返回值从PHP函数

Question

我试图返回值作为数组，但未能阵列，

函数调用

$rates = getamount($Id, $days, $date, $code); 

功能

function getamount($Id, $days, $date, $code) { 
    // fetch data from database 
    // run whileloop 
    while() { 
     // Do other stuff 
     // Last get final output values 
     $rates['id'] = $finalid; 
     $rates['days'] = $finaldays; 
     // more and more 
     $rates['amount'] = $finalamount; 
     //If echo any variable here, can see all values of respective variable. 
    } 
    //If echo any variable here, can only see last value. 
    return $rates; 
} 

而在去年的功能外（需要这因此也将变量加载到会话中也作为阵列）

$compid = $rates['id']; 
$totaldays = $rates['days']; 
$totalamount = $rates['amount']; 

试过几个解决方案，从SO，但不能让我的头围绕这个

Answer 1

马丁是对他的回答，我只想补充一些解释。

在你while循环，要覆盖同一阵列，所以在最后，它将包含从数据库查询结果的最后一行的数据。

要解决这个问题，您需要一个数组数组。每个$rate数组代表从数据库中一行，$rates是那些行的数组：

function getamount($Id, $days, $date, $code) 
{ 
    // fetch data from database 

    $rates = []; // initialize $rates as an empty array 

    while (whatever) { 
     $rate = []; // initialize $rate as an empty array 

     // fill $rate with data 
     $rate['id'] = $finalid; 
     $rate['days'] = $finaldays; 
     // more and more 
     $rate['amount'] = $finalamount; 

     // add $rate array at the end of $rates array 
     $rates[] = $rate; 
    } 

    return $rates; 
} 

现在尝试检查有什么$rates数组内：

$rates = getamount($Id, $days, $date, $code); 
var_dump($rates); 

为了摆脱$rates阵列的数据，你需要一个循环再次（与你创建它的方式相同）：

foreach ($rates as $rate) { 
    var_dump($rate); 

    // now you can access is as you were trying it in your question 
    $compid = $rate['id']; 
    $totaldays = $rate['days']; 
    $totalamount = $rate['amount']; 
} 

Answer 2

你想要做的事，如：

function getamount($Id, $days, $date, $code) { 
    $rates = []; 
    while (whatever) { 
     $rate = []; 
     $rate['id'] = $finalid; 
     $rate['days'] = $finaldays; 
     // more and more 
     $rate['amount'] = $finalamount; 

     $rates[] = $rate; 
    } 

    return $rates; 
}