功能名稱必須是PHP函數的計算串

Question

我試圖做一個計算在我的代碼，並將其與

Fatal error: Function name must be a string on line 3 


function compound_interest($compound_interest, $compound_interest_f, $investment, $interest_rate, $years, $compoundMonthly, $future_value_f, $future_value) { 
    if (isset($compoundMonthly)) { 
     $compoundMonthly = 'Yes'; 
     $compound_interest = ($investment(1 + $interest_rate/12)^(12 * $years) - $investment); 
     $compound_interest_f = '$' . number_format($compound_interest, 2); 
     return $compound_interest_f; 
    } else { 
     $compoundMonthly = 'No'; 
     $future_value = ($future_value + ($future_value * $interest_rate * .01)); 
     $future_value_f = '$' . number_format($future_value, 2); 
     return $future_value_f; 
    } 
} 

一個錯誤出現只有在該行的代碼試圖做一個計算。不打印字符串。我在這裏錯過的任何東西？

Answer 1

您錯過了用於乘法的*運算符;沒有這個，你試圖撥打$investment作爲一個函數，但它包含一個數字。你需要使用pow()函數來取冪; ^是按位異或。

$compound_interest = $investment * pow((1 + $interest_rate/12), 12 * $years) - $investment;