html - How can i search for a specific row in a Database using PHP? -

how can make search form in html page display result in same html page? having error when try , run code says inputpackageid not defined. how can solve this. seems variable not being recognized form. please help

here code. if dont understand question can ask me not clear.

<div class="main">     <div class="col-md-12">         <div class="row">             <div class="container">          </div>          <div class="search center-block col-md-6" align="center">         <p>             want track package? <br />             enter package id(pid)         </p>         <form method="post" action="trackshipment.php">         <div class="form-group">              <label class="sr-only" for="inputpackageid"></label>             <input type="search" name="inputpackageid" class="form-control edit-form-control" id="inputpackageid" placeholder="pid">         </div>         <div class="modal-footer">           <input type="submit" class="btn btn-primary" name="submit" />         </div>         </form>     </div>           <div class="col-md-8" align="center">         <h2>well, here package!</h2>                     <?php                         include '../includes/connection.php';                         //include 'phpscripts/trackshipment.php';                         $packageid = $_get['inputpackageid'];                          $sql = "select * package p_id='$packageid'";  ?>                   <table class="table table-striped">                 <tr>                     <th><span>pid</span></th>                     <th><span>customer id</span></th>                     <th>name</th>                     <th>description</th>                     <th>destination per km</th>                     <th>type</th>                     <th>price/kg</th>                 </tr>                 </th>                 <?php                        foreach ($db->query($sql) $count)?>                     <tr>                         <td>                             <?php echo $count['p_id'];  ?>                         </td>                         <td>                             <?php echo $count['cus_id']; ?>                         </td>                         <td>                             <?php echo $count['package_name']; ?>                         </td>                         <td>                             <?php echo $count['package_description']; ?>                         </td>                         <td>                             <?php echo $count['package_type']; ?>                         </td>                         <td>                             <?php echo $count['destination_per_km']; ?>                         </td>                         <td>                             <?php echo $count['price_per_kg']; ?>                         </td>                 <?php } ?>             </table>       </div> </div> 

it's showing it's not defined because form uses post method, , you're using array.

$packageid = $_get['inputpackageid']; 

change to

$packageid = $_post['inputpackageid']; 

plus, use isset()

if(isset($_post['inputpackageid'])){     $packageid = $_post['inputpackageid']; } 

since you're using entired code inside same page.

or !empty()

if(!empty($_post['inputpackageid'])){     $packageid = $_post['inputpackageid']; } 

if you're using same code inside same file, can use action="" , conditional statement submit button.

if(isset($_post['submit'])){...}

seeing comment of yours, isn't case , using seperate file, can scratch that.

however, if method want use, need change method method="get" in form. using method unsafe, consult footnotes.

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footnotes:

your present code open sql injection. use mysqli prepared statements, or pdo prepared statements, they're safer.

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"how can make search form in html page display result in same html page?"

• you can either use ajax this, or use action="".