PHP PDO: Search array value inside of MySQL's JSON object - php

Question

I got the following data in MySQL's JSON-type column:
{"2": [2, 3], "3": [29], "71": "test"}
I need to search array value inside of attribute "2", which works just fine when variables are placed inside the query, but not when using PHP's PDO arrays.
$field_id = 2;
$option_id = 2;
$query = "SELECT id FROM notes WHERE JSON_CONTAINS(data, '{\"$field_id\": $option_id }')";
try {
$stmt = $dbh->prepare($query);
$stmt->execute();
$used_qty = $stmt->rowCount();
} catch(PDOException $ex) {
echo 'Query failed: ' . $e->getMessage();
exit;
}
// $used_qty returns 1 which is correct;
Binding through array returns 0:
$query = "SELECT id FROM notes WHERE JSON_CONTAINS(data, '?')";
try {
$stmt = $dbh->prepare($query);
$stmt->execute(array('{"' . $field_id . '": ' . $option_id . '}"'));
$used_qty = $stmt->rowCount();
} catch(PDOException $ex) {
echo 'Query failed: ' . $e->getMessage();
exit;
}
Can't figure out what I missed here.
Please help. Thanks.

Answer 1

You quoted your placeholder:
$query = "SELECT id FROM notes WHERE JSON_CONTAINS(data, '?')";
^-^--
which means it's NOT a placeholder, it's a string containing a question mark.
Remove those quotes.

Answer 2

After enabling PDO exceptions:
$dbh = new PDO($dsn, $dsn_user, $dsn_password, array(PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));
I got clear error message saying that only value can be accepted as a second parameter:
Query failed: SQLSTATE[22032]: <>: 3141 Invalid JSON
text in argument 2 to function json_contains: "The document root must
not follow by other values." at position 8.
https://dev.mysql.com/doc/refman/5.7/en/json-search-functions.html#function_json-contains
But there is a third optional [path] parameter available: JSON_CONTAINS(json_doc, val[, path])
Which is exactly what I needed:
$query = "SELECT id FROM notes WHERE JSON_CONTAINS(data, ?, ?)";
try { $stmt = $dbh->prepare($query); $stmt->execute(array($option_id, '$."' . $field_id . '"')); $used_qty = $stmt->rowCount(); } catch(PDOException $ex) { echo 'Query failed: ' . $ex->getMessage(); exit; }
Thanks to Marc B.