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Home/ Questions/Q 3976098
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Editorial Team
  • 0
Editorial Team
Asked: May 20, 20262026-05-20T04:45:50+00:00 2026-05-20T04:45:50+00:00

I am using php query to retrieve information from a mysql database with mysql_fetch_array

  • 0

I am using php query to retrieve information from a mysql database with mysql_fetch_array and use the returned results to query another database? Im pretty novice. Is this possible?

$search =
"SELECT `a`,`b`

FROM 'DB'
WHERE `x` LIKE '$var'
LIMIT 0 , 30";

$result = mysql_query($search, $connection);
if($result){

while ($row = mysql_fetch_array($result)){
echo $row["a"]." ".$row["b"]."<br />";
                        }
} else {
echo "<p> Subject search failed. <p>";
echo "<p>" . mysql_error() . "</p>";
    }

<!-- SECOND QUERY -->   
$search2 = 
"SELECT `a`,`b`
FROM `DB`
WHERE `x` = **$row['a']**  *trying to grab the result from the first query and use as search query in second?*

LIMIT 0 , 100";

$result2 = mysql_query($search2, $connection);
if($result2){

while ($row2 = mysql_fetch_array($result2)){
echo $row2["a"]." ".$row2["b"]."<br />";
                        }
} else {
echo "<p> Subject search failed. <p>";
echo "<p>" . mysql_error() . "</p>";
}

Could someone point me in the right direction?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-20T04:45:51+00:00Added an answer on May 20, 2026 at 4:45 am

    $search = "SELECT a,b FROM DB WHERE x LIKE '$var' LIMIT 0 , 30";
    $result = mysql_query($search, $connection);
    if($result){
        while ($row = mysql_fetch_array($result)){
            echo $row["a"]." ".$row["b"]." ";
            $search2 = "SELECT a,b FROM DB WHERE x = '".$row['a']."' LIMIT 0 , 100";
            $result2 = mysql_query($search2, $connection);
            if($result2){
                while ($row2 = mysql_fetch_array($result2)) {
                    echo $row2["a"]." ".$row2["b"]." ";
                }
            } else {
                echo "Subject search failed.";
                echo mysql_error();
            }
        }
    } else {
        echo "Subject search failed. ";
        echo mysql_error();
    }
    

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