Php 警告：call_user_func_array（）要求参数1是有效的回调mysqli

我有一个脚本试图将数组放入mysqli_stmt_bind_param，我搜索了一下，找到了这个解决方案，但是，代码似乎工作不正常。我没有收到来自数据库的任何结果和警告

Warning: call_user_func_array() expects parameter 1 to be a valid callback, first array member is not a valid class name or object 

这是我的密码：-

$sql_param = array();
    $params = array();

    $query = "SELECT campname,url,views,leads,subs,country,affid FROM camp WHERE";

if (isset($_POST['fromdatetime']) && isset($_POST['todatetime'])) {
    $datefrom = $_POST['fromdatetime'].":00";
    $dateto = $_POST['todatetime'].":59";

    $sql_param[] = $datefrom;
    $sql_param[] = $dateto;
    $bindparam = "ss";
    $query .= "datevisit between (?) AND (?)";
}

if (isset($_POST['affidcode'])) {
    $affidcode = $_POST['affidcode'];
    $sql_param[] = $affidcode;
    $bindparam .= "s";
    $query .= " AND affid = (?)";
}

$sql_param_count = count($sql_param);
$params[] = &$bindparam;

for($i = 0; $i < $sql_param_count; $i++) {
      $params[] = &$sql_param[$i];
    }

    /* create a prepared statement */
    $stmt = mysqli_prepare($mysqli, $query);

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

    /* execute query */
    $status = mysqli_stmt_execute($stmt);

确保$stmt在此块之后是有效的mysqli\u stmt对象，且不为false：

/* create a prepared statement */
$stmt = mysqli_prepare($mysqli, $query);

然后试试这个：

array_unshift($params, $stmt);
call_user_func_array('mysqli_stmt_bind_param', $params); 

---

我尝试过的@PhiterFernandes可能重复，但不起作用。请尝试将mysqli_stmt_bind_param更改为bind_param。@brevis我也尝试过，仍然是相同的警告，并且没有显示结果。下面是另一个示例，但出现了另一个问题。。。。mysqli_stmt_bind_param期望参数1是mysqli_stmt，布尔givenCheck$stmt在$stmt=mysqli_prepare$mysqli$query之后；线如果它是错误的-SQL中有错误。很好！查询缺少查询结尾的空格。谢谢