为什么我的if ... else语句无法解决我的错误以及如何修复它？

Question

这里我如何调用我的查询，我想使用简单的if ... else语句，但我在某个地方犯了错误，有人可以解释我在哪里以及为什么？

 $result = mysqli_query($con, $sql) or die(mysqli_error($con,$sql));
    echo mysqli_error($con);
    $number=mysqli_num_rows($result);
    while ($row= mysqli_fetch_array($result)){
        $brand = $row['brand'];
        $model = $row['model'];
        $reg_num = $row['reg_num'];
        $horse_powers = $row['horse_powers'];
        $color = $row['color'];
        echo $number;
        if($number<1){
            echo "No rented cars for this period !";
        }
        else{
        echo $brand;
        }

    }

Answer 1

请尝试以下代码：

    $number = 0;
    $result = mysqli_query($con, $sql);
    if (!$result) 
         exit(mysqli_error($con));
    else
         $number = mysqli_num_rows($result);
    if ($number==0) {
        echo "No rented cars for this period !";
    } else {
        while ($number){
            $row = mysqli_fetch_row($result);
            $brand = $row['brand'];
            $model = $row['model'];
            $reg_num = $row['reg_num'];
            $horse_powers = $row['horse_powers'];
            $color = $row['color'];
            echo $brand;
            $number--;
        }
    }

Answer 2

此if语句的输出始终为false。

if($number<1){
   echo "No rented cars for this period !";
}
else{
   echo $brand;
}

如果$number < 1，则不会运行while循环。因此，如果$number < 1它将永远不会到达if语句。