Question

这第一种形式工作正常

            <div class="col-xs-6">
            <form action="categories.php" method="post">

            <div class="form-group">
               <input type="text" class="form-control" name="search">
                      </div>

        <div class="form-group">
         <input type="submit" class="btn btn-primary" name="search1" value="searchcategory">
                      </div>

                    </div>

当点击其中的任何链接或按钮时，第二种形式消失

                      <div class="col-xs-6">
                      <form action='categories.php' method='post';>
                      <table class='table table-bordered table-hover'>
                        <thead>
                        <tr>
                        <th>ID</th>
                        <th>categories</th>
                        </tr>   
                        </thead>
                   <?php
                        if(isset($_POST['search1'] )){
                        $searchval=$_POST['search'];
                        $querynew="SELECT * FROM categories WHERE cat_title='$searchval'";
                        $m1 = mysqli_query($connection1,$querynew);
                        $i=0;
                        while($row = mysqli_fetch_assoc($m1)) {

                        $cat_id1 = $row['cat_id'];
                        $cat_title2 = $row['cat_title'];

                        echo"
                        <tbody>
                        <tr>

                        <td>
                        $cat_id1
                        </td>";


                     if(isset($_GET['edit'.$i.''])){
                         $ab = $_GET['edit'.$i.''];
                         $query1="SELECT * FROM categories WHERE cat_id=$ab";
                        $m = mysqli_query($connection1,$query1);

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

                        $cat_id2 = $row['cat_id'];
                        $cat_title3 = $row['cat_title'];

                   echo"   
                        <td> 
                       <input type='text' class='form-control' name='txt".$i."' value='$cat_title3'>
                       </td>";

                        }
                     }

                        else{
                        echo"<td>
                        $cat_title2
                        </td>";
                        }
                        echo"
                        <td>
    <input type='submit' class='btn btn-primary' name='submit".$i."' value='Edit category'>
    </td>
    ";
                         if(isset($_POST['submit'.$i.''])){
                         $abc = $_POST['txt'.$i.''];
                         $query2="UPDATE categories SET cat_title='$abc' WHERE cat_id=$cat_id2";
                        $sm = mysqli_query($connection1,$query2);

                         }

                        echo"    <td>
                         <a href='categories.php?id1={$cat_id1}'>Delete</a>
                         </td>
                          <td>
                         <a href='categories.php?edit".$i."={$cat_id1}'>Edit</a>
                          </td>
                        </tr>";


                        $i++;
                        }
                    }

                    ?>
                    </tbody>
                    </table>
                    </form>
                    </div>


                     <?php 
                    // Delete category 


                        if(isset($_GET['id1'])){
                        $h = $_GET['id1'];
                        $query = "DELETE FROM categories WHERE cat_id = {$h} ";
                        $delete_query = mysqli_query($connection1,$query);






                          }


                    ?>

。 ////////////////////////////////////////////////// ////////////////////////////////////////////////// ////////////////////////////////////////////////// ////////////////////////////////////////////////// ///////////////////////////////////////

Answer 1

因为你只有一个名为＆＃34; search1＆＃34;的按钮。在表单1上但表单2没有，并且您已使用DelegatingFilterProxy包装整个代码块。

如果没有名为＆＃34; search1＆＃34;的按钮，则不会显示第二页上表单中的任何帖子。在第2页的表单中。或者，如果您单击名为＆＃34; search2＆＃34;在第1页，就此而言。

我实际上不确定你真正想要的是什么，因为代码很差，所以我无法提供实际编码的解决方案，而只是指出你的错误。

在if(isset($_POST['search1']))区块之外的第二页上添加var_dump($_POST);，您应该看到这一点。

Answer 2

将整个代码放在一个文件中。像这样它会起作用。

<div class="col-xs-6">
                  <form action='categories.php' method='post';>
                      <div class="form-group">
                      <input type="text" class="form-control" name="search">
                      </div>

                      <div class="form-group">
                      <input type="submit" class="btn btn-primary" name="search1" value="searchcategory">
                      </div>
                         <div class="form-group">
                      <input type="submit" class="btn btn-primary" name="search2" value="searchcategory">
                      </div>
                  <table class='table table-bordered table-hover'>
                    <thead>
                    <tr>
                    <th>
                    ID
                    </th>
                    <th>
                    categories
                    </th>
                    </tr>   
                    </thead>
               <?php


                    if(isset($_POST['search1'] )){
                    $searchval=$_POST['search'];
                    $querynew="SELECT * FROM categories WHERE cat_title='$searchval'";
                    $m1 = mysqli_query($connection1,$querynew);
                    $i=0;
                    while($row = mysqli_fetch_assoc($m1)) {

                    $cat_id1 = $row['cat_id'];
                    $cat_title2 = $row['cat_title'];

                    echo"
                    <tbody>
                    <tr>

                    <td>
                    $cat_id1
                    </td>";


                 if(isset($_GET['edit'.$i.''])){
                     $ab = $_GET['edit'.$i.''];
                     $query1="SELECT * FROM categories WHERE cat_id=$ab";
                    $m = mysqli_query($connection1,$query1);

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

                    $cat_id2 = $row['cat_id'];
                    $cat_title3 = $row['cat_title'];

               echo"   
                    <td> 
                   <input type='text' class='form-control' name='txt".$i."' value='$cat_title3'>
                   </td>";

                    }
                 }

                    else{
                    echo"<td>
                    $cat_title2
                    </td>";
                    }
                    echo"
                    <td>
<input type='submit' class='btn btn-primary' name='submit".$i."' value='Edit category'>
</td>
";
                     if(isset($_POST['submit'.$i.''])){
                     $abc = $_POST['txt'.$i.''];
                     $query2="UPDATE categories SET cat_title='$abc' WHERE cat_id=$cat_id2";
                    $sm = mysqli_query($connection1,$query2);

                     }

                    echo"    <td>
                     <a href='categories.php?id1={$cat_id1}'>Delete</a>
                     </td>
                      <td>
                     <a href='categories.php?edit".$i."={$cat_id1}'>Edit</a>
                      </td>
                    </tr>";


                    $i++;
                    }
                }

                ?>
                </tbody>
                </table>
                </form>
                </div>


                 <?php 
                // Delete category 


                    if(isset($_GET['id1'])){
                    $h = $_GET['id1'];
                    $query = "DELETE FROM categories WHERE cat_id = {$h} ";
                    $delete_query = mysqli_query($connection1,$query);






                      }


                ?>

单击任何链接或按钮时表单消失如何解决？

2 个答案: