JSON parse Mysql query result in php pdo - php

Question

I am trying to do following
$statement = $conn->prepare('SELECT * FROM myTable');
$statement->execute();
if(!($row = $statement->fetchAll(PDO::FETCH_ASSOC)))
{
return false;
}
$conn = null;
} catch(PDOException $e) {
throw $e;
return false;
}
return return json_encode(array('Result'=>$row);
Works and fetches all entries in a table make then JSON Array and send them,
However I want to make a query where selected ids must be send in a JSON Array
E.g 10, 20, 30
I assume that this will be done in a For loop perhaps
$statement = $conn->prepare('SELECT * FROM myTable WHERE id = :id');
$statement->bindParam(':id', $id, PDO::PARAM_STR);
$statement->execute();
$row = $statement->fetch(PDO::FETCH_OBJ)))
Now suppose i have id's = 10,20,30 i want to append all of them in a JSON Array How can i do that?
just like return json_encode(array('Result'=>$row);
Edited Code
function GetMyMembers()
{
$myId = trim($_REQUEST['myId']);
try {
$conn = $this->GetDBConnection();
$statement = $conn->prepare('SELECT valId FROM memList WHERE myId=:myId' );
$statement->bindParam(':myId', $myId, PDO::PARAM_INT);
$statement->execute();
if(!($row = $statement->fetchAll(PDO::FETCH_ASSOC)))
{
return false;
}
// $row contains ALL THE ID'S
$placeholders = str_repeat('?,', count($row));
$placeholders = substr($placeholders, 0, -1);
$sql = "SELECT id, * FROM players WHERE id IN ($placeholders)";
$statement = $conn->prepare($sql);
$statement->execute($row);
$rows = $sth->fetchAll(PDO::FETCH_ASSOC|PDO::FETCH_GROUP);
$conn = null;
} catch(PDOException $e) {
throw $e;
return false;
}
return $rows;
}

Answer 1

$statement = $conn->prepare('SELECT * FROM myTable');
$statement->execute();
$data = array();
while($row = $statement->fetch(PDO::FETCH_ASSOC)))
{
$data[$row['id']] = $row;
}
return json_encode(array('Result'=>$data));
Btw, using raw API is not convenient. With Database abstraction library your code can be as short as 2 following lines:
$data = $db->getInd("id",'SELECT * FROM myTable');
return json_encode(array('Result'=>$data));
Edit:
if you have an array of ids, you need more complex code
$ids = array(1,2,3);
$data = array();
$statement = $conn->prepare('SELECT * FROM myTable WHERE id = :id');
foreach ($ids as $id) {
$statement->bindValue(':id', $id, PDO::PARAM_STR);
$statement->execute();
$data[] = $statement->fetch(PDO::FETCH_OBJ);
}
return json_encode(array('Result'=>$data));
But while using Database abstraction library, there will be the same 2 lines:
$data = $db->getAll('SELECT * FROM myTable where id IN (?a)', $ids);
return json_encode(array('Result'=>$data));
Edit2:
if you need ids only
$statement = $conn->prepare('SELECT id FROM myTable');
$statement->execute();
$data = array();
while($row = $statement->fetch(PDO::FETCH_ASSOC)))
{
$data[] = $row['id'];
}
return json_encode(array('Result'=>$data));
while using Database abstraction library, it's still 2 lines:
$ids = $db->getCol('SELECT id FROM myTable');
return json_encode(array('Result'=>$ids));

Answer 2

$ids = array(1,2,3);
$placeholders = str_repeat('?,', count($ids));
$placeholders = substr($placeholders, 0, -1);
$sql = "SELECT id, * FROM table WHERE id IN ($placeholders)";
$sth = $dbh->prepare($sql);
$sth->execute($ids);
$rows = $sth->fetchAll(PDO::FETCH_ASSOC|PDO::FETCH_GROUP);
echo json_encode(array('Result' => $rows));
Based on additional comments:
Best option:
$sql = '
SELECT *
FROM table1 AS t1
INNER JOIN table2 t2
ON t2.foreign_key = t1.id
';
or
$sql = 'SELECT id FROM table1';
$sth = $dbh->prepare($sth);
$sth->execute();
$ids = $sth->fetchColumn();
//next look above