While Loop Code problem

[wanda46]

<?php
$query="select from table1";
$result=mysql_query($query) or die(mysql_error());
while($row=mysql_fetch_array($result))
{
$teamid=$row["team_id"];
//team_id is a field of table 1;
$query2="select from table2 where id='$teamid'";
//I want to open another table in this loop; id is a field of table 2;
$result2=mysql_query($query2) or die(mysql_error());
if (!mysql_fetch_array(mysql_query($query2))) echo "die";
else
{
$row2=mysql_fetch_array(mysql_query($query2));
echo $row2["anyfield"];
}
}
?>

I have 4 records in table 2. Finally I can correctly output the first 4 records' anyfield, but the last record simply output "die".

Any idea?

[ahundiak]

This is causing the first record to be skipped:
if (!mysql_fetch_array(mysql_query($query2))) echo "die";

Just do another while loop:
$result2=mysql_query($query2) or die(mysql_error());
while($row2 = mysql_fetch_array(mysql_query($query2))
{
echo $row2["anyfield"];
}

[wanda46]

I tried your codes this way, but it did not output anything:

<?php
$query="select from table1";
$result=mysql_query($query) or die(mysql_error());
while($row=mysql_fetch_array($result))
{
$teamid=$row["team_id"];
//team_id is a field of table 1;
$query2="select from table2 where id='$teamid'";
//I want to open another table in this loop; id is a field of table 2;
$result2=mysql_query($query2) or die(mysql_error());

while($row2 = mysql_fetch_array(mysql_query($query2))
{
echo $row2["anyfield"];
}

}
?>

[ahundiak]

There is an extra mysql_query in there.
Change
while($row2 = mysql_fetch_array(mysql_query($query2))
To
while($row2 = mysql_fetch_array($result2))

[wanda46]

Thanx. Works now!