php script not updating the database

saloo · February 17, 2007

Hi. i found this script online and i thought all i need to do is just change the database name, username and password but when i try to run the script i got lots of notices. i got 4 fields and a button by the name of (enter information). once the admin fills out the form and click on the button(enter information), the information is stored in the database which does not. i dont know whats going on any help will be really appreciated. and here is the guy url who created the script and the test page.

http://www.widgetmonkey.com/quiz/editquiz.php

and here is my script

<HTML>
<link rel="stylesheet" href="quiz.css" type="text/css">
<body><center>
<P> </P>

<B>Admin area - edit the quiz</B>
<?php
include("contentdb.php");
$submit = isset($_POST['submit']);
$id = "";
$update = "";
if($submit)
{
$sql = "INSERT INTO $table (question, opt1, opt2, opt3, answer) VALUES ('$question','$opt1','$opt2','$opt3','$answer')";
$result = mysql_query($sql);
echo "<br><br>Question added to quiz.<br><br>";
include "qinsert.php";
}
else if($update)
{
$sql = "UPDATE $table SET question='$question',opt1='$opt1',opt2='$opt2',opt3='$opt3',answer='$answer' WHERE id=$id";
$result = mysql_query($sql);
echo "<br><br>The quiz has been succesfully updated.<br><br>\n";
}

else if($id)
{
$result = mysql_query("SELECT * FROM $table WHERE id=$id",$db);
$myrow = mysql_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']
?>">
<input type="hidden" name="id" value="<?php echo $myrow["id"]?>">
    <b>Question:</b><br>
    <input type="Text" name="question" value="<?php echo $myrow["question"]?>" size="50">
    <br>
    <b>Option 1:</b><br>
    <input type="Text" name="opt1" value="<?php echo $myrow["opt1"]?>" size="30">
    <br>
    <b>Option 2:</b><br>
    <input type="Text" name="opt2" value="<?php echo $myrow["opt2"]?>" size="30">
    <br>
    <b>Option 3:</b><br>
    <input type="Text" name="opt3" value="<?php echo $myrow["opt3"]?>" size="30">
    <br>
    <b>Answer</b> (must be identical to correct option):<br>
    <input type="Text" name="answer" value="<?php echo $myrow["answer"]?>" size="30">
    <br>
    <br>

<input type="Submit" name="update" value="Update information"></form>
<?

}
else
{
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']
?>">
    <p><br>
      <b>Question:</b><br>
      <input type="Text" name="question" size="50">
      <br>
      <b>Option 1:</b><br>
      <input type="Text" name="opt1" size="30">
      <br>
      <b>Option 2:</b><br>
      <input type="Text" name="opt2" size="30">
      <br>
      <b>Option 3:</b><br>
      <input type="Text" name="opt3" size="30">
      <br>
      <b>Answer</b> (must be identical to correct option):<br>
      <input type="Text" name="answer" size="30">
      <br>
      <br>
      <input type="Submit" name="submit" value="Enter information">
    </p>
    </form>
<?
}
?>
<a href="editquizlist.php">Back to list of quiz questions</a>
</center>
</body>
</HTML>

emehrkay · February 17, 2007

try this

$result = mysql_query($sql) or die (mysql_error());

saloo · February 17, 2007

hi thanks for the reply bro. let me show you my output when i click on the button this is what i get:

Admin area - edit the quiz 
Notice: Undefined variable: question in c:\program files\easyphp1-7\www\quizv1.0\editquiz.php on line 14

Notice: Undefined variable: opt1 in c:\program files\easyphp1-7\www\quizv1.0\editquiz.php on line 14

Notice: Undefined variable: opt2 in c:\program files\easyphp1-7\www\quizv1.0\editquiz.php on line 14

Notice: Undefined variable: opt3 in c:\program files\easyphp1-7\www\quizv1.0\editquiz.php on line 14

Notice: Undefined variable: answer in c:\program files\easyphp1-7\www\quizv1.0\editquiz.php on line 14


Question added to quiz.

Back to list of quiz questions

it says that the question has been added to the quiz but when i click on (back to list of quiz questions) all i see is just one question which i entered manually through mysql and 3 blank fields. here is the output.

Admin area - edit the quiz 

1: who is the president of USA? [ edit ] [ delete ] 
2:  [ edit ] [ delete ] 
3:  [ edit ] [ delete ] 
4:  [ edit ] [ delete ] 
5:  [ edit ] [ delete ]

here is my editquizlist.php script

<HTML>
<link rel="stylesheet" href="quiz.css" type="text/css">
<body><center>
<P> </P>

<B>Admin area - edit the quiz</B>
<br><br>
  <table width="300" border="0" cellspacing="0" cellpadding="0">

        <?php

include("contentdb.php");

$result = mysql_query("SELECT id, question FROM $table ORDER BY id",$db);

echo "<table>";

while ($row = mysql_fetch_array($result))
{
$alternate = "";

$id = $row["id"];
$question = $row["question"];
if ($alternate == "1") {
$color = "#ffffff";
$alternate = "2";
}
else {
$color = "#efefef";
$alternate = "1";
}
echo "<tr bgcolor=$color><td>$id:</td><td>$question</td><td>[ <a href='editquiz.php?id=$id'>edit</a> ]</td><td>[ <a href='deletequiz.php?id=$id' onClick=\"return confirm('Are you sure?')\">delete</a> ]</td></tr>";
}
echo "</table>";
?>
        <br>
        <br>
        <a href="editquiz.php">Add a new question to the quiz</a></td>
    </tr>
  </table>
  <br>

and y am i getting so many notices ? i know it says undefined variable how can i get rid of those also.

Thankyou .

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php script not updating the database

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