Question

我有一个表“pharma_companies”，其中有一个名为“is_approved”的列名。如果is_approved列包含0值

，我想应用样式表

 <select name="current_company" id="current_company" class="validate[required]">
             <option value="">Select Company</option>
           <?php
               $current_sql = "SELECT pharma_companies FROM pharma_companies ORDER BY pharma_companies";
               $currenex = mysqli_query($db,$current_sql);

              while($pharama = mysqli_fetch_assoc($currenex))
              {
                   if($pharama['is_approved'] == '0')
                   {                                                
                       $style = 'style = "color:red;"';
                   }
                   else
                   {                                                
                       $style = '';
                   }



                  echo '<option value="'.$pharama['pharma_companies'].'" '.$style.'>'.$pharama['pharma_companies'].'</option>';
              }
          ?>
       </select>

但它不起作用。提前谢谢。

Answer 1

您实际上从不从数据库中抓取is_approved列。

将您的SQL更改为：

$current_sql = "SELECT pharma_companies, is_approved FROM pharma_companies ORDER BY pharma_companies";

Answer 2

1）您需要获取的第一件事是批准的字段

$current_sql = "SELECT pharma_companies, is_approved FROM pharma_companies ORDER BY pharma_companies";

2）尝试条件为

if($pharama['is_approved'] == '0' || $pharama['is_approved']==0 )

当is_approved等于0时，样式不适用

2 个答案: