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POpassing and receiving string variables using php
 

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    <p>I am passing the string value through link in the URL to the next page like this <code>&lt;a href="ApplicationRegister.php?plan=trial"&gt;</code> In the ApplicationRegister.php page, i am getting this value like this <code>$plan = $_GET["plan"];</code> and i will put this into a session variable like this <code>$_SESSION['plans'] = $plan;</code> Here i am getting the value. but after the if statement i am not getting the value for this plan even after using Session variable. </p> <p>My complete code is like this</p> <pre><code> $plan = $_GET["plan"]; echo $plan; $_SESSION['plan'] = $plan; $plans = $_SESSION['plan']; echo $_SESSION['plans']; include('connect.php'); If (isset($_POST['submit'])) { $CompanyName = $_POST['CompanyName']; $CompanyEmail = $_POST['CompanyEmail']; $CompanyContact = $_POST['CompanyContact']; $CompanyAddress = $_POST['CompanyAddress']; $StoreName = $_POST['StoreName']; echo $plans; $myURL ="$_SERVER[HTTP_HOST]"; $myURL =$StoreName.'.'.$myURL; if (stripos($myURL, 'www.') !== 0) { $myURL = 'www.' . $myURL; } if (stripos($myURL, 'http://') !== 0) { $myURL = 'http://' .$myURL; } if(stripos($myURL, '.com') !== 0) { $myURL = $myURL . '.com'; } echo $plans; $RegistrationType = $_POST['RegistrationType']; $Status = "Active"; $sql = "select * from plans where planname = '$plans'"; echo $sql; mysql_query($sql) or die (mysql_error()); $planID = $row['planid']; $query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ; $result1 = mysql_query($query1) or die ("ERROR: " . mysql_error()); $msg = ""; while ($row = mysql_fetch_array($result1)) { if($row['count(CompanyEmail)'] &gt; 0) { $msg = "&lt;font color='red'&gt; &lt;b&gt;This E-mail id is already registered &lt;/b&gt;&lt;/font&gt; "; break; } } if($msg == "") { $query2 = "select count(URL) from ApplicationRegister where URL = '$myURL' "; $result2 = mysql_query($query2) or die ("ERROR: " . mysql_error()); $msg = ""; while ($row = mysql_fetch_array($result2)) { if($row['count(URL)'] &gt; 0) { $msg = "&lt;font color='red'&gt; &lt;b&gt;This Stroename is already registered &lt;/b&gt;&lt;/font&gt; "; break; } } if($msg == "") { $sql = "INSERT INTO ApplicationRegister(planid, CompanyName, CompanyEmail, CompanyContact, CompanyAddress, RegistrationType, ApplicationPlan, ApplicationStatus, URL, CreatedDate) VALUES ('$planID', '$CompanyName', '$CompanyEmail', '$CompanyContact', '$CompanyAddress', '$RegistrationType', '$plans', '$Status', '$myURL', NOW() )"; mysql_query($sql) or die(mysql_error()); $id = mysql_insert_id(); $_SESSION['application_id'] = $id; if($plans == "trail") { header("Location: userRegister.php"); exit(); } else { header("Location : PaymentGateway.php"); exit(); } } } } ?&gt; </code></pre> <p>Only in the beginning it holds the value , if i try to display it within theIf <code>(isset($_POST['submit']))</code> it shows blank value for plans. Do not know what to do. Plz suggest </p> <p><strong>EDITED</strong> Even after using like this, its the same. i do not know what may be the problem :( <pre><code> $plan = $_GET["plan"]; echo $plan; $_SESSION['plans'] = $plans; echo $_SESSION['plans']; // $plan = +$plan; include('connect.php'); If (isset($_POST['submit'])) { $CompanyName = $_POST['CompanyName']; $CompanyEmail = $_POST['CompanyEmail']; $CompanyContact = $_POST['CompanyContact']; $CompanyAddress = $_POST['CompanyAddress']; $StoreName = $_POST['StoreName']; echo $_SESSION['plans']; </code></pre> <p><strong>EDITED</strong></p> <p>In ApplicationRegister.php, i have passed the hiddenvalue which i got fro\m previous page like this</p> <pre><code>&lt;input type="hidden" name="plan" value="&lt;?php echo $plan ?&gt;"/&gt; </code></pre> <p>then POST method i have used this. Now i am getting the value for it. Thanks to all</p> <p><strong>EDITED</strong></p> <pre><code>if($PlanName == "trail") { header("Location: userRegister.php"); exit(); } else { header("Location : PaymentGateway.php"); exit(); } </code></pre>
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    1. COAre you calling `session_start()` before any output is sent? As a side note, when you post code to SO, it is a good idea to simplify your code to the simplest possible example that still shows your problem (most people won't read through that much code).
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      1. USAndrew
    2. COYou have no SQL injection protection, [Bobby Tables](http://xkcd.com/327/). You should really consider using [PDO](http://php.net/pdo) or at least [`mysql_real_escape_string()`](http://php.net/mysql_real_escape_string).
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      1. USBailey Parker
    3. COand you are setting $_SESSION['plan'] and trying to echo $_SESSION['plans'].. (added s on plan)
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