php - 使用 PDO fetch 查询 Mysql 返回 false

Question

我正在尝试在 PDO 中查询我的数据库并将输出显示在表单上。但是 fetch 语句不起作用。代码是

try {
include '../../config/database.php';
$database = new Database();
$db = $database->getConnection();


//prepare query
$query = "select 
            payment_id, payment_supplier, payment_ref, payment_cost_rating, payment_amount 
        from 
            payments 
        where 
            payment_id = ? 
        limit 0,1";

$stmt = $db->prepare( $query );

//this is the first question mark
$stmt->bindParam(1, $_REQUEST['myData']);

//execute our query
if($stmt->execute()){
var_dump($stmt->fetch());
    //store retrieved row to a variable
    $row = $stmt->fetch(PDO::FETCH_ASSOC); 

    //values to fill up our form
    $payment_id = $row['payment_id'];
    $payment_supplier = $row['payment_supplier'];
    $payment_ref = $row['payment_ref'];
    $payment_cost_rating = $row['payment_cost_rating'];
    $payment_amount = $row['payment_amount'];

}else{
    echo "Unable to read record.";
}

var_dump ($stmt); prints

object(PDOStatement)#3 (1) { ["queryString"]=> string(157) "select payment_id, payment_supplier, payment_ref, payment_cost_rating, payment_amount from payments where payment_id = ? limit 0,1"}

但是 fetch() 总是返回 false。如果有帮助，这是包含的 database.php 文件

class Database{

// database credentials
private $host = "localhost";
private $db_name = "test-project";
private $username = "root";
private $password = "";
public $conn;

// get the database connection
public function getConnection(){

    $this->conn = null;

    try{
        $this->conn = new PDO("mysql:host=" . $this->host . ";dbname=" . $this->db_name, $this->username, $this->password);
    }catch(PDOException $exception){
        echo "Connection error: " . $exception->getMessage();
    }

    return $this->conn;
}}

我在这里错过了什么？

Answer 1

问题在于提取被使用了两次。

可以这样做:

if($stmt->execute()){
    //store retrieved row to a variable
    $row = $stmt->fetch(PDO::FETCH_ASSOC);

     // display the row after fetching the contents.
    var_dump($row); 

    //values to fill up our form
    $payment_id = $row['payment_id'];
    $payment_supplier = $row['payment_supplier'];
    $payment_ref = $row['payment_ref'];
    $payment_cost_rating = $row['payment_cost_rating'];
    $payment_amount = $row['payment_amount'];

}else{
    echo "Unable to read record.";
}

这应该可以正常工作。