CODEWITHSUNDEEP
phphtmlmysqlsqlmysqli
user13111897
user13111897

Reputation:

cant display specific row from database using PHP though no errors is shown

When I click the go to the form I want to see all other information about that id number. enter image description here Here is the code behind the go to the form option:

        <td width="174"  class="centertext"><a href="form.php?formid=<?php echo $row["formid"]; ?>"> Go to the form</a></td>

This is the code that I used to retrieve a specific row from my database. the code does not show any errors but it also does not show the result. All I see is empty page. What is the problem here? I have been trying to solve for 1 week and can't find anything. I am beginner at PHP. Thanks in advance.

<?php
$conn = mysqli_connect("localhost","root","","son_fbe");

if (mysqli_connect_error()) {

     echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();

    }


$formid = isset($_GET['formid ']) ? $_GET['formid '] : ''; 

    if($sql= "SELECT formid,gonderilen,gonderen FROM derssaydirma WHERE formid = ?")
                  {  
                     $stmt = $conn->prepare($sql); 
                     $stmt->bind_param("i", $formid);
                     $stmt->execute();
                     $result = $stmt->get_result();
                    while ($row = $result->fetch_assoc()) {
                         echo $row[ 'formid'];   
                         echo  $row['gonderilen'];
                         echo $row['gonderen'];   }

                     $stmt->close();
                 }
                 // show an error if the query has an error
                 else
                 {
                     echo "Error: could not prepare SQL statement";
                 }


?>

Upvotes: 1

Views: 95

Answers (1)

davidev
davidev

Reputation: 8517

I think I found the mistake.. you having a blank inside your name when accessing the get array. $formid will be "" and that's why your query is not working.

Change this line..

$formid = isset($_GET['formid ']) ? $_GET['formid '] : '';

to 

$formid = isset($_GET['formid']) ? $_GET['formid'] : '';

Upvotes: 1

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