在 PHP 中生成动态开关结构

Generate a dynamic switch structure in PHP

有没有办法创建动态生成的 switch 语句?我会进一步解释,我有一个 table 和所有可能的硬币。每个人都有一个 table 和他们自己的硬币。如果你点击一个硬币,php 中会发生一个新的动作,你将被直接转到 index.php?actie=Ripple

我的代码:

case "Ripple":
    if ($_SESSION["name"] == "d"){
        $dataFromTransactions = $pol->toonAllesD("XRP/BTC");
    }
    else{
        $dataFromTransactions = $pol->toonAlles("XRP/BTC");
    }
    Uitvoer::toonDeRippleTable($dataFromTransactions,"Ripple");
    break;
case "LiteCoin":
    if ($_SESSION["name"] == "d"){
        $dataFromTransactions = $pol->toonAllesD("LTC/BTC");
    }
    else{
        $dataFromTransactions = $pol->toonAlles("LTC/BTC");
    }
    Uitvoer::toonDeRippleTable($dataFromTransactions,"LiteCoin");
    break;

有没有我需要做的方法:

$alleCoins = $pol->getAlleCoinsYouBuyed()
foreach($alleCoins as $coinInfo){
case $coinInfo->Coinname :
if ($_SESSION["name"] == "d"){
        $dataFromTransactions = $pol->toonAllesD($coinInfo->Market);
    }
    else{
        $dataFromTransactions = $pol->toonAlles($coinInfo->Market);
    }
    Uitvoer::toonDeRippleTable($dataFromTransactions,$coinInfo->Coinname);
    break;
}

此代码必须在 index.php

这就是解决方案

if ($_SESSION["name"] == "d"){
    $alleCoins = dataBaseTrade::getPoloniexInstantie()->getAlleCoinsYouBuyedD();
}
else {
    $alleCoins = dataBaseTrade::getPoloniexInstantie()->getAlleCoinsYouBuyed();
}
$searchNameArray = json_decode(json_encode($alleCoins), true);
$searchNameArray = array_column($searchNameArray, "Coinname" );
switch ($actie){
    case ($actie != "home" || "user"):
        $valueOfThePlaceInTheArray = array_search($actie,$searchNameArray);
        $thisCoin = $alleCoins[$valueOfThePlaceInTheArray];
        if ($_SESSION["name"] == "d"){
            $dataFromTransactions = $pol->toonAllesD($thisCoin->Market);
        }
        else{
            $dataFromTransactions = $pol->toonAlles($thisCoin->Market);
        }
        Uitvoer::toonDeRippleTable($dataFromTransactions,$thisCoin->Coinname);
        break;
}