I am passing the string value through link in the URL to the next page like this <a href="ApplicationRegister.php?plan=trial">
In the ApplicationRegister.php page, i am getting this value like this $plan = $_GET["plan"];
and i will put this into a session variable like this $_SESSION['plans'] = $plan;
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.
My complete code is like this
$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)'] > 0)
{
$msg = "<font color='red'> <b>This E-mail id is already registered </b></font> ";
break;
}
}
if($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)'] > 0)
{
$msg = "<font color='red'> <b>This Stroename is already registered </b></font> ";
break;
}
}
if($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();
}
}
}
}
?>
Only in the beginning it holds the value , if i try to display it within theIf (isset($_POST['submit'])) it shows blank value for plans. Do not know what to do. Plz suggest
EDITED
Even after using like this, its the same. i do not know what may be the problem :(
$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'];
EDITED
In ApplicationRegister.php, i have passed the hiddenvalue which i got fro\m previous page like this
<input type="hidden" name="plan" value="<?php echo $plan ?>"/>
then POST method i have used this. Now i am getting the value for it. Thanks to all
EDITED
if($PlanName == "trail")
{
header("Location: userRegister.php");
exit();
}
else
{
header("Location : PaymentGateway.php");
exit();
}
It's because you're not calling session_start() at the top of the page. You need that for your sessions to persist across requests (which is the point of sessions)
As well as not calling session_start();, this code is wrong:
$plan = $_GET["plan"];
echo $plan;
$_SESSION['plan'] = $plan;
$plans = $_SESSION['plan'];
echo $_SESSION['plans'];
It should be:
$plan = $_GET["plan"];
echo $plan;
$_SESSION['plan'] = $plan;
$plans = $_SESSION['plans'];
echo $_SESSION['plans'];
You are setting $_SESSION['plan'] and then trying to access $_SESSION['plans'].
Also, are you clicking a link or submitting a form? You say that you have a link, yet your code tries to access values passed from a form.
If you are using a form, don't use links. Instead, use a select element to select a plan, and then change $plan = $_GET["plan"]; to $plan = $_POST["plan"];.
EDIT:
For the redirection problem, try this code:
echo "<pre>** Plan Name: **\n";
var_dump($PlanName);
echo "</pre>";
if($PlanName == "trail")
{
header("Location: userRegister.php");
exit();
}
else
{
header("Location: PaymentGateway.php");
exit();
}
and see what it outputs.
When someone clicks the link, it's going to set the variable properly. However, it's not going to hit the $_POST['submit'] logic, because it's not a post, just a get. Then, assuming your actually posting to that page at a later point, trying to access anything in $_GET will be null, and will then reset the session variable to null.
Your first page should have code something like this
<form action="ApplicationRegister.php" method="post">
<select name="plan">
<option value="trial">Trial</option>
</select>
<input type="submit"/>
</form>
Then, you check for $_POST['plan'] and $_POST['submit']
Related
<?php
if(isset($_POST['confirm'])){
$name = $db->validation($_POST['name']);
$id = "";
for($i= strlen($name) ; $i>=1 ; $i--){
if($name[$i-1] == 'a'){
break;
}else{
$id = $id.$name[$i-1];
}
}
$userid = strrev($id);
$query = "SELECT * FROM bluekolar_user WHERE token = '$name' AND user_id='$userid'";
$result = $db->select($query);
if($result){
if(mysqli_num_rows($result) ==1 ){
$qu = "UPDATE bluekolar_user SET active = '1' WHERE user_id='$userid'";
$res = $db->update($qu);
if($res){
$url = "profile/index.php?id=".$userid ;
header('location:'.$url) ;
exit();
}else{
$error = "Provide correct confirmation code";
}
}else{
$error = "Provide correct confirmation code";
}
}
}
?>
The code is performing good but it is unable to redirect. I don't know why. I need help.
Please help me so that I can redirect my code command to another page.
Thank you.
You need to ensure the path you are trying to redirect to is correct. The code in your question will try to redirect to a "profile" subdirectory of the location where this script is running. If you want to go to "http://example.com/profile/index.php?id=" you need to change your $url to be "/profile/index.php"
i have the following information displayed
<?php
$my_query="SELECT * FROM games";
$result= mysqli_query($connection, $my_query);
if (mysqli_num_rows($result) > 0)
while ($myrow = mysqli_fetch_array($result))
{
$description = $myrow["game_description"];
$image = $myrow["gamepic"];
$game_id = $myrow["game_id"];
$gamename = $myrow["game_name"];
echo "<div class='cover'>
</div>";
}
?>
as you can see i have created a game_details page which will display that specific Game_id when the image is clicked
im having trouble understanding how to pull the data out from that game_id in sql on the other page.
here is my attempt on the game_details page
<?php
if (!isset($_GET['$game_id']) || empty($_GET['game_id']))
{
echo "Invalid category ID.";
exit();
}
$game_id = mysqli_real_escape_string($connection, $_GET['game_id']);
$sql1 = "SELECT * games WHERE game_id={$game_id}'";
$res4 = mysqli_query($connection, $sql1);
if(!$res4 || mysqli_num_rows($res4) <= 0)
{
while ($row = mysqli_fetch_assoc($res4))
{
$gameid = $row['$game_id'];
$title = $row['game_name'];
$descrip = $row['game_description'];
$genre = $row['genretype'];
echo "<p> {$title} </p>";
}
}
?>
This attempt is giving me the "invalid category ID" error
Would appreciate help
There are a few issues with your code.
Let's start from the top.
['$game_id'] you need to remove the dollar sign from it in $_GET['$game_id']
Then, $row['$game_id'] same thing; remove the dollar sign.
Then, game_id={$game_id}' will throw a syntax error.
In your first body of code; you should also use proper bracing for all your conditional statements.
This one has none if (mysqli_num_rows($result) > 0) and will cause potential havoc.
Rewrites:
<?php
$my_query="SELECT * FROM games";
$result= mysqli_query($connection, $my_query);
if (mysqli_num_rows($result) > 0){
while ($myrow = mysqli_fetch_array($result))
{
$description = $myrow["game_description"];
$image = $myrow["gamepic"];
$game_id = $myrow["game_id"];
$gamename = $myrow["game_name"];
echo "<div class='cover'>
</div>";
}
}
?>
Sidenote for WHERE game_id='{$game_id}' in below. If that doesn't work, remove the quotes from it.
WHERE game_id={$game_id}
2nd body:
<?php
if (!isset($_GET['game_id']) || empty($_GET['game_id']))
{
echo "Invalid category ID.";
exit();
}
$game_id = mysqli_real_escape_string($connection, $_GET['game_id']);
$sql1 = "SELECT * games WHERE game_id='{$game_id}'";
$res4 = mysqli_query($connection, $sql1);
if(!$res4 || mysqli_num_rows($res4) <= 0)
{
while ($row = mysqli_fetch_assoc($res4))
{
$gameid = $row['game_id'];
$title = $row['game_name'];
$descrip = $row['game_description'];
$genre = $row['genretype'];
echo "<p> {$title} </p>";
}
}
?>
Use error checking tools at your disposal during testing:
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/function.error-reporting.php
You want to be using $_GET['gameid'] as that's the parameter you passed.
You are calling for game_id when the link to go to game_details.php has the variable gameid. Either change the parameter in the link to game_id or call for gameid in your $_GET['$game_id'].
Also, as Fred -ii- said, take out the dollar sign in $_GET['$game_id']
I have the following code to check if a row exists in MySQL:
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT 1 FROM files WHERE id='$code' LIMIT 1");
if (mysql_fetch_row($result)) {
echo 'Exists';
} else {
echo 'Does not exist';
}
}
?>
This works fine. But I need to change it a bit. I have the following fields:
id, title, url, type. When someone uses the code above ^ to check if a row exists, I need a variable to get the url from the same row, so I can redirect the user to there.
Do you have any idea how I can do that?
Thanks in advance! :)
Try this:
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT * FROM files WHERE id=" . $code . " LIMIT 1");
if (mysql_num_rows($result) > 0) {
while($rows = mysql_fetch_array($result)) {
echo 'Exists';
$url = $rows['url'];
}
} else {
echo 'Does not exist';
}
}
?>
It is quite simple. I think you don't show any effort to find the solution by yourself.
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT url FROM files WHERE id='$code' LIMIT 1");
if ($result) {
$url = mysql_fetch_row($resultado);
} else {
echo 'Does not exist';
}
}
<?php
$sql_query = "SELECT * FROM test WHERE userid ='$userid'";
$result1 =mysql_query($sql_query);
if(mysql_num_rows($result1)>0){
while($post = mysql_fetch_array($result1))
{
$url = $post['url'];
}
}
?>
If mysql_num_rows($result1)>0 it means row is existed fir the given user id
My header is not working.
<?php
$name = mysql_prep($_POST['name']);
$pastor = mysql_prep($_POST['pastor']);
$head = mysql_prep($_POST['head']);
$schedule = mysql_prep($_POST['schedule']);
$venue = mysql_prep($_POST['venue']);
$id = mysql_prep($_GET['ministryid']);
$errors = array();
$required_field = array('name', 'pastor', 'address', 'schedule', 'venue');
foreach ($required_field as $fieldname) {
if(!isset($_POST[$fieldname]) || empty($_POST[$fieldname])) {
$errors[] = $fieldname;
echo "Sorry, you missed to complete {$fieldname} <br />";
}
else {
$query = "UPDATE ministry SET
name = '{$name}',
pastor = '{$pastor}',
head = '{$head}',
schedule = '{$schedule}',
venue = '{$venue}'
WHERE id = {$id}";
mysql_query($query);
if(mysql_affected_rows() == 1) {
header('location: editministry.php?');
exit;
} else {
echo "Updating Failed on {$s_ministry['name']} <b />".mysql_error();
exit;
}
}
}
require_once("include/footer.php");
Every time I have successful update, the link change its address.
For example, when I'm updating id = 3, the address will change to editministry.php?ministryid=3.
You dont send anything along your URL, thats why your header won't work. Check for yourself.
if(mysql_affected_rows() == 1) {
header('location: editministry.php?');
exit;
Your link will become effectivly baseUrl/editministry.php?.It searches than for a variable, which is not defined. I am not sure how you actually can pass on a variable that you didnt define in a link, yet it sends you there. Don't know. But if you just tell it to the hard link without the questionmark, it should go to that page. For me it works at least within my code. For you it would be:
if(mysql_affected_rows() == 1) {
header('location: editministry.php' );
exit;
In my code it looks like this:
header( "Location: http://" . strip_tags( $_SERVER ['HTTP_HOST'] ) . "/newHolo/index.php" );
The following is the email verification code for my site.
The verification url sent to the user's email is as follows:
http://www.mywebsite.com/valid.php?confr=2774405&userid=2
Extra notes :
1) key is a column in my database which gets a random value on registration.
2) if $verify == 1 and password_in_db=== user_entered_password, then login takes place in the login page.
<?php
include 'connect.php';
$query = mysql_query("SELECT verify,key FROM users WHERE id = '$_GET['userid']'");
$details = mysql_fetch_assoc($query);
$verify = $details['verify'];
$confirm2 = $details['key'];
if($verify == "1") {
echo "Link Expired . Go to our login page :";
} else {
if (isset($_GET["confr"]) && isset($_GET["userid"])) {
$confirm1 =$_GET["confr"];
if($confirm1 == $confirm2) {
mysql_query("INSERT INTO users (`verify`) VALUES ('1') WHERE id = '$_GET["userid"]' ;");
echo "Thank You For Registering with us . Go to your LOGIN PAGE Here ";
} else {
echo "Invalid link ";
echo "Go to your LOGIN PAGE Here ";
}
} // of if isset
} // of else part
?>
Code for connect.php
<?php
mysql_connect("host", "username", "pass"); //connects to the server
mysql_select_db("database_name"); //selects the database
?>
The problem is that it is giving me a blank screen .
i believe the error lies in the sql
when ever i use a "WHERE" statement i always define as a variable, try this
<?php
include 'connect.php';
$user_id = $_GET["userid"];
$query = mysql_query("SELECT verify,key FROM users WHERE id = '$user_id'");
$details = mysql_fetch_assoc($query);
$verify = $details['verify'];
$confirm2 = $details['key'];
if($verify == "1"){
echo "Link Expired . Go to our login page :";
}
else{
if (isset($_GET["confr"]) && isset($_GET["userid"]))
{
$confirm1 =$_GET["confr"];
if($confirm1 == $confirm2){
mysql_query("INSERT INTO users (`verify`) VALUES ('1') WHERE id = '$user_id'");
echo "Thank You For Registering with us . Go to your LOGIN PAGE Here ";
}
else {
echo "Invalid link ";
echo "Go to your LOGIN PAGE Here ";
}
} // of if isset
} // of else part
?>
also, you have a semi colon in the insert sql
Try this.......
<?php
include 'connect.php';
$user_id = $_GET["userid"];
$query = mysql_query("SELECT verify,key FROM users WHERE id = '$user_id'");
while ($details = mysql_fetch_assoc($query)){
$verify = $details['verify'];
$confirm2 = $details['key'];
}
if($verify == "1"){
echo "Link Expired . Go to our login page :";
}
else{
if (isset($_GET["confr"]) && isset($_GET["userid"]))
{
$confirm1 =$_GET["confr"];
if($confirm1 == $confirm2){
mysql_query("INSERT INTO users (`verify`) VALUES ('1') WHERE id = '$user_id'");
echo "Thank You For Registering with us . Go to your LOGIN PAGE Here ";
}
else {
echo "Invalid link ";
echo "Go to your LOGIN PAGE Here ";
}
} // of if isset
} // of else part
?>
Note: insert statement has no where - as long as you dont use "insert into select..."
http://dev.mysql.com/doc/refman/5.1/de/insert.html