gpt4 book ai didi

php - mysqli_stmt::bind_result():绑定(bind)变量的数量与准备好的语句中的字段数量不匹配(一个函数工作,另一个不工作)

转载 作者:行者123 更新时间:2023-11-28 23:57:12 25 4
gpt4 key购买 nike

这个问题以前有人回答过,但格式不一样。

我正在尝试创建一个对象方法来获取数据库行。我不断收到标题错误。奇怪的是,如果我直接在父类方法中调用函数的相同代码,它就可以工作。这是两个例子。

示例 1:作品从单独的 PHP 页面调用的函数,例如: $email = trim($_POST['email']); $password = trim($_POST['密码']);

$found_user = User::authenticate($email, $password);

这是正在运行的函数。 (如果我正在制作一个可重用的函数来测试我的代码是否错误,我会设置该函数)

public static function authenticate($username="", $password="") {

$db = Database::getInstance();
$mysqli = $db->getConnection();

$field = 'email';

$sql_query = "SELECT * ";
$sql_query .= "FROM `".static::$table_name."`";
$sql_query .= "WHERE {$field}=? ";
$sql_query .= "LIMIT 1";

$stmt = $mysqli->stmt_init();

if (!$stmt->prepare($sql_query)) {
$error = "Errormessage: " . $mysqli->error;
return $error;
}

if (!$stmt->prepare($sql_query)) {
$error = "Errormessage: " . $stmt->error;
return $error;
} else {
$stmt->prepare($sql_query);

$valueType = gettype($username);

if ($valueType == 'string') {
$stmt->bind_param('s', $username);
} else if ($valueType == 'integer') {
$stmt->bind_param('i', $username);
} else if ($valueType == 'double') {
$stmt->bind_param('d', $username);
} else {
$stmt->bind_param('b', $username);
}

$stmt->execute();
$result = $stmt->get_result();

return $result->fetch_object();
}
}

现在,如果我在相同的方法中调用该函数,但来自扩展类(父类是扩展了 DatabaseOptions 的 User)。函数是这样写的:

public static function authenticate($username="", $password="") {
$dbObject = static::get_row('email', $username);

if (isset($dbObject)) {
return $dbObject;
}
}

DatabaseObject 中的函数:

public function get_row($field, $value) {
$db = Database::getInstance();
$mysqli = $db->getConnection();

$sql_query = "SELECT * ";
$sql_query .= "FROM `".static::$table_name."`";
$sql_query .= "WHERE {$field}=? ";
$sql_query .= "LIMIT 1";

$stmt = $mysqli->stmt_init();

if (!$stmt->prepare($sql_query)) {
$error = "Errormessage: " . $stmt->error;
return $error;
} else {
$stmt->prepare($sql_query);

$valueType = gettype($value);

if ($valueType == 'string') {
$stmt->bind_param('s', $value);
} else if ($valueType == 'integer') {
$stmt->bind_param('i', $value);
} else if ($valueType == 'double') {
$stmt->bind_param('d', $value);
} else {
$stmt->bind_param('b', $value);
}

$stmt->execute();
$result = $stmt->get_result();

return $result->fetch_object();
}
}

最佳答案

public function get_row($field, $value) {
$db = Database::getInstance();
$mysqli = $db->getConnection();

$sql_query = "SELECT * ";
$sql_query .= "FROM `".static::$table_name."`";
$sql_query .= "WHERE {$field}=? ";
$sql_query .= "LIMIT 1";

if ($stmt = $mysqli->prepare($sql_query)) {
$valueType = gettype($value);

if ($valueType == 'string') {
$stmt->bind_param('s', $value);
} else if ($valueType == 'integer') {
$stmt->bind_param('i', $value);
} else if ($valueType == 'double') {
$stmt->bind_param('d', $value);
} else {
$stmt->bind_param('b', $value);
}

$stmt->execute();
$result = $stmt->get_result();
return $result->fetch_object();
} else {
$error = "Errormessage: " . $stmt->error;
return $error;
}
}

原始代码的问题在于我使用的是 bind_result($row),我从数据库中提取的行有多个需要绑定(bind)的值。如$result = $stmt->get_result 然后返回一个fetch_object();

关于php - mysqli_stmt::bind_result():绑定(bind)变量的数量与准备好的语句中的字段数量不匹配(一个函数工作,另一个不工作),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31345282/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com