Question

HEY GUYS我从复选框中删除了一个数据，这是代码：

To Show Checkboxes from Different tables:
  <?php
          $query23 = "SELECT * from tbl_occasions";
         $result23 = mysqli_query($conn, $query23)
            or die("Error in query: ". mysqli_error($conn));

        while ($row = mysqli_fetch_assoc($result23))
        {
         ?>
         <div class="col-md-12">
         <label class="checkbox" for="checkboxes"> 
                <input type="checkbox" name="reg_occassions[]"  value="<?php echo $row['name'];?>"><?php echo $row['name']?> 
         </label>
            </div>
         <?php
        }   
             ?>


  <?php
          $query24 = "SELECT * from tbl_categories";
         $result24 = mysqli_query($conn, $query24)
            or die("Error in query: ". mysqli_error($conn));

        while ($row = mysqli_fetch_assoc($result24))
        {
         ?>
         <div class="col-md-12">
         <label class="checkbox" for="checkboxes"> 
                <input type="checkbox" name="reg_categories[]"  value="<?php echo $row['name'];?>"><?php echo $row['name']?> 
         </label>
            </div>
         <?php
        }   
         ?>

使用implode插入数据库：

 $occasions = implode( ';' , $_POST['reg_occassions'] );
 $categories= implode( ';' , $_POST['reg_categories'] );

 $query = "INSERT INTO tbl_flower (flower_type, website_description,florist_description,name,image,enabled,florist_choice,categories,occasions)
             VALUES ('$flowertype', '$websitedescription', '$floristdescription','$name', '$upfile', '$enabled','$floristchoice', '$categories','$occasions')";

这很有效。我做了结果....所以输入需要是例如：

flower type       Occasions                            Categories

12 Red Roses      Birthday;Anniversary;Valentines      Bouqets
1 Red Rose        Birthday;Valentines               Bouqets;Flowerarrangment

现在我需要做一个下拉列表。如果用户选择周年纪念日，则仅显示12个红玫瑰，但如果用户选择生日，则需要显示两个花。

我尝试了这段代码，但它无法正常工作。他只向我展示了有一种类型场合的花，例如24朵玫瑰只有生日场合，用户选择生日。

这是我试过的代码：

  $query = "SELECT * FROM tbl_occasions";
   $result= mysqli_query($conn, $query)
    or die("Error in query: ". mysqli_error($conn));
    $option ="";

    while ($row = mysqli_fetch_assoc($result)){  

         $option .= '<option  name = "'.$row['name'].'" value = "'.$row['name'].'">'.$row['name'].'</option>';

    }

我在下拉列表中显示它。

  if(isset($_POST['submit'])){

    $_SESSION['selected'] = $_POST['reg_occasion'];





            $query2 = "SELECT occasions from tbl_flower";

            $result2= mysqli_query($conn, $query2)
                or die("Error in query: ". mysqli_error($conn));

            while ($row = mysqli_fetch_array($result2)){
                $data = $row["occasions"];
                $da = explode(";" , $data);




            }

    $query1= "SELECT * FROM tbl_flower WHERE occasions NOT LIKE '". $da ."'";

    $result1= mysqli_query($conn, $query1)
    or die("Error in query: ". mysqli_error($conn));

    $result1_rows = mysqli_num_rows($result1);




    while ($row = mysqli_fetch_array($result1)){




    ?>

           <tr>

            <td><?php echo $row['name']?></td>
            <td><?php echo $row['position']?></td>
            <td>  <a href="editoccasionposition.php?id= <?php
                echo $row['id']?>"><input type="button" class="btn btn-secondary" name="edit " value="Edit Position"></a></td>



         </tr>

         <?php
    }
   }

?>

如果有人可以帮助我，我真的很感激，谢谢你。

内爆并爆炸php mysql

0 个答案: