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Answer 1

+2 A:

You assigned the received data as $rows, but you are trying to oupput the vaeriable $row, which doesnt exist,

change this way

while($row = mysql_fetch_array($result))

marvin 2010-06-10 11:21:28

Answer 2

A:

Let me give 2 advises

To discover such a mistake in the future, always turn error reporting level at max by adding error_reporting(E_ALL); in your scripts. In this case PHP will tell you that $row variable diesn't exist.
Consider dividing your scripts into 2 parts: getting data part and displaying data part

like this:

<?php
error_reporting(E_ALL);
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="sabs"; // Database name
$tbl_name="doador"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql) or trigger_error(mysql_error().$sql);
while($row = mysql_fetch_array($result)){
  $DATA = $row[];
}
mysql_close();
?>
<html>
<body>
<h1>Lista de usuários</h1>
<table>
<? foreach($DATA as $row): ?>
  <tr>
    <td><?=$row['id']?></td>
    <td><?=$row['nome']?></td>
    <td><?=$row['sobrenome']?></td>
    <td><?=$row['email']?></td>
    <td><?=$row['login']?></td>
    <td><?=$row['senha']?></td>
    <td><?=$row['idade']?></td>
    <td><?=$row['peso']?></td>
    <td><?=$row['fuma']?></td>
    <td><?=$row['sexo']?></td>
    <td><?=$row['doencas']?></td>
  </tr>
<? endforeach ?>
</table>
</body>
</html>

Html part can be put into separate file to ease of operating.
Also note the trigger_error function which will help you to detect SQL errors

Col. Shrapnel 2010-06-10 11:46:51

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Print table data mysql php

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