i am trying to display html values from my database into a web page. i am using tinymce as text editor. the data already stored as html values but when i tried to display it to a web page, the values display like this. example:
<p>THIS IS <strong>APPLICATION MANAGEMENT</strong></p>
my code to display is
<?php
$user_name = "BLANK";
$password = "BLANK";
$database = "BLANK";
$server = "BLANK";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM service WHERE service_id=1";
$result = mysql_query($SQL);
while ( $db_field = mysql_fetch_assoc($result) ) {
print $db_field['contents'];
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
my question is how can i display that html values exactly like when i am using tinymce. what i mean is i want to display that html value like this:
<p>THIS IS <strong>APPLICATION MANAGEMENT</strong></p>
to this
THIS IS APPLICATION MANAGEMENT
hope anybody can assist me..
You're looking for the problem in the wrong place.
When you put the data into the database in the first place, it looks like your code is expecting plain text input and converting it to HTML before passing it to the query.
Don't do that.
As a hack, you could run the data through htmlspecialchars_decode when you pull it out of the database, but that's not solving the real problem.
Related
So, I have this system where you can make a post. So, how it works is, there is an input field, whatever you type in the input field, and click post, it will send to the database as en entry and get posted. The post will be displayed. However, with my current system, after entering something in the input field, and clicking post, the entry gets sent to the database, but the post doesn't actually display. For it to display, you need to refresh the page again, which it displays then, and two entries go to the database.
I don't want this to happen. Right when the user enters text into the input field and clicks post, the post should display on the go, you shouldn't have to refresh for the post to be displayed, and only one entry should be sent to the database, not two. Now, I also included my database connection and my insert statements, but here is the code to display the post:
<div class="textPost">
<?php
$sql = "SELECT * FROM posts";
$result = mysqli_query($connection, $sql);
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_assoc($result)) {
?>
<div class="textpostFormat">
// all the displayed post content
</div>
<?php
}
}
?>
</div>
Insert Statement (post.php):
<?php
session_start();
// Making Connection To The Database
$dbHost = "localhost";
$dbUser = "root";
$dbPass = "root";
$database = "feed";
$connection = mysqli_connect($dbHost, $dbUser, $dbPass, $database) or die ("Sorry, we could not connect to the database");
// Posting System
if (!empty($_POST['postContent'])) {
$post = $_POST['postContent'];
$firstname = $_SESSION['firstname'];
$lastname = $_SESSION['lastname'];
$sql = "INSERT INTO posts (firstname, lastname, body, date_posted) VALUES (?, ?, ?, NOW())";
$stmt = mysqli_stmt_init($connection);
// nested if statement
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "";
} else {
mysqli_stmt_bind_param($stmt, "sss", $firstname, $lastname, $post);
mysqli_stmt_execute($stmt);
}
} else {
echo "";
}
?>
So, since PHP is a server side language, to make the calls, you would require AJAX to accomplish this problem/situation. The only problem is, I don't know what the AJAX would be. So my question is, for this particular situation, what would the AJAX be? I don't want any tutorials, just the code. Also, I can't try on some AJAX code and ask for help, because I don't even know what it would be in the first place. I don't want to use the PHP approach to solve this problem, I want the AJAX approach, and for that, the AJAX code to solve this situation.
So, please, please spare a few minutes of your time to answer this question. Please. Thanks a lot! I greatly appreciate your help.
This system is based on invitation codes, if u have a code that is present in the database you can submit the input therefore change a value in a row. There are 2 inputs, 1) Invitation Code (key), if exist in the database the user can submit the value 2)Name (user). I done the following code but it doesn't work, any suggestions?
<?php
//get value pass from form in login.php
$username = $POST['user'];
$password = $POST['key'];
//connect to the server and select database
mysql_connect("localhost", "...","...");
mysql_select_db("...");
// Query the database for user
$result = mysql_query("UPDATE invitation_keys SET name ='$username' WHERE key = '$password'";)
or die("Failed to query database".mysql_error());
$row = mysql_fetch_array($result);
if ($row['key'] == $password) {
echo "Login success!!!".$row['key'];
} else {
echo "Failed to login";
}
?>
When you are coding in PHP, var_dump($var) is your best friend.
So the first thing to do here, is to print the query.
You will see, that your $username and $password vars are NULL, because you missed the syntax of $_POST[].
After, you can put in var_dump what you want, and that's why its interesting, because you will debug faster with this.
I have been trying to create a unique page for each row in my database. My plan is to create a dictionary.php?word=title url, where I can display the description and title of that specific ID. My datbase is contains id, term_title and term_description.
I'm fresh outta the owen when it comes to PHP, but I've managed to atleast do this:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dbname";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Cannot connect to database." . mysqli_connect_error());
}
if (isset($_GET['id']))
{
$id = (int) $_GET['id'];
$sql = 'SELECT * FROM dbname WHERE id = $id LIMIT 1 ';
}
$sql = "SELECT * FROM terms";
$result = $conn->query($sql);
mysqli_close($conn);
?>
I'm really stuck and I dont know what the next step is, I've added the <a href='dictionary.php?=".$row["id"]."'> to each word I want to be linked, and this is properly displayed in the main index.php file (where all my words are listed with <li>. This is my code for this:
<?php
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<a href='dictionary.php?=".$row["id"]."'><li class='term'><h4 class='term-title'>" . $row["term_title"]. "</h4></li></a>";
} else {
echo "No words in database.";
}
?>
How do I create this unique page, only displaying title and description for that id? How do I add ?word= to the url?
Thanks for taking your time to help me.
Update from years later: Please, please use parameters when composing your SQL queries. See Tim Morton's comment.
You're on the right track, and ajhanna88's comment is right, too: you want to be sure to include the right key ("word" in this case) in the URL. Otherwise, you're sending a value without telling the page what that value's for.
I do see a couple other issues:
When you click on one of the links you created, you're sending along $_GET["word"] to dictionary.php. In your dictionary.php code, however, you're searching for your word by "id" instead of by "word". I'm guessing you expect users to search your dictionary for something like "celestial" and not "1598", so try this instead:
if (isset($_GET['word'])) {
$word = $_GET['word'];
$sql = 'SELECT * FROM dbname WHERE word = $word LIMIT 1 ';
}
BUT! Also be aware of a security problem: you were letting the user put whatever they want into your query. Take a look at the classic illustration of SQL injection. To fix that, change the second line above to this:
`$word = $conn->real_escape_string($_GET['word']);`
Another problem? You're looking for the word exactly. Instead, you'll probably want to make it case insensitive, so "Semaphore" still brings up "semaphore". There are plenty of ways to do that. The simplest way in my experience is just changing everything to lowercase before you compare them. So that $word assignment should now look like this:
`$word = $conn->real_escape_string(strtolower($_GET["word"]));`
And your query should look something like this:
`$sql = "SELECT * FROM dbname WHERE word = LOWER('$word') LIMIT 1 ";`
Next! Further down, you overwrite your $sql variable with SELECT * FROM terms, which totally undoes your work. It looks like you're trying to show all the words if the user doesn't provide a word to look up. If that's what you're trying to do, put that line in an else statement.
Your $result looks fine. Now you just have to use it. The first step there is to do just like you did when you tested the connection query (if(!$conn)...) and check to see that it came back with results.
Once you have those results (or that one result, since you have LIMIT 1 in your query), you'll want to display them. This process is exactly what you did when printing the links. It's just that this time, you'll expect to have only one result.
Here's a real basic page I came up with from your code:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dbname";
$conn=new mysqli($servername,$username,$password,$dbname);
if($conn->connect_errno){
die("Can't connect: ".$conn->connect_error);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Dictionary!</title>
</head>
<body>
<?php
if(isset($_GET["word"])){
$word = $conn->real_escape_string(strtolower($_GET["word"]));
$sql = $conn->query("SELECT * FROM dictionary WHERE word=LOWER('".$word."') LIMIT 1");
if(!$sql){
echo "Sorry, something went wrong: ".$conn->error_get_last();
} else {
while($row=$sql->fetch_assoc()){
echo "<h2>".$row["word"]."</h2>";
echo "<p>".$row["definition"]."</p>";
}
}
} else {
$sql = $conn->query("SELECT word FROM dictionary");
if(!$sql){
echo "Sorry, something went wrong: ".$conn->error_get_last();
} else {
echo "<p>Here are all our words:</p><ul>";
while($row=$sql->fetch_assoc()){
echo "<li>".$row["word"]."</li>";
}
}
echo "</ul>";
}
?>
</body>
</html>
You should also take care to be consistent in your terminology. For this, my MySQL table had three columns: id, word, and definition. I dropped term since your URLs were using word. In my experience, it's best to keep the same terminology. It avoids confusion when your application gets more complicated.
Lastly, to answer your question about creating separate pages, you can see there that for a simple page like this, you may not need a separate page to display the definitions and the links -- just an if/else statement. If you want to expand what's in those if/else blocks, I'd suggest looking at PHP's include function.
You have a great start. Keep at it!
I'm trying to build a relatively simple PHP login script to connect to MySQL database running on my home server. I know the connection works as I've gotten some data returned as I would expect. However, I am having trouble getting the full script to work.
Essentially, I'm taking in a username/password from the user, and I first do a lookup to get the user_id from the users table. I then want to use that user_id value to do a comparison from user_pswd table (i'm storing usernames and passwords in separate database tables). At one point, I was able to echo the correct user_id based on the username input. But I haven't been able to get all the issues worked out, as I'm pretty new to PHP and don't really know where to see errors since I load this onto my server from a remote desktop. Can anyone offer some advice/corrections to my code?
The end result is I want to send the user to another page, but the echo "test" is just to see if I can get this much working. Thanks so much for the help!
<?php
ob_start();
$con = new mysqli("localhost","username","password","database");
// check connection
if (mysqli_connect_errno()) {
trigger_error('Database connection failed: ' . $con->connect_error, E_USER_ERROR);
}
$users_name = $_POST['user'];
$users_pass = $_POST['pass'];
$user_esc = $con->real_escape_string($users_name);
$pass_esc = $con->real_escape_string($users_pass);
$query1 = "SELECT user_id FROM users WHERE username = ?;";
if ($result1 = $con->prepare($query1)) {
$result1->bind_param("s",$user_esc);
$result1->execute();
$result1->bind_result($userid);
$result1->fetch();
$query2 = "SELECT user_pswd_id FROM user_pswd WHERE active = 1 AND user_id = ? AND user_pswd = ?;";
if ($result2 = $con->prepare($query2)) {
$result2->bind_param("is",$userid,$pass_esc);
$result2->execute();
$result2->bind_result($userpswd);
$result2->fetch();
echo "test", $userpswd;
$result2->free_result();
$result2->close();
} else {
echo "failed password";
}
$result1->free_result();
$result1->close();
}
$con->close();
ob_end_clean();
?>
I'm new to PHP and am trying to build my a website to display information on TV shows stored in a MySQL DB. I've currently got a webpage that will create a table to display the information in the DB, however I'd like each row to link to a dynamically populated page with more info on each show (also pulling from the DB). My question is how do I get the site to know which link has been clicked and then save that as a variable so it can then be recalled on a new to populate the correct information?
I'm currently using this to populate the page.
<!--Populate page with data from SQL-->
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "media_server";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT show_title, show_desc, thumbnail_path FROM tv_shows WHERE status = 'Y'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th></th><th></th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "
<tr>
<td>
<img src='../images/thumbnails/tv/".$row["thumbnail_path"]."'>
</td>
<td class='td_title'>
<a href='#' onclick='show_var_set();'>".$row["show_title"]."</a>
</td>
<td class='td_desc'>".$row["show_desc"]."</td>
</tr>";
}
echo "</table>";
} else {
echo "Error - 0 results were returned my the database. Please try again.";
}
$conn->close();
?>
One option is to change your href links to point to this page and pass a GET variable you can retreive. The added bonus to this approach is you could bookmark a particular show and come back to that page, since the bookmark will include that GET variable.
So you could change your links to something like this:
echo '', $row['show_title'],'';
Then you'd retrieve that variable by testing for, then reading the GET variable and performing a db query to populate the page with that show's data.
Here's how you'd test for and retreive that variable:
if (isset($_GET['show']))
{
$show = $_GET['show'];
// Perform database lookup using $show
}
Remember to never put user input directly into a query, but use prepared statements and bind the user data to avoid the risk of SQL injection.
There are many ways to pass values from page to page but one is to use session variables:
//Include this at the top of your php scripts that use session variables
session_start();
$_SESSION['your_variable_name_here'] = value_you_want_to_store;
Then on the page you would like to access this use:
$someVariable = $_SESSION['your_variable_name_here'];
Change the SQL to return the show_id,
$sql = "SELECT show_id, show_title, show_desc, thumbnail_path FROM tv_shows WHERE status = 'Y'";
and use that as a parameter to the show_var_set() function.
<a href='#' onclick='show_var_set(".$row["show_id"].");'>".$row["show_title"]."</a>
The show_var_set() function can then use that parameter to get the details for that show from the database.