CODEWITHSUNDEEP
phpmysql
Muhammad Sohail Arif
Muhammad Sohail Arif

Reputation: 158

QUERY AND ECHO Data INTO HTML INPUT fields

I've written a script that in short is supposed to query data from the database and echo a result into a HTML form field. However, I have been unsuccessful. Please see code below:

<?php


include("dbconfig.php");

$val = '6';

$result = mysqli_query("Select * from test where testid= '$val'");
$name = (mysqli_num_rows($result)==1) ? mysqli_fetch_assoc($result) : null;

if(is_array($name)){


?>

<html>

<body>

<form>
Name: <input type="text" id="firstname" value="<?php echo $name['firstname']; ?>"/>
</form>

<?php
} else {

echo "No such name exists";


}

?>



</body>
</html>

Can someone please tell me what I'm doing wrong. Because it won't echo anything into the field and I find it rather annoying because majority of the scripts I've come across are quite similar to this one.

Help will be much appreciated.

Thank You,

Sohail.

Upvotes: 1

Views: 3922

Answers (2)

Professor Abronsius
Professor Abronsius

Reputation: 33813

I have tested the below and it works OK. @Fred-ii- gave you loads of good info, especially using error debugging - but you do need to supply the connection object which you were missing.

<?php
    error_reporting( E_ALL );

    include("conn.php");

    $val = 6;

    /* What is the name of the $connection object ? */
    $result = mysqli_query( $conn, "Select * from `test` where `testid`='$val'" );
    $name=( $result ) ? mysqli_fetch_assoc( $result ) : false;

?>  
<html>
    <head>
        <title>Ya gotta have a title...</title>
    </head>
    <body>
        <?php
            if( !empty( $name ) ){
                echo "
                    <form>
                        Name: <input type='text' id='firstname' value='{$name['firstname']}'/>
                    </form>";
            } else {
                echo "No such name exists";
            }
        ?>
    </

Upvotes: 1

Funk Forty Niner
Funk Forty Niner

Reputation: 74217

You did not pass your db connection to your query, so it never gets executed.

  • Assuming a successful connection using mysqli_

This line of code:

$result = mysqli_query("Select * from test where testid= '$val'");

needs to have a connection parameter:

$result = mysqli_query($connection, "Select * from test where testid= '$val'");

and is unknown to us as to which MySQL API you're using to connect with.

Your query may have failed, so check for errors.

$result = mysqli_query("Select * from test where testid= '$val'")
    or die(mysqli_error($connection));

and replacing the $connection variable with the one that you have assigned in your dbconfig.php which is unknown to us.

Different MySQL APIs/functions do not intermix.

Consult these following links http://php.net/manual/en/mysqli.error.php and http://php.net/manual/en/function.error-reporting.php and apply that to your code.

You're also open to an SQL injection. Use a prepared statement.

References:

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Displaying errors should only be done in staging, and never production.

If you want to check if a row exists, see my other answer on Stack:

Upvotes: 0

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