I am working on something it has 2 pages. One is index.php and another one is admin.php.I am making CMS page where you can edit information on the page yourself. Then it will go to the database, where the information is stored. I also have to have it where the user can update the information on the page. I am getting a little bit confused here.For instance here I am calling the database and I am starting a function called get_content:
<?php
function dbConnect(){
$hostname="localhost";
$database="blank";
$mysql_login="blank";
$mysql_password="blank";
if(!($db=mysql_connect($hostname, $mysql_login, $mysql_password))){
echo"error on connect";
}
else{
if(!(mysql_select_db($database,$db))){
echo mysql_error();
echo "<br />error on database connection. Check your settings.";
}
else{
return $db;
}
}
function get_content(){
$sql = "Select PageID,PageHeading,SubHeading,PageTitle,MetaDescription,MetaKeywords From tblContent ";
$query = mysql_query($sql) or die(mysql_error());
while ($row =mysql_fetch_assoc($query,MYSQL_ASSOC)){
$title =$row['PageID'[;
$PageHeading =$row['PageHeading'];
$SubHeading = $row['SubHeading'];
$PageTitle = $row['PageTitle'];
$MetaDescription =$row['MetaDescription'];
$MetaKeywords = $row['MetaKeywords'];
?>
And then on the index page and I am going to echo it out in the spot that someone can change:
<h2><?php echo mysql_result($row,0,"SubHeading");?>A Valid XHTML and CSS Web Design by WG.</h2>
I do know that the function is not finished I am still working on that part. What I am wondering is am I echoing it out right or I am way off. This is my first time messing with CMS in php and I am still learning it. I am working with navicat and text pad on this, yes I know it is old school but that is what I am being shown with. But my index is a form not a blog. I have seen many of CMS pages for blogs not to many to be used with forms. Any input will be considered thanks for reading my question.
Your question is a bit confusing and your code very incomplete. I'ts hard to say if you do it the right way since I don't see the rest of the script. You need to connect to the database there as well and get your data. The $row variable only exists in the while statement inside you function get_content() though.
You could complete the get_content() and use it in the index.php as well. Remember that the variables you define inside a function only is available there though. If you need the data outside that function you need to return the values you need and save them to some other variable there. Put if you do the same as you've started doing in the get_content() function in index.php, then you just have to echo the variables you define. Like this:
<h2><?php echo $SubHeading; ?></h2>
or you could also do it like this somewhere inside the php tags:
echo '<h2>{$SubHeading}</h2>';
I hope that answers your question.
EDIT:
What you need in the index.php page is exactly what you seem to be doing in the admin file. You need to connect to db using mysql_connect() and select db with mysql_select_db(). You then need to select the data from the db using the appropriate query with $query = mysql_query($sql). If it's more then one row you want to display you need to put it in a while loop otherwise (which seems to be the case here) you just need to do one $row = mysql_fetch_assoc($query). After that you can get the data using $row['column_name']. If you have more than one row you can just use $row['column_name'] in side the while loop to get each consecutive row's data.
Here is an example index.php:
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password') or
die('Could not connect: ' . mysql_error());
mysql_select_db('database_name')) or die('Could not select database: ' .
mysql_error());
$sql = "SELECT SubHeading FROM tblContent WHERE PageID='1' LIMIT 1;";
$query = mysql_query($sql);
$row = mysql_fetch_assoc($query);
echo '<h2>{$row[\'SubHeading\']}</h2>';
mysql_close();
?>
This is just what you need to display the SubHeading from you database. You probably also need to handle your form and save the submitted data to the database in your admin.php file.
Related
I am using similar syntax in my blog. However, On my forum, nothing happens! This has been such an infuriating thing to tackle, as everything seems to be working exactly as my blog did. Here's my code I pass through and call the delete_post page
CHUNK FROM VIEWPOST.PHP
while($row = mysqli_fetch_array($result)){
echo '<tr>';
echo '<td class="postleft">';
echo date('F j, Y, g:i a', strtotime($row['forumpost_Date'])) . "<br>" .$row['user_Name']. "<br>" .$row['forumpost_ID'];
echo '</td>';
echo '<td class="postright">';
echo $row['forumpost_Text'];
echo '</td>';
if(isset ($_SESSION['loggedin']) && ($_SESSION['user_AuthLvl']) == 1){
echo '<td class="postright">';
echo '<a class= "btn btm-default" href="#">Edit</a>';
echo '<a class= "btn btm-default" href="delete_post.php?forumpost_ID='.$row['forumpost_ID'].'">Delete</a>';
echo '</td>';}
else if(isset ($_SESSION['loggedin']) && ($_SESSION['user_ID']) == $row['forumpost_Author']){
echo '<td class="postright">';
echo '<a class= "btn btm-default" href="#">Edit</a>';
echo '<a class= "btn btm-default" href="delete_post.php?forumpost_ID='.$row['forumpost_ID'].'">Delete</a>';
echo '</td>';}
echo '</tr>';
}echo '</table>';
DELETE POST FUNCTION
<?php
include ('header.php');
include ('dbconnect.php');
//A simple if statement page which takes the person back to the homepage
//via the header statement after a post is deleted. Kill the connection after.
if(!isset($_GET['forumpost_ID'])){
header('Location: index.php');
die();
}else{
delete('hw7_forumpost', $_GET['forumpost_ID']);
header('Location: index.php');
die();
}
/********************************************
delete function
**********************************************/
function delete($table, $forumpost_ID){
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
?>
Now it is showing the ID's as intended, it just simply does not delete the post. It's such a simple Query, I don't know where my syntax is not matching up!
EDIT FOR DBCONNECT.PHP
<?php
/*---------------------------------------
DATABASE CONNECT PAGE
A simple connection to my database to utilize
for all of my pages!
----------------------------------------*/
$host = 'localhost';
$user = 'ad60';
$password = '4166346';
$dbname = 'ad60';
$connectDB = mysqli_connect($host, $user, $password, $dbname);
if (!$connectDB){
die('ERROR: CAN NOT CONNECT TO THE DATABASE!!!: '. mysqli_error($connectDB));
}
mysqli_select_db($connectDB,"ad60") or die("Unable to select database: ".mysqli_error($connectDB));
?>
Ok, I saw this and I would like to suggest the following:
In general
When you reuse code and copy paste it like you have done there is always the danger that you forget to edit parts that should be changed to make the code work within the new context. You should actually not use code like this.
Also you have hard coded configuration in your code. You should move up all the configuration to one central place. Never have hard coded values inside your functional code.
Learn more about this in general by reading up about code smell, programming patterns and mvc.
To find the problem
Now to fix your problem lets analyse your code starting with delete_post.php
First check if we actually end up inside delete_post.php. Just place an echo "hello world bladiebla" in top of the file and then exit. This looks stupid but since I can't see in your code if the paths match up check this please.
Now we have to make sure the required references are included properly. You start with the include functionality of php. This works of course, but when inside dbconnect.php something goes wrong while parsing your script it will continue to run. Using require would fix this. And to prevent files from loading twice you can use require_once. Check if you actually have included the dbconnect.php. You can do this by checking if the variables inside dbconnect.php exist.
Now we know we have access to the database confirm that delete_post.php received the forumpost_ID parameter. Just do print_r($_GET) and exit. Check if the field is set and if the value is set. Also check if the value is actually the correct value.
When above is all good we can go on. In your code you check if the forumpost_ID is set, but you do not check if the forumpost_ID has an actual value. In the above step we've validated this but still. Validate if your if
statement actually functions by echoing yes and no. Then test your url with different inputs.
Now we know if the code actually gets executed with all the resources that are required. You have a dedicated file that is meant to delete something. There is no need to use a function because this creates a new context and makes it necessary to make a call and check if the function context has access to all the variables you use in the upper context. In your case I would drop the function and just put the code directly within the else statement.
Then check the following:
Did you connect to the right database
Is the query correct (echo it)
Checkout the result of mysqli_query
Note! It was a while ago since I programmed with php so I assume noting from the codes behavior. This is always handy. You could check the php versions on your server for this could also be the problem. In the long run try to learn and use MVC. You can also use frameworks like codeigniter which already implemented the MVC design pattern.
You have to declare $connectDB as global in function.
function delete($table, $forumpost_ID){
global $connectDB;
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
See the reference about variable scope here:
http://php.net/manual/en/language.variables.scope.php
please try to use below solution.
<?php
include ('header.php');
include ('dbconnect.php');
//A simple if statement page which takes the person back to the homepage
//via the header statement after a post is deleted. Kill the connection after.
if(!isset($_GET['forumpost_ID'])){
header('Location: index.php');
die();
}else{
delete('hw7_forumpost', $_GET['forumpost_ID'], $connectDB);
header('Location: index.php');
die();
}
/********************************************
delete function
**********************************************/
function delete($table, $forumpost_ID, $connectDB){
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
?>
I wish this solution work for you best of luck!
I want to use one single page with pre-defined divs, layout etc. as basis so that when a product is clicked on from elsewhere it loads that product info onto the page?
They way im doing it ill be sitting here till about 2020 still typing out product info onto pages.
EDIT*************
function product ()
{
$get = mysql_query ("SELECT id, name, description, price, imgcover FROM products WHERE size ='11'");
if (mysql_num_rows($get) == FALSE)
{
echo " There are no products to be displayed!";
}
else
{
while ($get_row = mysql_fetch_assoc($get))
{
echo "<div id='productbox'><a href=product1.php>".$get_row['name'].'<br />
'.$get_row['price']. '<br />
' .$get_row['description']. '<br />
' .$get_row['imgcover']. "</a></div>" ;
}
}
}
In addition one problem I have with that code is that the <a href> tag only goes to product1.php. Any ideas how I can make that link to blank product layout page that would be filled with the product info that the user has just clicked on, basically linking to itself on a blank layout page.
Thanks any help would be great!
Thanks Maxyy
Since you dont have code this is a general way of doing this. What you want is a template for the product page
Query the database
load the data into a variable
make a script that will print out the data from the variable into a product page
somescript.php
<?php
$productid = $_REQUEST['productid']; //Of course do sanitation
//before using get,post variables
//though you should be using mysqli_* functions as mysql_* are depreciated
$result = mysql_query("select * from sometable where id='{$productid}");
$product = mysql_fetch_object($result);
include("productpage.php");
productpage.php
<div class="Product">
<div class="picture"><img src="<?php echo $product->imghref;?>" /></div>
<div class="price"><?php echo $product->price;?></div>
</div>
so on and so fourth. Included scripts use whatever variables are currently in the scope of the calling function
If you are meaning to load the products into the same page without doing another page load you will need to use ajax. Which is javascript code that use XHR requests to return data from a server. You can either do pure javascript or a library like jQuery to simplify the process of doing a xhr request by using $.ajax calls.
I know this question has been asked over 4 years ago, but since there's been no answer marked as right, I thought I might chip in.
First, let's upgrade from mysql and use mysqli - my personal favorite, you can also use PDO. Have you tried using $_GET to pull the id of whatever product you want to see and then displaying them all together or one at a time?
It could look something like this:
<?php // start by creating $mysqli connection
$host = "localhost";
$user = "username";
$pass = "password";
$db_name = "database";
$mysqli = new mysqli($host, $user, $pass, $db_name);
if($mysqli->connect_error)
{
die("Having some trouble pulling data");
exit();
}
Assuming the connection was made successfully we move on to checking for an ID being set. In this case I check it via an URL param assumed to be id. You can make it more complex, or take a different approach here.
if(isset($_GET['id']))
{
$id = htmlentities($_GET['id']);
$query = "SELECT * FROM table WHERE id = " . $id;
}
else
{
// if no ID is set, just bring all the results down
// then you can modify how, and which table the results
// are being used.
$query = "SELECT * FROM table ORDER BY id"; // the query can be changed to whatever you would be prefer
}
Once we have decided on a query we go on to start querying the database for information. I have three steps:
Check query >
Check table for records >
Loop through roles and create an object for each.
if($result = $mysqli->query($query))
{
if($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
// you can set up your element here
// you can set it up in whatever way you want
// to see your product being displayed, by simply
// using $row->column_name
// each column becomes an object here. So your id
// column would be pulled using $row->id
echo "<h1>" . $row->name . "</h1>";
echo "<p>" . $row->description . "</p>";
echo "<img src=" . $row->image_path . ">";
// etc ...
}
}
else
{
// if no records match the selected ID
echo "Nothing to see here...";
}
}
else
{
// if there's a problem with the query
echo "A slight problem with your query.";
}
$mysqli->close(); // close connection for safety
?>
I hope this answers your question and can help you if you are still stuck on this problem. This is the bare skeleton of what you can do with MySQLi and PHP, you could always use some Ajax to make the page more interactive, and user-friendly.
Adding content to a page on click needs to be done in either Javascript or in JQuery.
You can use ajax call to retrive the needed data from php page, Syntax is here.
Or you can also load a php page to a div content with .load() function in JQuery, Syntax is here.
So I'm trying to write a temp way to login to the admin panel using an if else statement while I read up on PDO. If someone could tell me where the error lies here it would be much appreciated.
I've updated my code after looking around a little bit, but I still have the issue of nothing showing up where my code belongs and pulling the information it should.
<?php
$admin = $_SESSION['admin_login'];
$con=mysql_connect("$server","$user","$pass");
if
(!$con)
{
die('Could not Connect' .mysql_error());
}
mysql_select_db($webdb, $con);
$result=mysql_query("SELECT * FROM permissions WHERE username= '$admin' ");
$row = mysql_fetch_assoc($result);
if ($row['permissions']=="3")
{
echo 'Admin Panel';
}
elseif ($row['permissions']=="1")
{
echo 'include acp_error.php';
}
?>
Is what I've updated to; Does anyone see any issue here?
mysql_query returns a statement HANDLE, not the value(s)/row(s) you're trying to select. YOu need to FETCH a row of data to be able to get the values you need to compare.
$result = mysql_query(...) or die(mysql_error());
$row = mysql_fetch_assoc($result);
if ($row['somefield'] == 3) {
...
}
Please note that things like
"$webdb"
are pointless cargo-cult programming. A simple
$webdb
is all that's needed for such things. There is not point in creating a new string, whose sole contents are the contents of a variable - just use the variable itself.
As well, note that you're vulnerable to SQL injection via that $_SESSION value you're using in the query. If that's a text value, and contains user-supplied data, your server is trivial to pwn.
im very new to this and have a very basic knowledge php and SQL this is my first website, please could some one help.
I currently have information being pulled from my database to my website, which is good.
however i want to run another couple of queries on the same html page one that will display a list of the latest 5 entries in the database and sort them by 'location_id' descending, the other will display a list of 5 entries and display them random every time the page refreshes. see my code below:
?php
require "connect.php";
$query = "select * from location";
$result = #mysql_query($query, $connection)
or die ("Unable to perform query<br>$query");
?>
<?php
while($row= mysql_fetch_array($result))
{
?>
<?php echo $row['location_id'] ?>
?php } ?>
Also when i was experimenting trying to solve this myself, i duplicated the loop shown above in a different location, and it would not display anything, can sone one explain why?
Thanks Guys
It needs reset,
mysql_data_seek( $result, 0 );
i have just seen the syntax.
i guess ";" is missing in
should be
<?php echo $row['location_id'] ; ?>
You could reset the query to do this or
create a new query and result with different names and then when the page runs have it select one of the queries to run.
$query1 = "select * from location";
$result1 = #mysql_query($query1, $connection);
or die ("Unable to perform query<br>$query1");
resetting might be easier but this is an option too
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)
...