PHP if Statement mysqli_num_rows not working
Question
I have a code which asks if a row exists in a database and if it does, it will echo a message. However, nothing is echoed no matter what input I put in.
Here is my code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
$mysqlhost="host";
$mysqldatabase="b33_15887129_Accounts";
$mysqlusername="username";
$mysqlpassword="password";
$connect=new mysqli($mysqlhost,$mysqlusername,$mysqlpassword);
if (!$connect) {
die("Connection failed: " . $connect->connect_error);
}
echo "Connected Successfully";
mysqli_select_db($connect,"b33_15887129_Accounts");
$loginusername=$_POST['loginusername'];
$loginpassword=$_POST['loginpassword'];
$checkifexist="SELECT * FROM Users WHERE Username='$loginusername' AND 'Password=$loginpassword'";
$runquery=mysqli_query($connect,$checkifexist);
$numofrows=mysqli_num_rows($runquery);
if($numofrows==1){
echo "Successfully logged in!";
}else{
echo "Failed to log in";
}
?>
</body>
</html>
It manages to echo Connected successfully but Successfully logged in and Failed to log in does not appear no matter what the input. Any help?
2 answers
solution1
2 ACCPTED 2015-03-02 09:21:17
Your "failed to log in" isn't displaying because you've forgotten the echo.
if($numofrows==1){
echo "Successfully logged in!";
}
else{
echo "Failed to log in";
}
In order to get the sql to work, @Ranjith is correct, you'll need to alter the quote placement in your sql statement.
solution2
0 2015-03-02 09:24:09
1) Your else statement have no echo !!!
2) Your SQL Statement is wrong! Check the quotes at the password! --> Password='$loginpassword'
In fact you should have a look at http://php.net/manual/en/mysqli.real-escape-string.php .
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