My 2nd Try asking the same "PHP Programming with MySQL question. When I run my code,...
My 2nd Try asking the same "PHP Programming with MySQL question.
When I run my code, I get the following messages:
Notice: Undefined index: email in
/Applications/XAMPP/xamppfiles/htdocs/Week9/VerifyLogin.php
on line 71
Notice: Undefined index: password in
/Applications/XAMPP/xamppfiles/htdocs/Week9/VerifyLogin.php
on line 71
Warning: mysqli_num_rows() expects parameter 1 to
be mysqli_result, null given in
/Applications/XAMPP/xamppfiles/htdocs/Week9/VerifyLogin.php
on line 75
The e-mail address/password combination entered is not valid.
Please use your browser's BACK button to return to the form and fix the errors indicated.
Please use your browser's BACK button to return to the form and fix the errors indicated.
- Please help me fix my code and provide screenshot and revised code. Thank you :)
------
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN""http://www.w3.org/TR/html4/strict.dtd">
<html xmlns="https://www.w3.org/1999/xhtml">
<head>
<title>Verify Intern Login</title>
</head>
<body>
<h1>College Internship</h1>
<h2>Verify Intern Login</h2>
<?php
{
$errors = 0;
$DBConnect = @mysqli_connect("localhost", "root", "");
if ($DBConnect === FALSE) {
echo "
<p>
Unable to connect to the database
server. " . "Error code " . mysqli_errno() . ": " . mysqli_error() . "
</p>\n";
++$errors;
} else {
$DBName = "internships";
$result = @mysqli_select_db($DBName, $DBConnect);
if ($result === FALSE) {
echo "
<p>
Unable to select the database. " . "Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect) . "
</p>\n";
++$errors;
}
}
}
{
$TableName = "interns";
if ($errors == 0) {
$SQLstring = "SELECT internID, first, last FROM
$TableName" . " where email='" . stripslashes($_POST['email']) . "' and password_md5='" . md5(stripslashes($_POST['password'])) . "'";
$QueryResult = @mysqli_query($SQLstring, $DBConnect);
if (mysqli_num_rows($QueryResult) == 0) {
echo "
<p>
The e-mail address/password " . " combination entered is not valid.
</p>\n";
++$errors;
} else {
$Row = mysqli_fetch_assoc($QueryResult);
$InternID = $Row['internID'];
$InternName = $Row['first'] . " " . $Row['last'];
echo "
<p>
Welcome back, $InternName!
</p>\n";
}
}
}
{
if ($errors > 0) {
echo "
<p>
Please use your browser's BACK button
to return " . " to the form and fix the errors
indicated.
</p>\n";
if ($errors > 0) {
echo "
<p>
Please use your browser's BACK button
to return " . " to the form and fix the errors
indicated.
</p>\n";
}
}
}
?>
</body>
</html>
Homework Answers
I think the problem is that you haven't checked if the POST variable is set, which you can check by
'if(isset($_POST['email']) && isset($_POST['email']))'
so code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN""http://www.w3.org/TR/html4/strict.dtd">
<html xmlns="https://www.w3.org/1999/xhtml">
<head>
<title>Verify Intern Login</title>
</head>
<body>
<h1>College Internship</h1>
<h2>Verify Intern Login</h2>
<?php
if(isset($_POST['email']) && isset($_POST['email']))
{
$errors = 0;
$DBConnect = @mysqli_connect("localhost", "root", "");
if ($DBConnect === FALSE) {
echo "
<p>
Unable to connect to the database server. " . "Error code " . mysqli_errno() . ": " . mysqli_error() . "
</p>\n";
++$errors;
} else {
$DBName = "internships";
$result = @mysqli_select_db($DBName, $DBConnect);
if ($result === FALSE) {
echo "
<p>
Unable to select the database. " . "Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect) . "
</p>\n";
++$errors;
}
}
$TableName = "interns";
if ($errors == 0) {
$SQLstring = "SELECT internID, first, last FROM
$TableName" . " where email='" . stripslashes($_POST['email']) . "' and password_md5='" . md5(stripslashes($_POST['password'])) . "'";
$QueryResult = @mysqli_query($SQLstring, $DBConnect);
if (mysqli_num_rows($QueryResult) == 0) {
echo "
<p>
The e-mail address/password " . " combination entered is not valid.
</p>\n";
++$errors;
} else {
$Row = mysqli_fetch_assoc($QueryResult);
$InternID = $Row['internID'];
$InternName = $Row['first'] . " " . $Row['last'];
echo "
<p>
Welcome back,".$InternName."!
</p>\n";
}
}
if ($errors > 0) {
echo "
<p>
Please use your browser's BACK button to return " . " to the form and fix the errors indicated.</p>\n";
if ($errors > 0) {
echo "
<p>
Please use your browser's BACK button to return " . " to the form and fix the errors indicated.
</p>\n";
}
}
}
?>
</body>
</html>
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My 2nd Try asking the same "PHP Programming with MySQL
question.
When I run my code,...
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