PHP output array into SELECT mysql - php

Question

Need to find array, and then run a MYSQL SELECT where array values are present (or not present).
$symbol = "abc";
$sql = "SELECT * FROM around";
$results = $conn->query($sql);
foreach($results as $row) {
$stop = preg_replace("/[0-9]/", "", $row['tip']);
if ($stop == $symbol)
{$sword = $row['tip'];
}}
So we need $sword to serve as an array in the event that there are multiple outputs. After we have that array, we need to run a mysql query that shows only those that have $sword array.
$query = "
SELECT * FROM ms WHERE `big` = '$sword'";
$result = mysql_query( $query );
So then we can do something like:
while ( $row = mysql_fetch_assoc( $result ) ) {
echo '"time": "' . $row['time'] . '",'; }

Answer 1

Here's the code, converting $sword to an array and using it in your 2nd query:
$symbol = array("abc", "def", "ghi");
$sword = array();
$sql = "SELECT * FROM around";
$results = $conn->query($sql);
foreach($results as $row) {
$stop = preg_replace("/[0-9]/", "", $row['tip']);
if (in_array($stop, $symbol)) {
$sword[] = $row['tip'];
}
}
if ($sword) {
$in = '';
$sep = '';
foreach ($sword as $s) {
$in .= "$sep '$s'";
$sep = ',';
}
/* $in now contains a string like "'123abc123', '12abc12', '1abc1'" */
$query = "
SELECT * FROM ms WHERE `big` IN ($in)";
$result = mysql_query( $query );
while ( $row = mysql_fetch_assoc( $result ) ) {
echo '"time": "' . $row['time'] . '",';
}
}
Now I'm going to go wash my hands to get all that mysql_query off me... :-)