Question

我正在为我的网站上的新帐户启用激活队列。

系统将通过迭代每个新用户帐户详细信息的数组来工作，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数组中，我使用下面的代码，但是我收到错误“array_push（）期望参数1为数组，给定为null”。我不知道为什么会造成这种情况，我已经尝试了以前建议的各种事情。提前谢谢。

<?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

“你在$ arr_foundits = array_push上覆盖$ arr_foundits（$ arr_foundits，$ it-＆gt; ID）;.删除$ arr_foundits =，因为array_push不返回数组而是返回int。”芭蕉

“array_push（）期望参数1为数组”错误消息

2 个答案: