爲什麼如果statings失敗並帶有兩個參數

Question

Im包括一個html塊，並且包含一些基於if語句的echo語句。如果我使用

<? 
    if($rows['business_type'] !== "Hotel" || "Bed & Breakfast") 
    { 
?> 


    <div id="mini_titles"><h2>Opening times</h2></div> 

    <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Monday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_monday_from']; ?> - 
<?php echo $rows1['opening_monday_to']; ?> 
       </p> 
      </div> 
    </div> 
    <div id="complete_info_wrapper"> 
     <div id="complete_info_title"><p>Tuesday</p></div> 
     <div id="complete_info_details_opening"> 
      <p> 
<?php echo $rows1['opening_tuesday_from']; ?> - 
<?php echo $rows1['opening_tuesday_to']; ?> 
      </p> 
     </div> 
     </div> 

     <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Wednesday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_wednesday_from']; ?> - 
<?php echo $rows1['opening_wednesday_to']; ?> 
       </p> 
      </div> 
     </div> 

     <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Thursday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_thursday_from']; ?> - 
<?php echo $rows1['opening_thursday_to']; ?> 
       </p> 
      </div> 
     </div> 

     <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Friday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_friday_from']; ?> - 
<?php echo $rows1['opening_friday_to']; ?> 
       </p> 
      </div> 
     </div> 

     <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Saturday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_saturday_from']; ?> - 
<?php echo $rows1['opening_saturday_to']; ?> 
       </p> 
      </div> 
     </div> 

     <div id="complete_info_wrapper"> 
      <div id="complete_info_title"><p>Sunday</p></div> 
      <div id="complete_info_details_opening"> 
       <p> 
<?php echo $rows1['opening_sunday_from']; ?> - 
<?php echo $rows1['opening_monday_to']; ?> 
       </p> 
      </div> 
     </div> 
<?php 
    } 

?> 

的，如果statment工作只有一個，即：

if($rows['business_type'] !== "Hotel"){} 

但只要我添加||，or，&&，或and它失敗，無論顯示第二。

我確定它有些愚蠢和簡單的即時通訊丟失。

請幫助，謝謝。

Answer 1

在，如果如果條件$rows['business_type'] !== "Hotel"爲真，那麼條件將永遠是正確的。因爲if ("Bed & Breakfast")是true。所以true || true將是true。

它應該是 -

if($rows['business_type'] !== "Hotel" || $rows['business_type'] != "Bed & Breakfast") 

或者你也可以做到 -

if (!in_array($rows['business_type'], array("Hotel", "Bed & Breakfast"))) 

Answer 2

嘗試：

if($rows['business_type'] !== "Hotel" || $rows['business_type'] !== "Bed & Breakfast") 

Answer 3

它，因爲它有一些錯字：

變化

<? 
if($rows['business_type'] !== "Hotel" || "Bed & Breakfast") 
// This //statement will always return true, because even if first condition 
// satisfies or not, 'Bed & Breakfast' will always return true (it is a 
//string which is not empty. 
{ 
?> 
// TRUE || TRUE --> TRUE 
// FALSE || TRUE --> TRUE 
// TRUE || FALSE --> TRUE 
// FALSE || FALSE --> FALSE 

到

<? 
if($rows['business_type'] !== "Hotel" || $rows['business_type'] !== "Bed & Breakfast") // PUT YOUR CONDITION CORRECTLY. 
{ 
?> 

Answer 4

通過這種方式無法與兩個值進行比較。使用in_array：

if (!in_array($rows['business_type'], array("Hotel", "Bed & Breakfast"))) {...} 

或者usign ||

if ($rows['business_type'] !== "Hotel" || $rows['business_type'] !== "Bed & Breakfast") {...} 

Answer 5

「牀&早餐」 是一個字符串不是一個比較，它總是返回true。所以x or true總是等於true。這是你的if語句的問題。如你預期 這可能會實現：

if($rows['business_type'] !== "Hotel" || $rows['business_type'] !== "Bed & Breakfast") 

Answer 6

嘗試陣列方法

if (in_array(array('Hotel', 'Bed'), $rows['business_type'])) { 
    echo "'Bed' was found\n"; 
}