PHP: Undefined shown on blank page - php

Question

I'm trying to print/echo values from previous file however instead of showing error, all that is shown is 'Undefined' on a blank page. I've researched and tried several method but nothing works. Please help.
<?PHP
$user_name = "root";
$password = "";
$database = "leadership_program";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
if (isset($_POST['survey_id'])) {
$survey_id = $_POST['survey_id'];
echo $survey_id;
}
if (isset($_POST['marks'])) {
foreach ($_POST['marks'] as $value) {
echo"$value";
}
}
if (isset($_POST['id'])) {
$id = $_POST['id'];
echo $id;
}
// $SQL2 = "UPDATE answer_table SET marks='$value' WHERE survey_id= '$survey_id' AND student_id= '$id'";
//$result2 = mysql_query($SQL2);
//mysql_close($db_handle);
} else {
print "Database NOT Found ";
mysql_close($db_handle);
// header("Location: surveyView.php");
}
?>
Here is displayresult.php
<form action="student_mark_save.php" method="POST"> //<?php...more codes here
if ($strucrow["qns$i"] === 'radio' || $strucrow["qns$i"] === 'checkbox') {
foreach ($arr as $b) {
echo "<br/>";
if (strpos($b, '%#%') !== false) {
$c = substr($b, 3, -2);
//echo $c;
$d = str_replace("$arr[0] :-", ':', $c);
echo $d, "<br/>";
echo "<br/>";
echo "<tr> Marks : <input type=\"text\" name=\"marks[]\"></tr><br />";
//echo $b;
} else {
echo $b;
// echo "is not with comment qns";
}
}
} else if ($strucrow["qns$i"] === 'comment') {
foreach ($arr as $b) {
echo $b;
echo "<tr> Marks : <input type=\"text\" name=\"marks[]\"></tr><br />";
// echo "is not with comment qns";
}
} else {
}
echo "<p/>";
}
$marksquery = sprintf(
"SELECT marks FROM answer_table WHERE survey_id = '%d' AND student_id = '$studentid' ", mysql_real_escape_string($survey_id)
);
$marksQuer = mysql_query($marksquery) or die(mysql_error());
$marksrow = mysql_fetch_assoc($marksQuer);
echo "<td><input type=\"hidden\" value= \"$survey_id\" name=\"survey_id\"></td><br />" ;
echo "<td><input type=\"hidden\" value= \"$studentid\" name=\"id\"></td><br />" ;
echo "<p><input type=\"submit\" value=\"Update\"></p>";
?>
</form>

Answer 1

for undefined mean, there is something which is not existing in global $_POST[]
check try to print $_POST in your first if statement as:
if ($db_found) {
print_r($_POST);
exit();
if (isset($_POST['survey_id'])) {
$survey_id = $_POST['survey_id'];
echo $survey_id;
}
now just see what do you have in $_POST, It might help you...Thanks

Answer 2

Most likely some Post data is not set. Try vardump($_POST); die('happy');.
It would also be better to handle missing Post data instead of just ignoring it so maybe if ($db_found) { should be
<?php
if ( $db_found
&& isset($_POST['survey_id'])
&& isset($_POST['marks'])
&& isset($_POST['id'])
)
{
$survey_id = $_POST['survey_id']; //maybe ensure int
$id = $_POST['id']; //maybe ensure int
$values = array(); //this is important $value will not be available outside the for loop or contain only the last value
foreach ((array)$_POST['marks'] as $value) { //ensure $_POST['marks'] is an array
$values[] = $value; //maybe mysql escape each $value here
}
$SQL2 = sprintf(
"UPDATE answer_table SET marks='%s' WHERE survey_id=%d AND student_id=%d",
impolde(',' $values), //assuming comma separated list here
$survey_id,
$id
);
$result2 = mysql_query($SQL2);
mysql_close($db_handle);
} else { ...
Please also consider to escape the values you read from POST. Doing it this way makes it very easy to SQL inject into your script!