SQL / PHP如果为NULL显示其他显示
Finally after spending two days on this and with help from everyone from Stack Overflow we've manage to get this working. Thank you for all your hard work guys and talking me through step by step!
We are trying to get the following code to work currectly,
It needs to A. Check Activation code is NULL and if so move the user to one of the forms B. If Check Activation comes back other than NULL it should tell the user to try another activation code. I know this is pretty simple but we can't seem to see the issue.
<?php
$username = $_POST['username'];
$activation_code = $_POST['activation_code'];
$activation_codeurl = $activation_code;
$usernameurl = $username;
$db_host = "localhost";
$db_name = "aardvark";
$db_use = "aardvark";
$db_pass = "aardvark";
$con = mysql_connect("localhost", $db_use, $db_pass);
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db_name, $con);
$checkcustomer = mysql_query("SELECT `Check_Activation` FROM `members` WHERE `Username` = '".mysql_real_escape_string($username)."' AND `Activation` = '".mysql_real_escape_string($activation_code)."'; ");
$array = mysql_fetch_array($checkcustomer);
if (!$array === false)
{
$username = substr($username, 0, 1);
if($username == '1') {
$redirect_url='form-one.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '2') {
$redirect_url='form-two.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '3') {
$redirect_url='form-three.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
}
header("Location:". $redirect_url);
}
else
{
?>
<html>
<head>
<link rel='stylesheet' id='style-css' href='css/style.css' type='text/css' media='all' />
<meta name="viewport" content="width=960, initial-scale=0.32">
<META NAME="ROBOTS" CONTENT="NOINDEX, NOFOLLOW">
<link rel="shortcut icon" href="http://welovebarrio.com/favicon.gif">
<link rel="icon" href="http://welovebarrio.com/favicon.gif" type="image/gif">
<title>Friends of BARRIO</title>
<script type="text/javascript">
var _gaq = _gaq || [];
_gaq.push(['_setAccount', 'UA-35015193-1']);
_gaq.push(['_setDomainName', 'welovebarrio.com']);
_gaq.push(['_trackPageview']);
(function() {
var ga = document.createElement('script'); ga.type = 'text/javascript'; ga.async = true;
ga.src = ('https:' == document.location.protocol ? 'https://ssl' : 'http://www') + '.google-analytics.com/ga.js';
var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(ga, s);
})();
</script>
</head>
<body>
<div class="inner-wrapper stage-one">
<div class="barrio-logo">Friends of Barrio</div>
<div class="barrio-wel-message">
<h1>Welcome Friends of Barrio</h1>
<span>-</span>
<h2>Enter a valid membership number<br/> and activation code to continue</h2>
</div>
<form name="form1" method="post" action="check-activation.php" class="membership-form">
<h3>Your membership number</h3>
<input name="username" type="text" id="username" value="<?php echo $username; ?>" class="membership-number">
<h3>our activation code</h3>
<input name="activation_code" type="text" id="activation_code" value="<?php echo $activation_code; ?>" class="activation-code">
<input type="submit" name="Submit" value="Continue" class="membership-continue">
</form>
</div>
<div class="error-message">
<span>Your membership number & activation code <br/>is not valid, please check and re-enter</span>
</div>
<div class="background-tl"></div>
<div class="background-tr"></div>
<div class="background-bl"></div>
<div class="background-br"></div>
</body>
</html>
<?php
}
$con->close();
?>
The problem is that your query most likely doesn't return anything to begin with.
You are using the & operator, which is a bitwise operator.
Instead of
SELECT `Check_Activation` FROM `members`
WHERE `Username` = '".$username."' & `Activation` = '".$activation_code."';
use
SELECT `Check_Activation` FROM `members`
WHERE `Username` = '".$username."' AND `Activation` = '".$activation_code."'
Also, remove the ; at the end of the query.
To check if your query actually returns data, use
$array = mysql_fetch_array($checkcustomer);
if ($array === false)
{
// Do something if the query failed to return anything, i.e.
echo "Invalid username/activation code
}
Another note: Don't use $_POST values in queries, make sure you use mysql_real_escape on them first. Or even better, use prepared statements with PDO or mysqli.
replace if (is_null($array['Check_Activation']) for if(!$array['Check_Activation'])
If the activation row is not found, mysql_fetch_array will return FALSE. Try this:
if (empty($array['Check_Activation'])) {
It'll check if any row has been found and if the Check_Activation field is not empty.
Put AND instead of & in your query.
You should process $username and $activation_code through some escaping function to prevent SQL injection.
... WHERE `Username` = '" . mysql_real_escape_string($username) . "' ...
You have to match only name to check if the value is null or not
$checkcustomer = mysql_query("SELECT `Check_Activation` FROM `members`
WHERE `Username` = '".$username."'");
Then you have to check
$array = mysql_fetch_array($checkcustomer);
if (is_null($array['Check_Activation'])) {
$username = substr($username, 0, 1);
if($username == '1') {
$redirect_url='form-one.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '2') {
$redirect_url='form-two.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '3') {
$redirect_url='form-three.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
}
header("Location:". $redirect_url);
}
elseif($array['Check_Activation']==$activation_code )
{
/*your code */
}else{
?>