MYSQL order limit with PHP variable - php

Question

This is my code below
$query = mysql_query("SELECT * FROM table ORDER BY id LIMIT 10");
while($f = mysql_fetch_array($query)){
$match = 0; //In reality it's an array search function which returns 1 on match
if($match == 1) {
echo"Show content!";
}
}
Im trying to make an list with 10 rows, and i have a function which uses "name" from table to run an search query with an array generated by twitter API. In example if i get 3 matching records, i still want to show a list with 10 rows but hide the matching elements from there.
At the moment the script hides the matching elements and shows 7 rows instead of 10.
That is what i need help with, cheers :)

Answer 1

Process the data you are getting from the twitter api, collect the data you need in an array and query the database afterwards.
Something like that:
<?php
$collectArray = array();
foreach ($twitterData as $index => $data) {
if ($someCriteria === TRUE) {
$collectArray[] = $data;
}
}
$implodedCollectArray = "'" . implode("', '", $collectArray) . "'";
$sql = "SELECT * FROM `table_name` WHERE `some_column` IN (" . $implodedCollectArray . ")";
$query = mysql_query($sql) or die(mysql_error());
while($f = mysql_fetch_array($query)){
echo $f['column'];
}
?>

Answer 2

reinitilize match variable as the code:
$query = mysql_query("SELECT * FROM table ORDER BY id LIMIT 10");
while($f = mysql_fetch_array($query)){
if($f['name']!='variable_name_twiter'){
$match = 1; //In reality it's an array search function which returns 1 on match
}
if($match == 1) {
echo"Show content!";
}
$match = 0;
}