當沒有圖像時顯示默認圖像

Question

當CMS中沒有圖像時，我想顯示默認圖像。

到目前爲止，我得到這個代碼：

<li><img alt="" class="img-responsive" src=" 
         <? if($artikel[0]['images'] == ''){ ?> 
           images/defaultimg.jpg 
         <? }else{ ?> 
         cms/images/<? 

           $plaatje = $artikel[0]['images']; 
           $plaatje = explode('\/', $plaatje); 
           $plaatje = $plaatje[1]; 
           $plaatje = explode('"', $plaatje); 
           $plaatje = $plaatje[0]; 
           echo $plaatje; 

         } 
         ?>"></li> 

但這不顯示圖像。我知道圖像在那裏，所以我認爲輸出的html代碼有些問題。

有人看到了嗎？根據要求

感謝

輸出代碼：

<img alt="" class="img-responsive" src=" 
                   cms/images/"> 

我得到了正確的代碼：

   <?php 
        $image = "images/defaultimg.jpg"; 

        $pl = $artikel[0]['images']; 
        $pl = explode('\/', $pl); 
        $pl = $pl[1]; 
        $pl = explode('"', $pl); 
        $pl = $pl[0]; 
        $pl = explode('.', $pl); 
        $pl = $pl[1]; 

        echo $pl; 

        if($pl == 'jpg'){ //Check if string is empty 
         $plaatje = $artikel[0]['images']; 
         $plaatje = explode('\/', $plaatje); 
         $plaatje = $plaatje[1]; 
         $plaatje = explode('"', $plaatje); 
         $plaatje = $plaatje[0]; 
         // If the string supplied in the if condition isn't empty i assume that there's an image 
         // so i prepend the imagefolder as source 
         $image = 'cms/images/'.$plaatje; 
        } 

        $html = '<li><img alt="" class="img-responsive" src="'.$image.'"></li>'; 

        echo $html 

       ?> 

Answer 1

爲什麼不分開，如果從您的標記else邏輯，而不是混亂起來在該來源屬性上。例如：

<?php 

$image = ''; 
if($artikel[0]['images'] == '') { 
    $image = 'images/defaultimg.jpg'; // wrap it with quotes 
} else { 
    $plaatje = $artikel[0]['images']; 
    $plaatje = explode('\/', $plaatje); 
    $plaatje = $plaatje[1]; 
    $plaatje = explode('"', $plaatje); 
    $image = 'cms/images/' . $plaatje[0]; // append the filename with that particular path 
} 

?> 

<li><img alt="" class="img-responsive" src="<?php echo $image; ?>"></li> 

Answer 2

嘗試

此

<li><img alt="" class="img-responsive" src=" 
         <?php if(empty($artikel[0]['images'])){ 
           echo "images/defaultimg.jpg"; 
         }else{ 


           $plaatje = $artikel[0]['images']; 
           $plaatje = explode('\/', $plaatje); 
           $plaatje = $plaatje[1]; 
           $plaatje = explode('"', $plaatje); 
           $plaatje = $plaatje[0]; 
           echo "cms/images/".$plaatje; 

         } 
         ?>"></li> 

Answer 3

AS @Ghost說，從圖separe邏輯，另一種解決方案更乾淨：

<?php 

    if (empty($artikel[0]['images']){ 
     $artikel[0]['images'] = 'images/defaultimg.jpg'; 
    }else{ 
     $artikel[0]['images'] = 'cms/images/' . basename($artikel[0]['images']); 
    } 

?> 

<li><img alt="" class="img-responsive" src="<?php echo $artikel[0]['images']; ?>"></li> 

Answer 4

的問題似乎是在佈局你的代碼，正如你在輸出中清楚地看到的那樣，你得到了換行符，可能還有更多未知的未知數。

更好地重構它，所以像這樣。這一切

<?php 
    $image = "images/defaultimg.jpg"; 
    if($artikel[0]['images'] != ''){ //Check if string is empty 
     $plaatje = $artikel[0]['images']; //Do your magic here 
     $plaatje = explode('\/', $plaatje); 
     $plaatje = $plaatje[1]; 
     $plaatje = explode('"', $plaatje); 
     $plaatje = $plaatje[0]; 
     // If the string supplied in the if condition isn't empty i assume that there's an image 
     // so i prepend the imagefolder as source 
     $image = 'cms/images/'.$plaatje; 
    } 

    $html = '<li><img alt="" class="img-responsive" src="$image"></li>'; 

    echo $html 

?> 

首先將提高你的代碼的所有你能不頭痛你想在那裏做什麼看的可讀性和第二。

這樣你不必關心2個案例，你有一個案例需要照顧，其餘部分由默認變量處理。

但是請注意，在您的$ artikel [0] ['images']中也可能存在一些錯誤，所以最終您會添加一個錯誤字符串到您的imagesource中，而這並不是你想要的。

無論何時向用戶顯示某些內容，都應檢查該文件是否存在。你再一次得到了你未知的未知數。

您的變量的賦值會再看看這樣

if(file_exists('cms/images/'.$plaatje)){ 
    $image = 'cms/images/'.$plaatje; 
} 

這將防止上述問題，既然你有退出的一個點，該代碼工作的人之後，你會很感激。

Answer 5

我得到了正確的代碼。

     <?php 
          $image = "images/defaultimg.jpg"; 

          $pl = $artikel[0]['images']; 
          $pl = explode('\/', $pl); 
          $pl = $pl[1]; 
          $pl = explode('"', $pl); 
          $pl = $pl[0]; 
          $pl = explode('.', $pl); 
          $pl = $pl[1]; 

          echo $pl; 

          if($pl == 'jpg'){ //Check if string is empty 
           $plaatje = $artikel[0]['images']; 
           $plaatje = explode('\/', $plaatje); 
           $plaatje = $plaatje[1]; 
           $plaatje = explode('"', $plaatje); 
           $plaatje = $plaatje[0]; 
           // If the string supplied in the if condition isn't empty i assume that there's an image 
           // so i prepend the imagefolder as source 
           $image = 'cms/images/'.$plaatje; 
          } 

          $html = '<li><img alt="" class="img-responsive" src="'.$image.'"></li>'; 

          echo $html 

         ?>