Fetching An Array With MySQL / PHP - php

Question

I have an existing database which has a table called PERSON with a field called NAME. What I’m trying to do is select all of the rows in the table where the NAME is "bill". And then I want the result to be stored in an array that I can step through at a later point.
Now, the problem is my code will only select the FIRST row with the name "bill" and ignore the rest of the rows where the NAME is "bill". At least that’s how it appears when I print out the array contents with print_r(). My code below:
<?php
$getAllPreview = "SELECT * from PERSON where NAME = 'bill'";
$getAllResult = #mysql_query( $getAllPreview );
$getAllRows = #mysql_fetch_assoc( $getAllResult );
print "<pre>";
print_r($getAllRows);
print "</pre>";
?>

Answer 1

<?php
$getAllPreview = "SELECT * from PERSON where NAME = 'bill'";
$getAllResult = #mysql_query( $getAllPreview );
while ($row = #mysql_fetch_assoc( $getAllResult ) ) {
$getAllRows[] = $row;
}
print "<pre>";
print_r($getAllRows);
print "</pre>";
?>

Answer 2

while($row = mysql_fetch_array($getAllResult, MYSQL_ASSOC)) {
$data[] = $row;
}

Answer 3

You just keep looping over mysql_fetch_assoc until no further rows are returned. If you want to output or process them, just do so in each iteration of the loop, as it's more efficient than placing it in an array first. But here you go anyway:
$allRows = array ();
while ($row = mysql_fetch_assoc( $getAllResult)) $allRows [] = $row;

Answer 4

<?php
$getAllPreview = "SELECT * from PERSON where NAME = 'bill'";
$getAllResult = #mysql_query( $getAllPreview );
$num_rows = mysql_num_rows($getAllResult);
while ($row = #mysql_fetch_assoc($getAllResult))
{
for($i=0;$i<$num_rows;$i++)
{
$array[] = $row;
}
}
?>