For循環在第一次迭代後不會繼續

所以我寫了for循環，如下所示，出於某種原因，第一次循環後，循環停止。在這種情況下，我試圖通過發送值4，並且它失敗了從checkRes函數檢查字符串「true」的if語句。就是那個函數所做的就是返回字符串。這是我對循環的理解，它將繼續循環直到頂部的語句被滿足，或者退出。我在下面做錯了什麼？
For循環在第一次迭代後不會繼續

// assume $avaliable = 4; 
for ($i=$avaliable;10>$i;$i++) { 
    $check = checkRes($i, $people_no, $booking_date); 
     if ($check === "true") { 
      switch($i) { 
       case 0: $newResTime = "6 PM"; break; 
       case 1: $newResTime = "6:30 PM"; break; 
       case 2: $newResTime = "7 PM"; break; 
       case 3: $newResTime = "7:30 PM"; break; 
       case 4: $newResTime = "8 PM"; break; 
       case 5: $newResTime = "8:30 PM"; break; 
       case 6: $newResTime = "9 PM"; break; 
       case 7: $newResTime = "9:30 PM"; break; 
       case 8: $newResTime = "10 PM"; break; 
       case 9: $newResTime = "10:30 PM"; break; 
       case 10: $newResTime = "11 PM"; break; 
      } 
      // Replace next line with your return from the chatbot 
      echo "We're sorry, that time isn't avaliable, but a reservation at $newResTime has been made!"; 
      exit; 
     } 
    }

來源

2017-10-13 Adam McGurk

刪除您'退出;'' – aldrin27

$ check'是真實的，所以它進入'if'並在其上你有一個'退出;'所以這使得環路停止 – GEPD

這段代碼很瘋狂。爲什麼你的check res函數只返回一個bool？爲什麼不直接返回時間？迭代指數與某人的預訂有什麼關係？ – Rafael

您正在用exit爲「默認」的解決方案。但它總是終止程序的執行（因爲這是exit所做的）。

而是做到這一點

switch($i) { 
    case 0: $newResTime = "6 PM"; break; 
    case 1: $newResTime = "6:30 PM"; break; 
    case 2: $newResTime = "7 PM"; break; 
    case 3: $newResTime = "7:30 PM"; break; 
    case 4: $newResTime = "8 PM"; break; 
    case 5: $newResTime = "8:30 PM"; break; 
    case 6: $newResTime = "9 PM"; break; 
    case 7: $newResTime = "9:30 PM"; break; 
    case 8: $newResTime = "10 PM"; break; 
    case 9: $newResTime = "10:30 PM"; break; 
    case 10: $newResTime = "11 PM"; break; 
    default: echo "..."; break;

來源

2017-10-13 02:48:15

這不起作用，因爲程序從未將它傳遞給switch語句，因爲$ check沒有計算爲真，所以它從來沒有使它在if語句中退出。即使它確實如此陳述，我也沒有看到回聲 –

For循環在第一次迭代後不會繼續

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