form in html and php not showing up on the screen

Question

I am writing out a form in php,html... witch should normally update to my database but the form isnt showing up on the screen. Help!!! I am not a very expirianced coder so pleases if you could tell meif there is any other probleme with my code. Thanks :-)

<html>

    <head>
        <title>help</title>
    </head>

    <body>
        <?php

            if(isset($_POST['update'])) {
                $dbhost = 'localhost';
                $dbuser = '*********';
                $dbpass = '*****';
                $conn = mysqli_connect($dbhost, $dbuser, $dbpass);
                if(! $conn ) {
                   die('Could not connect: ' . mysqli_error());
                }
                $Userid = $_POST['UserID'];
                $TableID = $_POST['tableID'];
                $Life_points = $_POST['Life_points'];
                $xp_points = $_POST['xp_points'];
                $sql = "UPDATE points SET TableID = " . $TableID . " WHERE UserID = ". $Userid . " AND life_points = " . $Life_points . " AND xp_points= " . $xp_points;
                mysqli_select_db('womath');

                $retval = mysqli_query(  $conn, $sql );
                <?php echo $_SERVER['PHP_SERVER'] ?>            

                if(! $retval ) {
                    die('Could not update data: ' . mysqli_error());
                }
                echo "Updated data successfully\n";
                mysqli_close($conn);
                } else {
                ?>
                    <form method = "post" action = "<?php $_PHP_SELF ?>">
                        <table width = "400" border =" 0" cellspacing = "1" cellpadding = "2">
                            <tr>
                                <td width = "100">UserID</td>
                                <td><input name = "UserID" type = "number" id = "UserID"></td>
                            </tr>
                            <tr>
                                <td width = "100">TableID</td>
                                <td><input name = "TableID" type = "number" id = "TableID"></td>
                            </tr>
                            <tr>
                                <td width = "100">life_points</td>
                                <td><input name = "life_points" type = "number" id = "life_points"></td>
                            </tr>
                            <tr>
                               <td width = "100">xp_points</td>
                               <td><input name = "xp_points" type = "number" id = "xp_points"></td>
                            </tr>
                            <tr>
                                <td width = "100"> </td>
                                <td> </td>
                            </tr>
                            <tr>
                                <td width = "100"> </td>
                                <td><input name = "update" type = "submit" id = "update" value = "Update"></td>
                            </tr>
                       </table>
                   </form>
               <?php 
            } 
        ?>

    </body>
</html>

Please learn about prepared statements and use them. always. — Tschallacka, Feb 13 '17 at 11:53
when you have already declared your writing in PHP? Remove the !! — DamiToma, Feb 13 '17 at 12:02
@AnwenEvans http://php.net/manual/en/mysqli.quickstart.prepared-statements.php and read http://stackoverflow.com/questions/24988867/when-should-i-use-prepared-statements/24989031 — Tschallacka, Feb 13 '17 at 12:03

score 3 · Accepted Answer · edited May 23 '17 at 11:53

You have issued a query to the database for compilation and execution before you have told MYSQL which database you are trying to gain access to.

The mysqli_select_db('womath'); must happen before your first query

In fact that function is more for use when you want to switch from one database to another during a scripts execution, you can add the database name to the

$conn = mysqli_connect($dbhost, $dbuser, $dbpass);

like this

$conn = mysqli_connect($dbhost, $dbuser, $dbpass,'womath');

line if you like and forget about the line

mysqli_select_db('womath');

completely.

<html>
<head>
  <title>help</title>
</head>
<body>
<?php
    ini_set('display_errors', 1); 
    ini_set('log_errors',1); 
    error_reporting(E_ALL); 
    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
     if(isset($_POST['update'])) {
        $dbhost = 'localhost';
        $dbuser = '*********';
        $dbpass = '*****';

        $conn = mysqli_connect($dbhost, $dbuser, $dbpass,'womath');
                            // --------------------------^^^^^^^^
        if(! $conn ) {
           die('Could not connect: ' . mysqli_error());
        }
        // move this calll to here or add database to the connect line
        //mysqli_select_db('womath');

        $Userid = $_POST['UserID'];
        $TableID = $_POST['tableID'];
        $Life_points = $_POST['Life_points'];
        $xp_points = $_POST['xp_points'];

        $sql = "UPDATE points SET TableID = '$TableID'
                WHERE UserID = '$Userid' 
                  AND life_points = '$Life_points' 
                  AND xp_points = '$xp_points'";

        $retval = mysqli_query(  $conn, $sql );

        // this following line also need a `;`
        echo $_SERVER['PHP_SERVER'];

        if(! $retval ) {
           die('Could not update data: ' . mysqli_error());
        }
        echo "Updated data successfully\n";

        mysqli_close($conn);

    // remove this else, as with it in place
    // you only show the form when you are NOT updating the database
    // }else {
    // just terminate the IF so the form will show after an update
    // AND when page is first loaded and there is no user input
    }
?>
           <form method = "post" action = "<?php $_PHP_SELF ?>">
              <table width = "400" border =" 0" cellspacing = "1" 
                 cellpadding = "2">

                <tr>
                    <td width = "100">UserID</td>
                    <td><input name = "UserID" type = "number" 
                       id = "UserID"></td>
                </tr>
                <tr>
                    <td width = "100">TableID</td>
                    <td><input name = "TableID" type = "number" id = "TableID"></td>
                </tr>
                <tr>
                    <td width = "100">life_points</td>
                    <td><input name = "life_points" type = "number" id = "life_points"></td>
                </tr>
                <tr>
                    <td width = "100">xp_points</td>
                    <td><input name = "xp_points" type = "number" id = "xp_points"></td>
                 </tr>
                 <tr>
                    <td width = "100"> </td>
                    <td>&nbsp;</td>
                 </tr>
                 <tr>
                    <td width = "100"> </td>
                    <td>
                       <input name = "update" type = "submit" id = "update" value = "Update">
                    </td>
                 </tr>
              </table>
           </form>
  <?php
     }
  ?>

Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements

Maybe one or more of your columns are defined as some sort of text datatype. — RiggsFolly, Feb 13 '17 at 12:09
For it to say could not update data I had to take away echo $_SERVER['PHP_SERVER'] — Anwen Evans, Feb 13 '17 at 12:11
I have done what you have asked but it is now showing  HTTP 500 error That’s odd... the website can’t display this page The site may be under maintenance or could have a programming error. — Anwen Evans, Feb 13 '17 at 16:14
Can you check the error log, or add the 4 new lines I added after the ` — RiggsFolly, Feb 13 '17 at 16:31
https://anwen.000webhostapp.com/new/index.php here is a link to what It is showing now.@RiggsFolly — Anwen Evans, Feb 13 '17 at 17:03

form in html and php not showing up on the screen

1 Answers1