i want to keep some variable alive so that it is available to all the pages of the site ;
i tried global but that don't work with these kind of problem ;
i use the following code :
while($result1 = mysql_fetch_array( $result))
{
$adm_no = $result1['adm_no'];
$adm_dt = $result1['adm_dt'];
$name = $result1['name'];
$dob = $result1['dob'];
$f_name = $result1['f_name'];
$f_office = $result1['f_office'];
$f_o_no = $result1['f_o_no'];
$m_name = $result1['m_name'];
$m_office = $result1['m_office'];
$addr = $result1['addr'];
$pho_no = $result['pho_no'];
these same variable in another page called tc.php . how can i do that ????
If you want to access all that data again in another page I would recommend storing the information needed to retrieve data from your mysql table in a session rather than the result of the query. This means you don't have a load of trivial data in your session space. For example.
Imagine I have a person table and want to get bits of information for that person on different pages I just store the person_id in a session like so:
//home.php
$_SESSION['personID'] = $personID;
Then on any page I want to retrieve person information on I just get the person id from the session and run the query to get the specific information I need.
//profile.php
$personID = $_SESSION['personID'];
//Get specific information here
If you really cant change the way that you are doing this which I really hope you can as it'll make your life a hell of a lot easier then just changing your code to this:
//make sure that you have started a session at the top of your page before you do anything else
session_start();
while($result1 = mysql_fetch_array($result)) {
$_SESSION['adm_no'] = $result1['adm_no'];
$_SESSION['adm_dt'] = $result1['adm_dt'];
$_SESSION['name'] = $result1['name'];
$_SESSION['dob'] = $result1['dob'];
//etc
}
Use
$_SESSION['myvar']= "your value";
echo $_SESSION['myvar'];
will can access any page
Fetch data again in tc.php - it is the best way in this case I think.
You can also set that data to the session, and in tc.php get it from there.
Related
On my website, I allow users to view a users information by simply clicking their name. Once they click the persons name, they can schedule the person to come to an event. When the user clicks "schedule me" I take the them full name from the "user_id" and send it as a "$_SESSION['speaker']" to the next file that pretty much checks if the user came from the last file and takes the name and uses it as the input value for the calendar. The problem I am having is that when the user didn't "click schedule" from the other file and goes to the calendar website alone, the name from the previous person they clicked stays there and I want it to be blank in case they want to put a different name. So pretty much i would access the calendar website just by typing the URL and the name would still be in the session. I want to clear the session without logging the user out so they don't see the name of the previous person they clicked. Here is some of my code
First file
$_GET['speaker'] = $_SESSION['speaker_id'];
$speaker_id = $_GET['speaker'];
$stmtSpeaker = $handler->prepare("SELECT * FROM formdata WHERE user_id= :speaker_id");
$stmtSpeaker->bindParam(':speaker_id', $speaker_id, PDO::PARAM_INT);
$stmtSpeaker->execute();
$formData = $stmtSpeaker->fetch();
if(isset($_POST['schedule_me'])){
$_SESSION['admin'] = $adminBoolean;
$_SESSION['speaker'] = $formData['fullname'];
$_SESSION['speaker_came'] = true;
header("Location: admincalendar.php");
exit;
}
Second file
$adminBoolean = $resultChecker['admin'];
if($_SESSION['speaker_came'] = true){
$speaker = $_SESSION['speaker'];
}else{
$speaker = "";
}
Unset will destroy a particular session variable whereas session_destroy() will destroy all the session data for that user.
It really depends on your application as to which one you should use. Just keep the above in mind.
unset($_SESSION['name']); // will delete just the name data
session_destroy(); // will delete ALL data associated with that user.
You can unset session variable
$adminBoolean = $resultChecker['admin'];
if($_SESSION['speaker_came'] = true){
$speaker = $_SESSION['speaker'];
}else{
unset($_SESSION['speaker']);
unset($_SESSION['speaker_came']);
$speaker = '';
}
You need to first get the tempkey of the element and then unset it. Try this:
if(($tempkey = array_search($speaker_id, $_SESSION['speaker'])) !== FALSE)
unset($_SESSION['speaker'][$tempkey]);
How to post values to loginchk_coustomer.php given in below code, not through Url by any other way.
Is there any other way to post these value to loginchk_coustomer.php becoz it is not secure.
<?php
include "include/connect.php";
$user_name = $_REQUEST['user_name'];
$password = $_REQUEST['password'];
//echo "select * from school_info where school_id='$user_name' and school_password='$password'";
$sql_query = mysql_fetch_assoc(mysql_query("select * from school_info where school_id='$user_name' and school_password='$password'"));
$db_username = $sql_query['db_username'];
$db_password = $sql_query['db_password'];
$db_databasename = $sql_query['db_databasename'];
echo "<script>";
echo "self.location='member/loginchk_customer.php?db_username=$db_username&db_password=$db_password&db_databasename=$db_databasename&user_name=$user_name&password=$password'"; // Comment this line if you don't want to redirect
echo "</script>";
?>
You need to create a session to store all that information.
Here's what they are - from http://php.net/manual/en/features.sessions.php:
Session support in PHP consists of a way to preserve certain data across subsequent accesses.
To start a session write at the beginning of your code:
session_start(); // needed in all pages that will use the variables below
and then after your assign the information this way:
$_SESSION['username'] = $sql_query['db_username'];
$_SESSION['password'] = $sql_query['db_password'];
$_SESSION['databasename'] = $sql_query['db_databasename'];
All the information will persist on those variables along the site until you do:
session_destroy();
I also recommend you not to redirect with javascript, but this way in PHP:
header('Location: member/loginchk_customer.php');
Possibly after checking this answer you will think about to change the way you check the login information. But that's okay. It's the way of learning.
More information about sessions: http://php.net/manual/en/book.session.php
I hope this helps.
I have am creating a Website that showes Visitors Info. Users are able to visit the page and use Textarea to pick a name for their URL, and the name will be saved as a table in mysql database..
I am using the $name variable in my first php file which is a replacement for the text "visitor_tracking". But today I noticed that there is also another php file and more sql codes, and once again I can see that this file also has the "visitor_tracking" text used in the sql code.
But I think I failed big time, because I simply dont know how to replace the "visitor_tracking" text with my the variable name called $name.
<?php
//define our "maximum idle period" to be 30 minutes
$mins = 30;
//set the time limit before a session expires
ini_set ("session.gc_maxlifetime", $mins * 60);
session_start();
$ip_address = $_SERVER["REMOTE_ADDR"];
$page_name = $_SERVER["SCRIPT_NAME"];
$query_string = $_SERVER["QUERY_STRING"];
$current_page = $page_name."?".$query_string;
//connect to the database using your database settings
include("db_connect.php");
if(isset($_SESSION["tracking"])){
//update the visitor log in the database, based on the current visitor
//id held in $_SESSION["visitor_id"]
$visitor_id = isset($_SESSION["visitor_id"])?$_SESSION["visitor_id"]:0;
if($_SESSION["current_page"] != $current_page)
{
$sql = "INSERT INTO visitor_tracking
(ip_address, page_name, query_string, visitor_id)
VALUES ('$ip_address', '$page_name', '$query_string', '$visitor_id')";
if(!mysql_query($sql)){
echo "Failed to update visitor log";
}
$_SESSION["current_page"] = $current_page;
}
} else {
//set a session variable so we know that this visitor is being tracked
//insert a new row into the database for this person
$sql = "INSERT INTO visitor_tracking
(ip_address, page_name, query_string)
VALUES ('$ip_address', '$page_name', '$query_string')";
if(!mysql_query($sql)){
echo "Failed to add new visitor into tracking log";
$_SESSION["tracking"] = false;
} else {
//find the next available visitor_id for the database
//to assign to this person
$_SESSION["tracking"] = true;
$entry_id = mysql_insert_id();
$lowest_sql = mysql_query("SELECT MAX(visitor_id) as next FROM visitor_tracking");
$lowest_row = mysql_fetch_array($lowest_sql);
$lowest = $lowest_row["next"];
if(!isset($lowest))
$lowest = 1;
else
$lowest++;
//update the visitor entry with the new visitor id
//Note, that we do it in this way to prevent a "race condition"
mysql_query("UPDATE visitor_tracking SET visitor_id = '$lowest' WHERE entry_id = '$entry_id'");
//place the current visitor_id into the session so we can use it on
//subsequent visits to track this person
$_SESSION["visitor_id"] = $lowest;
//save the current page to session so we don't track if someone just refreshes the page
$_SESSION["current_page"] = $current_page;
}
}
Here is a very short part of the script:
I really hope I can get some help to replace the "visitor_tracking" text with the Variable $name...I tried to replace the text with '$name' and used also different qoutes, but didnt work for me...
And this is the call that I used in my 2nd php file that reads from my first php file:
include 'myfile1.php';
echo $var;
But dont know if thats correct too. I cant wait to hear what I am doing wrong.
Thank you very much in advance
PS Many thanks to Prix for helping me with the first php file!
first you need to start session in both pages. it should be the first thing you do in page before writing anything to page output buffer.
In first page you need to assign the value to a session variable. if you don't start session with session_start you don't have a session and value in $_SESSION will not be available.
<?php
session_start(); // first thing in page
?>
<form action="" method="post" >
...
<td><input type="text" name="gname" id="text" value=""></td>
...
</form>
<?PHP
if (isset($_POST['submit'])) {
$name = $_POST['gname'];
//...
//Connect to database and create table
//...
$_SESSION['gname'] = $name;
...
// REMOVE THIS Duplicate -> mysql_query($sql,$conn);
}
?>
in second page again you need to start session first. Before reading a $_SESSION variable you need to check if it has a value (avoid errors or warnings). next read the value and do whatever you want to do with it.
<?php
session_start(); // first thing in page
...
if(isset($_SESSION['gname'])){
// Read the variable from session
$SomeVar = $_SESSION['gname'];
// Do whatever you want with this value
}
?>
By the way,
In your second page, I couldn't find the variable $name.
The way you are creating your table has serious security issue and least of your problems will be a bad table name which cannot be created. read about SQL injection if you are interested to know why.
in your first page you are running $SQL command twice and it will try to create table again which will fail.
Your if statement is finishing before creating table. What if the form wasn't submitted or it $_POST['gname'] was emptY?
there are so many errors in your second page too.
There are not really and direct answers on this, so I thought i'd give it a go.
$myid = $_POST['id'];
//Select the post from the database according to the id.
$query = mysql_query("SELECT * FROM repairs WHERE id = " .$myid . " AND name = '' AND email = '' AND address1 = '' AND postcode = '';") or die(header('Location: 404.php'));
The above code is supposed to set the variable $myid as the posted content of id, the variable is then used in an SQL WHERE clause to fetch data from a database according to the submitted id. Forgetting the potential SQL injects (I will fix them later) why exactly does this not work?
Okay here is the full code from my test of it:
<?php
//This includes the variables, adjusted within the 'config.php file' and the functions from the 'functions.php' - the config variables are adjusted prior to anything else.
require('configs/config.php');
require('configs/functions.php');
//Check to see if the form has been submited, if it has we continue with the script.
if(isset($_POST['confirmation']) and $_POST['confirmation']=='true')
{
//Slashes are removed, depending on configuration.
if(get_magic_quotes_gpc())
{
$_POST['model'] = stripslashes($_POST['model']);
$_POST['problem'] = stripslashes($_POST['problem']);
$_POST['info'] = stripslashes($_POST['info']);
}
//Create the future ID of the post - obviously this will create and give the id of the post, it is generated in numerical order.
$maxid = mysql_fetch_array(mysql_query('select max(id) as id from repairs'));
$id = intval($maxid['id'])+1;
//Here the variables are protected using PHP and the input fields are also limited, where applicable.
$model = mysql_escape_string(substr($_POST['model'],0,9));
$problem = mysql_escape_string(substr($_POST['problem'],0,255));
$info = mysql_escape_string(substr($_POST['info'],0,6000));
//The post information is submitted into the database, the admin is then forwarded to the page for the new post. Else a warning is displayed and the admin is forwarded back to the new post page.
if(mysql_query("insert into repairs (id, model, problem, info) values ('$_POST[id]', '$_POST[model]', '$_POST[version]', '$_POST[info]')"))
{
?>
<?php
$myid = $_POST['id'];
//Select the post from the database according to the id.
$query = mysql_query("SELECT * FROM repairs WHERE id=" .$myid . " AND name = '' AND email = '' AND address1 = '' AND postcode = '';") or die(header('Location: 404.php'));
//This re-directs to an error page the user preventing them from viewing the page if there are no rows with data equal to the query.
if( mysql_num_rows($query) < 1 )
{
header('Location: 404.php');
exit;
}
//Assign variable names to each column in the database.
while($row = mysql_fetch_array($query))
{
$model = $row['model'];
$problem = $row['problem'];
}
//Select the post from the database according to the id.
$query2 = mysql_query('SELECT * FROM devices WHERE version = "'.$model.'" AND issue = "'.$problem.'";') or die(header('Location: 404.php'));
//This re-directs to an error page the user preventing them from viewing the page if there are no rows with data equal to the query.
if( mysql_num_rows($query2) < 1 )
{
header('Location: 404.php');
exit;
}
//Assign variable names to each column in the database.
while($row2 = mysql_fetch_array($query2))
{
$price = $row2['price'];
$device = $row2['device'];
$image = $row2['image'];
}
?>
<?php echo $id; ?>
<?php echo $model; ?>
<?php echo $problem; ?>
<?php echo $price; ?>
<?php echo $device; ?>
<?php echo $image; ?>
<?
}
else
{
echo '<meta http-equiv="refresh" content="2; URL=iphone.php"><div id="confirms" style="text-align:center;">Oops! An error occurred while submitting the post! Try again…</div></br>';
}
}
?>
What data type is id in your table? You maybe need to surround it in single quotes.
$query = msql_query("SELECT * FROM repairs WHERE id = '$myid' AND...")
Edit: Also you do not need to use concatenation with a double-quoted string.
Check the value of $myid and the entire dynamically created SQL string to make sure it contains what you think it contains.
It's likely that your problem arises from the use of empty-string comparisons for columns that probably contain NULL values. Try name IS NULL and so on for all the empty strings.
The only reason $myid would be empty, is if it's not being sent by the browser. Make sure your form action is set to POST. You can verify there are values in $_POST with the following:
print_r($_POST);
And, echo out your query to make sure it's what you expect it to be. Try running it manually via PHPMyAdmin or MySQL Workbench.
Using $something = mysql_real_escape_string($POST['something']);
Does not only prevent SQL-injection, it also prevents syntax errors due to people entering data like:
name = O'Reilly <<-- query will bomb with an error
memo = Chairman said: "welcome"
etc.
So in order to have a valid and working application it really is indispensible.
The argument of "I'll fix it later" has a few logical flaws:
It is slower to fix stuff later, you will spend more time overall because you need to revisit old code.
You will get unneeded bug reports in testing due to the functional errors mentioned above.
I'll do it later thingies tend to never happen.
Security is not optional, it is essential.
What happens if you get fulled off the project and someone else has to take over, (s)he will not know about your outstanding issues.
If you do something, finish it, don't leave al sorts of issues outstanding.
If I were your boss and did a code review on that code, you would be fired on the spot.
Please could someone help im building my first website that pulls info from a MySQL table, so far ive successfully managed to connect to the database and pull the information i need.
my website is set up to display a single record from the table, which it is doing however i need some way of changing the URL for each record, so i can link pages to specific records. i have seen on websites like facebook everyones profile ends with a unique number. e.g. http://www.facebook.com/profile.php?id=793636552
Id like to base my ID on the primary key on my table e.g. location_id
ive included my php code so far,
<?php
require "connect.php";
$query = "select * from location limit 1";
$result = #mysql_query($query, $connection)
or die ("Unable to perform query<br>$query");
?>
<?php
while($row= mysql_fetch_array($result))
{
?>
<?php echo $row['image'] ?>
<?php
}
?>
Thanks
Use $_GET to retrieve things from the script's query (aka command line, in a way):
<?php
$id = (intval)$_GET['id']; // force this query parameter to be treated as an integer
$query = "SELECT * FROM location WHERE id={$id};";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) == 0) {
echo 'nothing found';
} else {
$row = mysql_fetch_assoc($result);
echo $row['image'];
}
There are many things to consider if this is your first foray into MsSQL development.
SQL Injection
Someone might INSERT / DELETE, etc things via using your id from your url (be careful!, clean your input)
Leaking data
Someone might request id = 1234924 and you expected id = 12134 (so some sensitive data could be shown, etc;).
Use a light framework
If you haven't looked before, I would suggest something like a framework (CodeIgniter, or CakePHP), mysql calls, connections, validations are all boilerplate code (always have to do them). Best to save time and get into making your app rather than re-inventing the wheel.
Once you have selected the record from the database, you can redirect the user to a different url using the header() function. Example:
header('Location: http://yoursite.com/page.php?id=123');
You would need to create a link to the same (or a new page) with the URL as you desire, and then logic to check for the parameter to pull a certain image...
if you're listing all of them, you could:
echo "" . $row['name'] . ""
This would make the link.. now when they click it, in samepage.php you would want to look for it:
if (isset($_GET['id']) && is_numeric($_GET['id'])) {
//query the db and pull that image..
}
What you are looking for is the query string or get variables. You can access a get variable through php with $_GET['name']. For example:
http://www.facebook.com/profile.php?id=793636552
everything after the ? is the query string. The name of the variable is id, so to access it through your php you would use $_GET['id']. You can build onto these this an & in between the variables. For example:
http://www.facebook.com/profile.php?id=793636552&photo=12345
And here we have $_GET['id'] and $_GET['photo'].
variables can be pulled out of URL's very easily:
www.site.com/index.php?id=12345
we can access the number after id with $_GET['id']
echo $_GET['id'];
outputs:
12345
so if you had a list of records (or images, in your case), you can link to them even easier:
$query = mysql_query(...);
$numrows = mysql_num_rows($query);
for ($num=0;$num<=$numrows;$num++) {
$array = mysql_fetch_array($query);
echo "<a href=\"./index.php?id=". $row['id'] ."\" />Image #". $row['id'] ."</a>";
}
that will display all of your records like so:
Image #1 (links to: http://www.site.com/index.php?id=1)
Image #2 (links to: http://www.site.com/index.php?id=2)
Image #3 (links to: http://www.site.com/index.php?id=3)
...