比较两个不同表中的值，如果相等则显示[关闭]

Question

I have a variable called $row['mykey'] that was genterated from a query on table one, I have echoed it to my page and it does display the value I'm looking for. Now I need to use this varible to set my new query on table two, to only display results if it matches contractor_lock from table two. $stmt1 one works fine to display the entire table, but I want to use $stmt2 but get no data results. Thanks for any help!

try {
 $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
 $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);


 $stmt1 = $conn->prepare("SELECT client_id, companyname, contact, contractor_lock FROM table_two");


 $stmt2 = $conn->prepare("SELECT client_id, companyname, contact, contractor_lock FROM table two WHERE contractor_lock = $row['mykey']"); 

 $stmt->execute();
 $result = $stmt->setFetchMode(PDO::FETCH_ASSOC); 
 foreach(new TableRows(new RecursiveArrayIterator($stmt->fetchAll())) as $k=>$v) { 
     echo $v;
 }
 }

catch(PDOException $e) {
 echo "Error: " . $e->getMessage();
 }
 $conn = null;
 }

Answer 1

You have to concatenate $row['mykey'] to stmt2 instead of including it directly because it might be interpreted as a string.

It should be like this

$stmt2 = $conn->prepare("SELECT client_id, companyname, contact, contractor_lock FROM table two WHERE contractor_lock = ?");
$stmt2->execute($row['mykey']);