基於兩個輸入文件的數據庫更新

Question

我正在使用具有多個輸入文本和兩個輸入文件的表單來更新數據庫表。

我設置了兩個隱藏着圖像名稱作爲值的輸入。

如果圖像設置爲imageFile，我需要更新該數據，對於imageFileBanner也是如此。

如果兩者都設置，我需要更新所有字段。

如何更新只有一個圖像？

if(!empty($_FILES['imageFile']['name']) && empty($_POST['imageIsSet']) !empty($_FILES['imageFileBanner']['name'])){ 
     $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', logo='%s', vizibil='%d', ordine='%d', banner='%s' WHERE id='%d'", $nume, $img, $status, $ordine, $img_banner, $id)); 

    }else{ 
     $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', vizibil='%d', ordine='%d' WHERE id='%d'", $nume, $status, $ordine, $id)); 

    } 

有了這個說法，如果我選擇ImageFile一個文件，它會爲我設置爲imageFileBanner空的更新，反之亦然。

這裏是我的形式

 <form id="editTabForm" name="editTabForm" enctype="multipart/form-data"> 
     <div class="row"> 
      <div class="col-sm-12 col-md-12"> 
      <div class="info"><i class="fa fa-info"></i> Completati numele tab-ului apoi selectati o imagine pentru afisare.</div> 
      </div> 
     </div> 
     <div class="row"> 
      <div class="col-sm-6 col-md-6"> 
      <div class="control-group form-group"> 
       <label for="tab-name">Nume tab</label> 
       <input class="form-control" type="text" name="tab-name" id="tab-name" placeholder="nume tab" value="{$tab.nume}" /> 
      </div> 
      <div class="control-group form-group"> 
       <label>Status</label> 
       <select name="status" class="category-tree"> 
       {if $tab.vizibil == 1} 
       <option value="1" selected="selected">Activ</option> 
       <option value="0">Inactiv</option> 
       {else} 
       <option value="1">Activ</option> 
       <option value="0" selected="selected">Inactiv</option> 
       {/if} 
       </select> 
      </div> 
      <div class="control-group form-group"> 
       <label for="tab-order">Ordinea afisarii in meniul orizontal</label> 
       <input class="form-control" type="text" name="tab-order" id="tab-order" style="width:20%;" value="{$tab.ordine}"/> 
      </div> 
      </div> 
      <div class="col-sm-6 col-md-6"> 
      <div class="control-group form-group"> 
       <label for="image">Selectati o imagine pentru afisarea tab-ului</label> 
       <div id="imagePreview"> 
       {if $tab.logo} 
       <img src="{$website}/media/taburi/{$tab.logo}" class="imagePreview thumbnail" style="max-width:50px; max-height:50px;" /> 
       {else} 
       <img src="/admin/mmadmin/template/assets/img-upload-50_50.png" class="imagePreview thumbnail" style="max-width:50px; max-height:50px;" /> 
       {/if} 
       </div> 
       <input type="file" name="imageFile" id="image" /> 
      </div> 
      <div class="control-group form-group"> 
       <label for="image">Selectati o imagine default pentru tab</label> 
       <div id="imagePreviewBanner"> 
       {if $tab.banner} 
       <img src="{$website}/media/cat/{$tab.banner}" class="imagePreviewBanner thumbnail" style="max-width:150px; max-height:150px;" /> 
       {else} 
       <img src="/admin/mmadmin/template/assets/img-upload-150_150.png" class="imagePreviewBanner thumbnail" style="max-width:150px; max-height:150px;" /> 
       {/if} 
       </div> 
       <input type="file" name="imageFileBanner" id="imageBanner" /> 
       <input type="hidden" name="id" value="{$tab.id}" /> 
       <input type="hidden" name="imageIsSet" value="{$tab.logo}" /> 
       <input type="hidden" name="bannerIsSet" value="{$tab.banner}" /> 
      </div> 
      </div> 
     </div> 
     <div class="modal-footer"> 
      <button type="button" class="btn btn-default" data-dismiss="modal">Anuleaza</button> 
      <button type="button" class="btn btn-primary" id="update-tab" data-id="{$tab.id}">Salveaza</button> 
     </div> 
     </form> 

Answer 1

解決::
有沒有什麼更好的辦法？

if(!empty($_FILES['imageFile']['name'])){ 

      $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', logo='%s', vizibil='%d', ordine='%d' WHERE id='%d'", $nume, $img, $status, $ordine, $id)); 

     }else if(!empty($_FILES['imageFileBanner']['name'])){ 

      $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', vizibil='%d', ordine='%d', banner='%s' WHERE id='%d'", $nume, $status, $ordine, $img_banner, $id)); 

     }else if(!empty($_FILES['imageFile']['name']) && !empty($_FILES['imageFileBanner']['name'])){ 

      $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', logo='%s', vizibil='%d', banner='%s', ordine='%d' WHERE id='%d'", $nume, $img, $status, $ordine, $img_banner, $id)); 

     }else{ 

      $sql = $database->update(sprintf("UPDATE taburi SET nume='%s', vizibil='%d', ordine='%d' WHERE id='%d'", $nume, $status, $ordine, $id)); 

     }