PHP：我自定义了一个函数，用来查取数据库内容，但应该如何返回 mysqli_fetch_array结果的数组

index.php:


<?php
    include('index.ajax.php')
    ?>



<html>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    </html>

您好，企业管理员:<?php get_admin_username() ?></br>

企业名称：</br>
<?php 
    function get_admin_username(){
        global $link;
        $admin_information_array = get_from_database("users","isadmin=1")
        
        echo mysqli_real_escape_string($link,$admin_information_array['username']);
    }
    ?>

index.ajax.php：

<html>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    </html>


<?php
    include('mysql_link.php');
    ?>


<?php
    function get_from_database($form,$condition){

        global $link;
        $sql = "SELECT * from $form where $condition";
        echo $sql;
        $query = mysqli_query($link,$sql);
        $result = mysqli_fetch_array($query);
        return $result;
    //echo mysqli_real_escape_string($link,$admin_username['username']);
        }
    ?>

运行结果：

Parse error: syntax error, unexpected 'echo' (T_ECHO) in C:\PHPCUSTOM\qiyeguanwang\admin\index.php on line 19

为何无结果

$admin_information_array = get_from_database("users","isadmin=1") 後面需要放 分號 ;