"array_push() expects parameter 1 to be array"错误消息

Question

我正在研究网站上新帐户的激活队列。

该系统将通过在数组中迭代，并在每个新用户帐户中详细介绍，然后显示它们，以便管理员可以接受或拒绝该帐户。要收集帐户详细信息并将其显示在我使用以下代码的数组中，但是我会遇到错误" Array_push（）期望参数1为数组，null给出"。我不知道为什么会引起这一点，我尝试了以前建议的各种事情。预先感谢。

<?php
session_start();
require "classes.php";
$TF = new TF_Core ();
$ActQueueQuery = "SELECT username, surname, forename, joined FROM users
WHERE rank = 'Unactivated'";
if ($statement = TF_Core::$MySQLi->DB->prepare($ActQueueQuery)) {
    $statement->execute();
    $results = $statement->get_result();
}
if($results->num_rows == 0){
    $data = 1;
}
else{
    $_SESSION["ActQueue"] = "";
    while($row = $results->fetch_assoc()){
        $_SESSION["ActQueue"] = array_push($_SESSION["ActQueue"], array($row["username"], $row["surname"], $row["forname"], $row["joined"]));
    }
    $data = 0;
    }
echo $data;
?>

Answer 1

<?php
session_start();
$_SESSION["ActQueue"] = array(); // define an empty array
require "classes.php";
$TF = new TF_Core ();
$ActQueueQuery = "SELECT username, surname, forename, joined FROM users
WHERE rank = 'Unactivated'";
if ($statement = TF_Core::$MySQLi->DB->prepare($ActQueueQuery)) {
    $statement->execute();
    $results = $statement->get_result();
}
if($results->num_rows == 0){
    $data = 1;
}
else{
    while($row = $results->fetch_assoc()){
        $_SESSION["ActQueue"][] = array($row["username"], $row["surname"], $row["forname"], $row["joined"]); // check the change
    }
        $data = 0;
}
echo $data;
?>

Answer 2

"您正在覆盖$ arr_foundits上的$ arr_foundits = array_push（$ arr_foundits，$ it-> id）;。删除$ arr_foundits = array_push array_push a array_push dy array not and arnay not array and int and int。"穆萨