Php SQL命令的使用条件

我在编写PHP代码时遇到了一个问题。这就是错误：

!！警告：mysqli_fetch_数组期望参数1为 mysqli_结果，布尔值在

这是我的PHP代码：

if(!empty($_GET['KEY'])){
    $CAT=$_GET['CAT'];
    $KEY=$_GET['KEY'];

    $content = mysqli_query($con,"SELECT * FROM model WHERE cat1 ='$CAT' UNION
                                SELECT * FROM model WHERE cat2 ='$CAT' UNION
                                SELECT * FROM model WHERE cat3 ='$CAT' UNION
                                SELECT * FROM model WHERE keyword1 ='$KEY' UNION
                                SELECT * FROM model WHERE keyword2 ='$KEY' UNION
                                SELECT * FROM model WHERE keyword3 ='$KEY' UNION
                                SELECT * FROM model WHERE keyword4 ='$KEY' UNION
                                SELECT * FROM model WHERE keyword5 ='$KEY' 
                                ORDER BY rate DESC;");

    }else {
$CAT=$_GET['CAT'];
$content = mysqli_query($con,"SELECT * FROM model WHERE cat1 ='$CAT' UNION
                            SELECT * FROM model WHERE cat2 ='$CAT' UNION
                            SELECT * FROM model WHERE cat3 ='$CAT' 
                            ORDER BY rate DESC;");
}

    $output = array();
while($row = mysqli_fetch_array($content)){
     $record = array();
     $record['rate'] = $row['rate'];
     $output[] = $record;

}

---

当然，我承认我是PHP初学者，希望大家能帮助我。这表明SQL不会返回任何记录。$content变量为空

if (!empty($content)) {
 while($row = mysqli_fetch_array($content)){
     $record = array();
     $record['rate'] = $row['rate'];
     $output[] = $record;
 }
}

这将帮助您

可能重复的