PDO - 将两个选择值放入数组中

Question

我有这个 PDO：

$id = 1;
$title = 'resourceName';
$url = 'resourceURL';
$result = array($title => $url);

include('../dbconnect.php');

$pdo = new PDO("mysql:host=$db_host;dbname=$db_name;", $db_user, $db_password);
$stmt = $pdo->prepare("SELECT resourceName, resourceURL FROM Resources WHERE categoryID = :id");
$stmt->bindParam(':id', $id);
$stmt->execute(array_values($result));
$row = $stmt->fetchAll();
print_r($row);

我刚刚收到此错误： 警告：PDOStatement::execute() [pdostatement.execute]: SQLSTATE[HY093]: 参数号无效：参数未定义

如何让 PDO 结果成为一个以 resourceName 为键、resourceURL 为值的数组？

Answer 1

您正在将不需要的参数绑定到查询。

$stmt->bindParam(':id', $id);
$stmt->execute(array_values($result)); //On this line. These parameters are 
                                         not needed

让我解释一下

$stmt->bindParam(':id', $id);

将 $id 的值绑定到 SQL 参数 :id 并再次绑定

$stmt->execute(array_values($result));

您正在绑定另一个没有任何索引的参数。

因此您的查询需要 1 个参数，而您要发送两个参数。

解决方案：使用其中一种

要么

$stmt->bindParam(':id', $id);

或者，直接这样

$stmt->execute(array(":id" => $id));

之后，从行中获取列并将它们转换为所需格式的新数组

$row = $stmt->fetchAll();
//Now assuming only one was returned from the database this might work
$new = array($row[0] -> resourceName => $row[0] -> resourceURL);

Answer 2

你正在混合参数绑定

$id = 1;
$title = 'resourceName';
$url = 'resourceURL';
$result = array($title => $url);

include('../dbconnect.php');

$pdo = new PDO("mysql:host=$db_host;dbname=$db_name;", $db_user, $db_password);
$stmt = $pdo->prepare("SELECT resourceName, resourceURL FROM Resources WHERE categoryID = :id");
$stmt->bindParam(':id', $id);
$stmt->execute(); //the array you put in execute will be used as parameter
$row = $stmt->fetchAll();
print_r($row);

http://www.php.net/manual/en/pdostatement.execute.php