {
$table .= "\t<td>$col umn</td>\n";
}
$table .= "</tr>\n";
}
$table .= "</table>\n\n";
$table .= "<!--End debug from SQL query-->\n\n";
return $table;
}

You could use it like this, using your $ergebnis variable:

print _mysql_result_a ll($ergebnis);

That is, copy the function above into your script, and then use the
preceding line to output the result.

Enjoy...

--
//Marius