PHP-如何从满足条件的表中获取所有行？

I've a table invitation with coloumns toemail, fromemail, send

I need to print toemail & fromemail from all rows having valid send=1

This is the code that I'm using

<?php
$query = mysql_query("SELECT fromemail,toemail FROM invitations WHERE sent = 1");
if(mysql_num_rows($query)!=0)
{
    $row = mysql_fetch_assoc($query);
    echo $row['toemail'];
    echo $row['fromemail];
}
?>

But this prints only the first row which satisfying the condition.
Help me to get the php code please.
Thanks in advance.

Use this,

$query = mysql_query("SELECT fromemail,toemail FROM invitations WHERE sent = 1");
if(mysql_num_rows($query)!=0)
{
    while($row = mysql_fetch_assoc($query)) {
           echo $row['toemail'];
           echo $row['fromemail];
    }
}

You can either use following -

$query = mysql_query("SELECT fromemail,toemail FROM invitations WHERE sent = 1");
if(mysql_num_rows($query)!=0)
{
    $result = mysql_result($query);
    foreach($result in $row) {
           echo $row->toemail;
           echo $row->fromemail;
    }
}

P.S. - MYSQL extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used.

Reference - http://www.php.net/manual/en/faq.databases.php#faq.databases.mysql.deprecated