获取选择选项php的值

Question

I want to get a value of select option php on the same page. I brought selection list from database and listed them on the select. But even though I choose something on the list, it's not posted. Is there any problem in the code below? Thanks in advance.

<div class="ibox-content m-b-sm border-bottom">
        <div class="row">
            <div class="col-sm-4">
                <div class="form-group">
                    <label class="control-label" for="status">Search by category</label>
                    <form action="" method="POST" >
                    <select name="category" class="form-control">
                        <option value="" selected>All</option>
                <?php
                    while($category = mysqli_fetch_array($result1)) {
                        // output data from each row
                        echo "<option value=\"{$category['categoryID']}\">{$category['name']}</option>";
                    }
                ?>
                    </select>
                        <input class="btn btn-primary" type="submit" name="submit" value="Search">
                        </input>
                    </form>
                </div>
            </div>
        </div>
    </div>
    <?php
    if (isset($_POST['submit'])) {
        echo "Got it!";
        echo $_POST['category'];
    }?>

Answer 1

You should specify a name for the option:

<option name='category' value=\"{$category['categoryID']}\"...

And try adding action='#' , and then it works:

array (size=2)
  'category' => string '1' (length=1)
  'submit' => string 'Search' (length=6)

Got it!1

And take care, in your code the option for 'all' has no value, and won't be displayed.

Answer 2

You have:

<?php
 while($category = mysqli_fetch_array($result1)) {
     // output data from each row
     echo "<option value=\"{$category['categoryID']}\">{$category['name']}</option>";
 }
?>

An array key -> inside a template var -> inside an option -> in a double-quoted string (which will parse vars), -> inside a while loop (which may or may not be running). That level of nested complexity will certainly decrease your time-to-debug and simplicity-to-debug.

I would first ignore the html entirely and:

$category = mysqli_fetch_array($result1)
var_dump($category, $category['categoryID'], $category['name']);

in naked php, to make sure that you have exactly what you need before you put anything into the html.

Also, form action='' sometimes causes problems, so make that and everything else you can explicit and fixed just while debugging.

I have to say that you're making things harder for yourself by not using a template engine, especially at the early stages. Your html+php is low-readability, and that decreases the ease of debugging. Use a clean template and it'll be easier to debug echoing issues, and as a side benefit you'll get easy escaping.