CODEWITHSUNDEEP
phphtml
Anwen Evans
Anwen Evans

Reputation: 17

form in html and php not showing up on the screen

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>

Upvotes: 0

Views: 111

Answers (1)

RiggsFolly
RiggsFolly

Reputation: 94642

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

Upvotes: 3

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