Question

我在编写PHP代码时遇到了问题。这是错误：

（！）警告：mysqli_fetch_array（）期望参数1为 mysqli_result，在...中给出的布尔值。

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;

}

课程我承认我是PHP的初学者并且希望你能帮助我，感谢大家。

Answer 1

这表明，SQL不会返回任何记录。您$content变量为空。

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

这会对你有所帮助

使用SQL命令的条件

1 个答案: