Question

我想从数据库中获取数据。下面我得到了这个错误。问题是这个代码在我的其他项目中使用并且它工作得很好。有人知道问题出在哪里吗？

我收到了这个错误：

Fatal error: Call to a member function execute() on a non-object in /public_html/website/index.php on line

CODE：

include 'scripts/db_conn.php';
include 'scripts/functions.php';

$prep_stmt = "SELECT date FROM blocked_dates WHERE date >= CURDATE();";
    $stmt = $mysqli->prepare($prep_stmt);
    $stmt->execute();
    if (!$stmt) {
        die('There was an error running the query [' . $mysqli->error . ']');
        exit();
    }
    $meta = $stmt->result_metadata();
    while ($field = $meta->fetch_field()) {
        $parameters[] = & $row[$field->name];
    }

    call_user_func_array(array($stmt, 'bind_result'), $parameters);

    while ($stmt->fetch()) {
        foreach ($row as $key => $val) {
            $x[$key] = $val;
        }
        $results[] = $x;
    }

Answer 1

您无法将对象绑定到新变量。

尝试像这样开始你的代码：

$stmt = new Mysqli();

连接数据库后，

$stmt->prepare($prep_stmt);
$debug = $stmt->execute();

if (!$debug) {
    die('There was an error running the query [' . $mysqli->error . ']');
    exit();
}

$meta = $stmt->result_metadata();

while ($field = $meta->fetch_field()) {
    $parameters[] = & $row[$field->name];
}

call_user_func_array(array($stmt, 'bind_result'), $parameters);

while ($stmt->fetch()) {
    foreach ($row as $key => $val) {
        $x[$key] = $val;
    }
    $results[] = $x;
}

致命错误：在/public_html/website/index.php上的非对象上调用成员函数execute（）

1 个答案: