I have looked everywhere and cannot seem to find out the answer to fix my problem.
I have an online messaging system and i would like to implement Online and Offline status to the users. But no matter what i do with this code... I cannot seem to get it working. Any help would be greatly appreciated. Keep in mind I dont want to alter any of my existing code other than the ones that have to do with $online and the q2 variables.
my code is:
<?php
//show all the users expect me
$q = mysqli_query($con, "SELECT * FROM `users` WHERE id!='$user_id'");
$q2 = mysqli_query($con, "SELECT status FROM `users`");
//display all the results
while($row = mysqli_fetch_assoc($q2)){
$online = $row['status'];
if($online == "online"){
$tcolor = "green";
}
elseif($online=="offline"){
$tcolor = "red";
}
elseif($online ==""){
$tcolor= "gray";
}
}
while($row = mysqli_fetch_assoc($q)){
echo "<a href='/users/pm?id={$row['id']}'><li style='border-left:{$tcolor} 15px solid;'><img src='{$row['pic']}'> {$row['name']}</li></a>";
}
?>
Your logic for setting $online is reversed. It should be: $online = $row['status'];
EDIT BELOW:
Your original $q2 was unnecessary as it was just retrieving the status column and nothing was linked to the users. Also, you have all the data you need in $q.
<?php
//show all the users expect me
$q = mysqli_query($con, "SELECT * FROM `users` WHERE id!='$user_id'");
//display all the results
while($row = mysqli_fetch_assoc($q)){
$online = $row['status'];
if($online == "online"){
$tcolor = "green";
}
elseif($online=="offline"){
$tcolor = "red";
}
elseif($online ==""){
$tcolor= "gray";
}
echo "<a href='/users/pm?id={$row['id']}'><li style='border-left:{$tcolor} 15px solid;'><img src='{$row['pic']}'> {$row['name']}</li></a>";
}
?>
Related
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']
Im trying to make a code where it displays all names in database where "clanwars" is set to 1 (int) and if all of them is 0 echo an message like: "Noone has signed up yet!"
This is some of the code i have:
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
while($row = mysqli_fetch_array($result))
{
if ($row['clanwars'] != '0') {
echo $row['mantra']."<br>";
} else {
echo 'Noone has signed up yet!';
}
}
mysqli_close($con);
First, in your example row['clanwars'] will never equal to 0 because you already specified WHERE clanwars = 1 in your query, so MySQL will return only those that have clanwars=1. If I understand well, you need to do something like:
<?
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
if (mysqli_num_rows($result)==0) echo 'Noone has signed up yet';
else {
while ($row = mysqli_fetch_array($result)) {
//do what you need
}
}
?>
So basically, you retrieve everyone who has CLANWARS set to 1 in the database. If there are records, process them, if there are no records, it means that nobody has signed up.
Is this what you need?
Try this:
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
if (mysqli_num_rows($result) == 0)
{
echo 'Noone has signed up yet!';
}
else
{
while ($row = mysqli_fetch_array($result))
{
echo $row['mantra']."<br>";
}
}
mysqli_close($con);
I am making an online quiz, in this quiz, one question is asked at a time. If the answer is correct, I want to add that point, to the total score earned. I want the $_SESSION to show how many points the user has earned so far. The code I have so far is this:
$sql = "SELECT * FROM quiz_questions WHERE Question='$question' AND Answer='$answer'";
$result = mysql_query($sql);
$num_rows = mysql_num_rows($result);
if ($num_rows != 0){
$score++;
$total_score += $score;
session_start();
$_SESSION['score'] = $total_score;
} else {
echo "Incorrect";
}
}
I already have $score equal to 0 to start in the beginning of the code. Thank you in advance for all your help!
session_start(); // should be at the beginning of your script(s);
if (!isset($_SESSION['score'])) { $_SESSION['score'] = 0; }
$sql = "SELECT * FROM quiz_questions WHERE Question='$question' AND Answer='$answer'";
$result = mysql_query($sql);
$num_rows = mysql_num_rows($result);
if ($num_rows != 0){
// $row = fetch row to get the actual score for this question, if answered right?
$row = mysql_fetch_row($result);
//$score++; ??????
//$total_score += $score;
$_SESSION['score'] += $row['score_index'];
//$_SESSION['score']++; // or just this, dunno
echo 'Your current score is: '.$_SESSION['score'];
}
else {
echo "Incorrect";
}
I am trying to echo out all of the user rows in my database as a select in a form.
It is only showing a blank space. Nothing else.
Here is my code.
<?php
session_start();
require('../../config.php');
$user = $_SESSION['user'];
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
$logged = $_SESSION['loggedin'];
if ($logged == true) {
if ($rank >= 3) {
echo "Succesful, $user.<br />
<form method='POST' action='delete.php'>
<select><option>Please select</option>";
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
}
echo "<option>$lol</option>";
echo "</select>
</form>";
} else {
echo "Your not an admin. $user";
}
} else {
echo "Please login.";
}
?>
First, please stop using mysql_ functions as they are being deprecated. Look into mysqli_ or PDO. Be aware that your script is vulnerable to SQL injection.
The reason your script is not working is because it appears you are calling mysql_fetch_assoc twice. When calling it the second time, there won't be any output if your query only returns a single row.
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
You will need to resubmit a query (something like below) and call that result separately to populate the drop down, or store the result in an array.
$qry=("SELECT `uname` FROM users");
$result=mysql_query($qry);
while ($row = mysql_fetch_assoc($result)) {
echo '<option>' . ucwords($row['uname']) . '</option>';
}
It looks like your while loop is set up badly, I think you need to change
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
}
echo "<option>$lol</option>";
to this
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
echo "<option>$lol</option>";
}
UPDATE:
I reread your question and it seems that you want to display all unames from your table.
The original query was to check if the user is an admin.
Here's how it will look like. But it's just a duplicate of njk's answer.
<?php
session_start();
require('../../config.php');
$user = $_SESSION['user'];
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
$logged = $_SESSION['loggedin'];
if ($logged == true) {
if ($rank >= 3) {
echo "Succesful, $user.<br />
<form method='POST' action='delete.php'>
<select><option>Please select</option>";
// added these 2 lines
$qry=("SELECT `rank`, `uname` FROM users");
$result=mysql_query($qry);
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
echo "<option>$lol</option>";
}
echo "</select>
</form>";
} else {
echo "Your not an admin. $user";
}
} else {
echo "Please login.";
}
?>
Are you sure your database has more than 1 row? You are calling the fetch 1 time so the second time it will try to fetch the second row.
Try also running the query directly into mysql and see if you get the rows you're expecting back, may be its not returning what you expected
I am trying to echo out all of the user rows in my database as a select in a form.
It is only showing a blank space. Nothing else.
Here is my code.
<?php
session_start();
require('../../config.php');
$user = $_SESSION['user'];
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
$logged = $_SESSION['loggedin'];
if ($logged == true) {
if ($rank >= 3) {
echo "Succesful, $user.<br />
<form method='POST' action='delete.php'>
<select><option>Please select</option>";
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
}
echo "<option>$lol</option>";
echo "</select>
</form>";
} else {
echo "Your not an admin. $user";
}
} else {
echo "Please login.";
}
?>
First, please stop using mysql_ functions as they are being deprecated. Look into mysqli_ or PDO. Be aware that your script is vulnerable to SQL injection.
The reason your script is not working is because it appears you are calling mysql_fetch_assoc twice. When calling it the second time, there won't be any output if your query only returns a single row.
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
You will need to resubmit a query (something like below) and call that result separately to populate the drop down, or store the result in an array.
$qry=("SELECT `uname` FROM users");
$result=mysql_query($qry);
while ($row = mysql_fetch_assoc($result)) {
echo '<option>' . ucwords($row['uname']) . '</option>';
}
It looks like your while loop is set up badly, I think you need to change
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
}
echo "<option>$lol</option>";
to this
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
echo "<option>$lol</option>";
}
UPDATE:
I reread your question and it seems that you want to display all unames from your table.
The original query was to check if the user is an admin.
Here's how it will look like. But it's just a duplicate of njk's answer.
<?php
session_start();
require('../../config.php');
$user = $_SESSION['user'];
$qry=("SELECT `rank`, `uname` FROM users WHERE `uname` = '$user'");
$result=mysql_query($qry);
$row = mysql_fetch_assoc($result);
$rank = $row['rank'];
$logged = $_SESSION['loggedin'];
if ($logged == true) {
if ($rank >= 3) {
echo "Succesful, $user.<br />
<form method='POST' action='delete.php'>
<select><option>Please select</option>";
// added these 2 lines
$qry=("SELECT `rank`, `uname` FROM users");
$result=mysql_query($qry);
while ($row = mysql_fetch_assoc($result)) {
$users = $row['uname'];
$lol = ucwords($users);
echo "<option>$lol</option>";
}
echo "</select>
</form>";
} else {
echo "Your not an admin. $user";
}
} else {
echo "Please login.";
}
?>
Are you sure your database has more than 1 row? You are calling the fetch 1 time so the second time it will try to fetch the second row.
Try also running the query directly into mysql and see if you get the rows you're expecting back, may be its not returning what you expected