mysqli 中的动态变量绑定 bind_param

Question

当我尝试低于代码时它给我一个警告

mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables

$stmt = $mysqli->prepare('SELECT * FROM users WHERE lname = ? AND fname = ?');
    $type = "ss";
    $param = array("Maq","bool");
    $params[] = &$type;
    $params[] = &$param;

    call_user_func_array( array($stmt, 'bind_param'), $params );

我真的不知道为什么会发生这种情况我在 Stack Overflow 中阅读了很多相同标题的问题，但有些问题使用逗号分隔类型或需要更多类型但在我的情况下我认为它是正确的但为什么我不断收到警告？

Answer 1

$stmt = $mysqli->prepare('SELECT * FROM users WHERE lname = ? AND fname = ?');
$stmt->bind_param('ss', $param[0], $param[1]);
$stmt->execute();

//other version
$stmt = $mysqli->prepare('SELECT * FROM users WHERE lname = ? AND fname = ?');
$type = "ss";
$names = array("Maq", "bool");
$params = array(&$type, &$names[0], &$names[1]);

call_user_func_array(array($stmt, 'bind_param'), $params);

Answer 2

这是一个迟到的答案，希望能对某人有所帮助。

它是我的 mysqliLayer class 的一部分，所以我提取了需要的变量。

// establish mysqli connection
$conn = new mysqli(.....); 
$tableName = 'users';
// Types to bind
$type = 'isss';
$fields = ['id','name', 'email', 'created'];
$values = [1, 'angel', 'email@test.com', '2018-1-12'];

$sql = "INSERT INTO " . $tableName . " (" . join(',', $fields) . ") VALUES (?,?,?,?)";

$stmt = $conn->prepare($sql);
// Using ...token introduced in php v.5.6 instead of call_user_func_array
// This way references can be omitted, like for each value in array
$stmt->bind_param($type, ...$values);

$stmt->execute();

$stmt->close();

mysqli 中的动态变量绑定 bind_param

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