SQL / PHP如何從表中提取數據並將其放在變量中

Question

假設我的表中有一個唯一鍵，並且該鍵是通過get方法發送到我的站點的，那么我該如何拉出特定的鍵並將表中的所有數據分配給變量。 這是我到目前為止所掌握的，但似乎無法弄清楚。

$query1 = "SELECT * 
       FROM todo_item2 as ti INNER JOIN todo_category2 as tc ON ti.todo_id = tc.todo_id'
       WHERE todo_id = :todo_id";

$statement1 = $db->prepare($query1);
$statement1 -> execute(array(
    'todo_id' =>$id
));

while ($row = $statement1->fetch()) 
{
    $text = $row['todo'];
    $cat = $row['category'];
    $percent = $row['precent'];
    $date = $row['due_date'];
}

Answer 1

您應該准備好execute實際操作..要execute的參數（我假設您正在使用PDO或類似的東西）是查詢的標記。 您想要的是這樣的：

$query = " ... WHERE todo_id = ?"
$stmt = $db->prepare($query);
$stmt->execute(array($id));
while ($row = $stmt->fetch()) {
   //$row is now an associative array of row values.
}

Answer 2

// Start the Load
$query1 = "SELECT * 
           FROM todo_item2 as ti INNER JOIN todo_category2 as tc ON ti.todo_id = tc.todo_id
           WHERE ti.todo_id = :todo_id";

$statement1 = $db->prepare($query1);
$statement1 -> execute(array(
    'todo_id' =>$id
));

// Make Sure the Data Exists
if( $statement1->rowCount() == 0 )
{
    die('Please Enter a Valid ID Tag - (id)');
}


while($row = $statement1->fetch())
{
    $text = $row['todo'];
    $cat = $row['category'];
    $percent = $row['percent'];
    $date = $row['due_date'];
}