Php Sqli数组问题

我试图将数据库中的值推送到数组中，但收到一条警告

 $goal=array();
                //goal
                $sql="";
                //!mysqli_query($con,$sql))
                $result = mysqli_query($con,"SELECT * FROM metrics where ini_name = '$ini'");/*table name*/
                while($row = mysqli_fetch_array($result))
                {
                                array_push($goal,$row['metric_desc']);/*column name*/
                                array_push($goal_id,$row['metric_id']);
                }
                                $matrix=array();
                //goal
                $sql="";

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您不需要2个数组。您可以（也应该）将id存储为结果数组的键

Warning: array_push() expects parameter 1 to be array, null given in C:\xampp\htdocs\xampp\Testing\Int\sq2.php on line 195

或者，借助于，整体您的代码可以更改为

while($row = mysqli_fetch_array($result))
{
    $goal[$row['metric_id']] = $row['metric_desc']);
}

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什么是

$goal\u id

？您尚未在发布的代码中将其定义为数组。请尝试将目标和目标id初始化为数组…类似于

$goal=array（）@YueYuan谢谢你解决了这个问题problem@MichaelBerkowski谢谢你，迈克尔
$goal = $db->getIndCol('metric_id',"SELECT * FROM metrics where ini_name = ?i",$ini);