Php posting script not getting data from database
Question
I have a php posting script and I need it to grab the data from the database. Here's the script:
<?php
error_reporting(E_ALL);
session_start();
// If the session vars aren't set, try to set them with a cookie
if (!isset($_SESSION['user_id'])) {
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Cheesecake Productions - Post Topic</title>
<link rel="stylesheet" type="text/css" href="include/style/content.css" />
</head>
<body>
<?php
include ("include/header.html");
include ("include/sidebar.html");
?>
<div class="container">
<?php
require_once('appvars.php');
require_once('connectvars.php');
// Make sure the user is logged in before going any further.
if (!isset($_SESSION['user_id'])) {
echo '<p class="login">Please <a href="login.php">log in</a> to access this page.</p>';
exit();
}
else {
echo('<p class="login">You are logged in as ' . $_SESSION['username'] . '. <a href="logout.php">Log out</a>.</p>');
}
// Connect to the database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die('could not connect to mysql '.mysqli_connect_error());
// Grab the profile data from the database
$query = "SELECT first_name FROM ccp2_user WHERE first_name = '" . $_SESSION['user_id'] . "'";
$data = mysqli_query($dbc, $query);
///////////////////////////
///What must I do after////
//getting the data from////
//database. I am new to////
//PHP//////////////////////
//////////////////////////
$row = mysqli_fetch_array($data);
$first_name = mysqli_real_escape_string($dbc, trim($_POST['first_name']));
if (isset($_POST['submit'])) {
// Grab the profile data from the POST
$post1 = mysqli_real_escape_string($dbc, trim($_POST['post1']));
// Update the profile data in the database
if (!$error) {
if (!empty($post1)) {
// Only set the picture column if there is a new picture
$query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`) VALUES ('$first_name', NOW(), '$post1')";
mysqli_query($dbc, $query);
// Confirm success with the user
echo '<p>Your post has been successfully added. Would you like to <a href="viewpost.php">view all of the posts</a>?</p>';
mysqli_close($dbc);
exit();
}
else {
echo '<p class="error">You must enter information into all of the fields.</p>';
}
}
} // End of check for form submission
else {
echo '<p>Grr</p>';
}
mysqli_close($dbc);
?>
<form enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="MAX_FILE_SIZE" value="<?php echo MM_MAXFILESIZE; ?>" />
<fieldset>
<legend>Post Here:</legend>
<label type="hidden" for="post1">Post Content:</label><br />
<textarea rows="4" name="post1" id="post1" cols="50">Post Here...</textarea><br />
</fieldset>
<input type="submit" value="Save Post" name="submit" />
</form>
</div>
<?php
include ("include/footer.html");
?>
</body>
</html>
This script is supposed to grab first_name from the database and it is not. Help?
Edit: There's the whole code.
3 answers
solution1
2 ACCPTED 2014-01-26 02:09:48
Many things are strange with your code
I believe it's blank because one of the if/else is messed up:
if (isset($_POST['submit'])) {
....
}
else {//here
else {
echo '<p class="error">There was a problem accessing your profile.</p>';
}
}
then you have $error variable that have no meaning
$error = false;
Then you have in your form :
<input type="text" id="first_name" name="first_name" value="" /><br />
but you dont want to grab it from there, but the database:
$query = "SELECT first_name FROM ccp2_user
WHERE user_id = '" . $_SESSION['user_id'] . "'";
Then your wanna grab $last_name From the post
$last_name = mysqli_real_escape_string($dbc, trim($_POST['last_name']));
but you don't have it in your form
Also this part:
if (!empty($first_name) && !empty($post1)) {
// Only set the picture column if there is a new picture
if (!empty($new_picture)) {
$query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`)
VALUES ('$first_name', NOW(), '$post1')";
}
else {
$query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`)
VALUES ('$first_name', NOW(), '$post1')";
}
}
You you have a condition on new_picture Where did you initialize that. Why is it the same insert query again?
Don't you need quote around that?
you have so many issues here, I advice you to trouble shoot step by step. and redesign tis whole thing.
solution2
1 2014-01-26 02:58:58
I put something real quick together that works on my system.
This is a basic method and I mean basic, so you'll need to do the rest.
Just change the DB credentials to your own, and the the_user_id assigned to $_SESSION['user_id']
It's the best I can do to help.
<?php
$DB_HOST = "xxx";
$DB_USER = "xxx";
$DB_PASS = "xxx";
$DB_NAME = "xxx";
$dbc = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($dbc->connect_errno > 0) {
die('Connection failed [' . $dbc->connect_error . ']');
}
session_start();
$_SESSION['user_id'] = "the_user_id"; // change this to the user's id
// You can use * also as the line from below
// $sql = $dbc->query("SELECT * FROM `ccp2_user` WHERE `user_id` = '" . $_SESSION['user_id'] . "'");
$sql = $dbc->query("SELECT `first_name` FROM `ccp2_user` WHERE `user_id` = '" . $_SESSION['user_id'] . "'");
while($row= mysqli_fetch_array($sql))
{
echo $row['user_id'];
}
// for testing purposes
// var_dump($_SESSION['user_id']);
// var_dump($_SESSION);
mysqli_close($dbc);
solution3
0 2014-01-26 01:41:53
Its here,
require_once('appvars.php');
require_once('connectvars.php');
One of these file must not be set or php cant find these file. So as it says 'require' which means till we dont get this file it will not proceed. so it halt the execution there itself.
try it with :
include('appvars.php');
include('connectvars.php');
It you see the page then problem is here itself.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.