Php 可缓存致命错误：类mysqli_result的对象无法转换为字符串

无法在第27行将类mysqli_result的对象转换为字符串 有人能解释一下如何修复这个错误吗，错误描述，我试图为一个朋友创建一个虚拟PHP商店，我试图做的是显示一个报价列表及其描述，我试图做的是从数据库获取数据以列出报价，代码一直在说这个错误，有人知道如何修复它吗

  <?php 
//Offer Wall
// Put your CPA Networks Virtual Currency Widget after the End of this first PHP
//Segment
include "mysqli_config.php";
?>
<table>
<tr>
<th>Offer Name</th>
<th>Description</th>
<th>Payout</th>
</tr>
</table>
<?php
$offername= "SELECT * FROM offers WHERE active = 1";
$exec= $mysqli->query($offername);
$array = array($exec);
if (mysqli_num_rows($exec) == 0){
echo "No Offers Yet";
}else{
while (list($x, $y, $z, $a) = $array){
echo " <tr>\n " .
" <td><a href=\"click.php?=$a\">Click Here to Open Offer</a></td>\n" .
" <td>$z</td>\n" .
" <td>$y</td>\n" .
" <td>$x</td>\n";
}}
?>

报盘名称
描述
支出

x，y，z，a和0，1，2，3必须更改以获得所需的结果，但这就是想法。错误在这一行：$array=array（$exec）

报盘名称
描述
支出
签出此文件-
<?php 
//Offer Wall
// Put your CPA Networks Virtual Currency Widget after the End of this first PHP
//Segment
include "mysqli_config.php";
?>
<table>
<tr>
<th>Offer Name</th>
<th>Description</th>
<th>Payout</th>
</tr>
</table>
<?php
$offername= "SELECT * FROM offers WHERE active = 1";
$exec= $mysqli->query($offername);
//$array = array($exec);
if (mysqli_num_rows($exec) == 0){
    echo "No Offers Yet";
}else{
   while ($array=mysqli_fetch_row($exec)){
     $a=$array[3];
     $x=$array[0];
     $y=$array[1];
     $z=$array[2];
     echo " <tr>\n " .
        " <td><a href=\"click.php?=$a\">Click Here to Open Offer</a></td>\n" .
        " <td>$z</td>\n" .
        " <td>$y</td>\n" .
        " <td>$x</td>\n";
}}