在PHP中回显两个变量的乘积

Question

我有以下代码：

    $sel_referrals1="SELECT SUM(amount) as da_sum FROM topup WHERE user_id IN(SELECT t1.referree
FROM referrals AS t1
LEFT JOIN referrals AS t2 ON t2.referrer = t1.referree
LEFT JOIN referrals AS t3 ON t3.referrer = t2.referree
LEFT JOIN referrals AS t4 ON t4.referrer = t3.referree
LEFT JOIN referrals AS t5 ON t5.referrer = t4.referree
LEFT JOIN referrals AS t6 ON t6.referrer = t5.referree
LEFT JOIN referrals AS t7 ON t7.referrer = t6.referree
LEFT JOIN referrals AS t8 ON t8.referrer = t7.referree
LEFT JOIN referrals AS t9 ON t9.referrer = t8.referree
LEFT JOIN referrals AS t10 ON t10.referrer = t9.referree
LEFT JOIN referrals AS t11 ON t11.referrer = t10.referree
WHERE t1.referrer = '{$_SESSION['user']}')
AND YEAR(date_time) = YEAR(CURRENT_DATE()) 
AND MONTH(date_time) = MONTH(CURRENT_DATE())";


    $selected1=mysqli_query($conn,$sel_referrals1);
    $sel_level1_array=mysqli_fetch_row($selected1);
    $level1=number_format($sel_level1_array[0],2);

$level1计算结果为1260.00.I有另一个变量$rate1，其设置如下：$rate1=0.02.I需要将$level1与$rate1相乘。但是，当代码执行$level1 * $rate1的乘法时，得到的结果是0.02。

我的预期结果是25.20。我做错了什么？

Answer 1

$level1的数据类型是3操作中的字符串(它将被转换为值等于1的整数)。

Answer 2

我想您使用$level1值声明了number_format(1260,2)！它的echo $level1."<br>";给你1,260.00！你应该宣布它是纯数字！

Answer 3

百事大吉。

试试这个：

<?php
 $rate1=0.02;
 $level1=1260;
 echo "Level1 = ".$level1.'<br>';
 echo "Rate1 = ".$rate1.'<br>';
 echo "Format Number using variables = ".number_format($level1 * $rate1,2).'<br>';
 echo "Format Number using numbers = ".number_format(1260 * 0.02,2);
?>