如何将 php 脚本中的文本/变量传递给输入值？

Question

I need to get id and name from URL when I click edit anchor and I get it actually but I can't post data to input value to make an update by the text input... here is my HTML form:

<form action="" method="post">
  <div class="form-group">
   <label for="cc-payment" class="control-label mb-1">Category Name</label>
   <input id="cc-pament" name="category-name" type="text" class="form-control" value="">
  </div>
  <div>
   <button id="payment-button" type="submit" name="submit" class="btn btn-lg btn-info">Save
   </button>
  </div>
</form>

here is HTML TABLE And have the Edit anchor to go if isset

<thead>
 <tr>
   <th>ID</th>
   <th>Name</th>
   <th>Edit</th>
   <th>Delete</th>
 </tr>
</thead>
<tbody action="" method="get">
    <?php 
      $query = " select * from category ";
      $result = mysqli_query($conn,$query);
      while($cat = mysqli_fetch_assoc($result)){
        echo "<tr>";
        echo "<td>{$cat['cat_id']}</td>";
        echo "<td>{$cat['cat_name']}</td>";
        echo "<td><a href='?edit={$cat['cat_id']}&category_name={$cat['cat_name']}'  name='edit' 
              class='btn 
        btn-info'> Edit</a></td>";
        echo "<td><a href='delete_category.php?id={$cat['cat_id']}' class='btn btn-danger'>Delete</a> 
        </td>";
        echo "</tr>";
        }
    ?>
</tbody>

here is the PHP code

 if (isset($_GET['edit'])) {
    # code...
    $name= $_GET['category_name'];
    echo $_POST['category-name']= $name;
    }

Answer 1

If I am understanding correctly, I believe you want to populate the category-name field on your form based the edit value in the URL.

If so, you have most of what you need, you simply need to insert it into the HTML(I am assuming the form is part of a PHP script as well).

   <input id="cc-pament" name="category-name" type="text" class="form-control" value="<?= htmlentities($_GET["edit"]??"") ?>">

Be sure to encode the value with htmlentities or similar to avoid XSS vulnerabilities.

Side note, what exactly is the intent of this line?

    echo $_POST['category-name']= $name;