...while connecting 2 table variables for use.
Here is what I have for query:
//GRABBING USER ID
if (isset($_SESSION['login_email'])) {
$session_entry = $_SESSION['login_email'];
$user_query = mysqli_query($connection, "SELECT id FROM users WHERE email = '$session_entry'");
$user_assoc = mysqli_fetch_assoc($user_query);
$user_id = $user_assoc;
}
//GRABING ACCESS ID
if (isset($_SESSION['login_email'])) {
$access_query = mysqli_query($connection, "SELECT access FROM user_access WHERE user_id = '$user_id'");
$access_assoc = mysqli_fetch_assoc($access_query);
$access = $access_assoc;
}
and the $access echo's "Array"... Don't know what to do about it.
Right now, in my user_access table I have 2 columns: 1. user_id and 2. access
I want it to echo out the access code, but like said above, all I get is "Array".
I attend on using this for the purpose of giving users access codes for accessing specific webpages that they have permission to access through these codes.
$access = $access_assoc;
must be
$access = $access_assoc['access'];
Also
$user_id = $user_assoc;
must be
$user_id = $user_assoc['id'];
Check mysqli_fetch_assoc Manual
Related
I am trying to select a record from my database, and I am return instead the first one in the table. No matter what I try, the first one gets returned.
Here's the query:
$query_task_owner = "select user_id from users where full_name = '$c_task_owner_name'";
$response = #mysqli_query($dbc, $query_task_owner);
Then I try a test to see the value that is returned as such:
echo $response or die(mysql_error());
This is where I see the user_id of the first row.
Even if I try to put a specific value in the query, as follow, I am getting the same result:
$query_task_owner = "select user_id from users where full_name = 'LeBron James'";
I do not understand because when I trying this query directly in PHPMyAdmin, I am getting the right result. So the query itself is correct.
Any idea?
Fetch $response using mysqli_fetch_array().
<?php
$query_task_owner = "select user_id from users where full_name = '$c_task_owner_name'";
$response = #mysqli_query($dbc, $query_task_owner);
$row = mysqli_fetch_array($response,MYSQLI_ASSOC);
echo $row['user_id'];
?>
If, users are more related to that full name. Then, use while loop to fetch all record.
<?php
$query_task_owner = "select user_id from users where full_name = '$c_task_owner_name'";
$response = #mysqli_query($dbc, $query_task_owner);
while($row = mysqli_fetch_array($response,MYSQLI_ASSOC))
{
echo $row['user_id']."<br>";
}
?>
Ok, so I am building on my first question here https://stackoverflow.com/questions/38102208/php-mysql-how-to-only-echo-links-with-search-bar-post-that-arent-already-echo
trying to only echo only usernames of people whose id is NOT in a mysql table called conversation along with a set id (the person who is signed in).
I echo the people who their id is user_two in a table conversation REGARDLESS if a search bar is posted here:
//$num = mysqli_query($con, "SELECT * FROM `pm_messages` WHERE user_from=".$account['id']."");
$numCon = mysqli_query($con, "SELECT * FROM `conversation` WHERE user_one=".$account['id']."");
$numrows = mysqli_num_rows($numCon);
while ($u = mysqli_fetch_assoc($numCon)) {
//get other users usernames to echo link
$getUserTwo = mysqli_query($con, "SELECT * FROM `accounts` WHERE id=".$u['user_two']."");
$s = mysqli_fetch_assoc($getUserTwo);
//echo $s['username'];
echo "<a href='message.php?id={$s['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/>{$s['username']} </li></a>";
}
This works well, meaning only individuals who a conversation has been started with (a row exists for this user and the one signed in conversation table) are echoed in a link.
Problem comes here with the search bar because it echoes all individuals even if a conversation has been started, resulting in duplicates:
(notice the 2 khusteds)
This does not make sense because here I select the row in conversation where user_one is the signed in user and user_two is the second user and only echo a link if the result is FALSE (meaning there's no conversation):
if (isset($_POST['searchbarpm'])) {
//$sess->getUsers();
$dbh = mysqli_connect("localhost","username","password","sqlserver");
$query = $_POST['searchbarpm'];
$q = mysqli_query($dbh, "SELECT * FROM sqlserver.accounts WHERE username LIKE '%".$query."%'");
//display all the results
while($row = mysqli_fetch_assoc($q)) {
$checkConvo = mysql_query("SELECT 'id' FROM sqlserver.conversation WHERE user_one=".$user_id." AND user_two=".$row['id']."");
//only output users they dont have convo going with because theyre already printed!!!
if ($checkConvo==false && $row['id']!= $user_id) {
echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
}
}
}
But it looks like the query is always false because again, all users are echoed. Why is this happening? How can I only echo users not in conversation table with the signed in user (user_one)?
EDIT:
new code (sorry for screenshot); :
#IanH -
$con = mysqli_connect("localhost","username","password","sqlserver");
//$num = mysqli_query($con, "SELECT * FROM `pm_messages` WHERE user_from=".$account['id']."");
$numCon = mysqli_query($con, "SELECT * FROM `conversation` WHERE user_one=".$account['id']."");
$numrows = mysqli_num_rows($numCon);
while ($u = mysqli_fetch_assoc($numCon))
{
//get other users usernames to echo link
$getUserTwo = mysqli_query($con, "SELECT * FROM `accounts` WHERE id=".$u['user_two']."");
$s = mysqli_fetch_assoc($getUserTwo);
//echo $s['username'];
if(isset($_POST['searchbarpm'])){
//$sess->getUsers();
$dbh = mysqli_connect("localhost","username","password","sqlserver");
$query = $_POST['searchbarpm'];
$q = mysqli_query($dbh, "SELECT * FROM sqlserver.accounts WHERE username LIKE '%".$query."%'");
//display all the results
while($row = mysqli_fetch_assoc($q)){
if($row['id']!= $user_id && $row['id']!=$s['id']) { //only output users they dont have convo going with because theyre already printed!!!
echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
}
}
}
else {
echo "<a href='message.php?id={$s['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/>{$s['username']} </li></a>";
}
}//
First of all, you need change the mysql_query( ... ) on line 9 of your second posted code block to mysqli_query( ... ), for API consistency and compatibility.
Also, you could have duplicate results if you are allowing multiple entries in the conversation table where users A and B can be entered as user1 = A, user2 = B in one conversation, and user1 = B, user2 = A in a different conversation.
Lastly, as Jay Blanchard said, you should use prepared statements to avoid SQL injection.
another question. I need to display a username and so what I'm doing is getting the author id (which is '1'), and then using a query saying get the username from the users table where the author id is the same as the user id.
My problem is that I'm getting the value of '1' returned as I've said above and the following is my code. Am I missing something here or...?
$users = mysqli_query($sql, "SELECT * FROM users WHERE userid = '$authorid'") or die($users . "<br/>" . mysqli_error($sql));
while($userData = mysqli_fetch_array($users)) {
$postAuthor = $usersData['username'];
}
Edit 1
I said also mention the above information should show the name 'Dan'.
The user 'Dan' has a user id of 1 should the author id should be 1 (which it is) from the post row. This is the only use of anything to do with the user table so it's not being overridden anywhere. I'm so confused.
This should have been an error:
while($userData = mysqli_fetch_array($users)) {
^^ no s
$postAuthor = $usersData['username'];
^^ used different variable name
}
Note: Use prepared statements also in this case, you're using mysqli anyways.
$select = $sql->prepare("SELECT * FROM users WHERE userid = ?");
$select->bind_param('i', $author_id);
$select->execute();
$results = $select->get_result();
while($userData = $results->fetch_assoc()) {
$postAuthor = $userData['username'];
}
I am trying to get four different values from my database. The session variable username and usernameto are working, but I want to get 4 different values -- two each from username and usernameto:
<?php
session_start(); // startsession
$Username=$_SESSION['UserID'];
$Usernameto= $_SESSION['UserTO'];
$db = mysql_connect("at-web2.xxxxxx", "yyyyy", "xxxxxxx");
mysql_select_db("db_xxxxxx",$db);
$result1 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
while($myrow1)
{
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
while($myrow2)
{
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
?>
Edit - just realized that you didn't tell us what wasn't working about the code you provided. Are you getting an error message or are you not getting the correct data back? You still should fix your query, but we'll need some more information to know what's wrong.
Your query statements shouldn't have "and" between the select parameters, so it should be:
Edit 2 - I just noticed that you had a while loop that you don't need, try this:
$result1 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
if (isset($myrow1)) {
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
if (isset($myrow2)) {
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
An example from the php manual echoing an html table
I don't know if you can derive what you need from this?
More specific: You can use:
$line = mysql_fetch_array($result, MYSQL_ASSOC);
I have a table (users) with various users. if you refer someone to register it creates a referal id in a field (refcode) similar to your (id) in his/her profile. now when he/she is logged in, i want to add value to the referal's profile.
So basically i need to add value to a user not logged in.
here is my code.
$sql="SELECT * FROM users";
$val=(2);
$result=mysql_query($sql,$bd);
$data=mysql_fetch_assoc($result);
while ($array = mysql_fetch_array($result)) {
// equate the value to a variable to use outside
// this while loop
$acc_balance = $array['com_balance'];
$comm = $array {$_SESSION['refcode']};
$commision = $array['id'];
}
$remainder = $acc_balance + $val;
$update_query = mysql_query("UPDATE users SET com_balance = '". mysql_real_escape_string($remainder) ."'
WHERE id=refcode");
if ($update_query) {
print ""
A bunch of changes, marked by numbers:
$sql="SELECT * FROM users where userID='$_SESSION['refcode']'"; //#1
$val = 2; //#2 : Why do u need the ()?
$result = mysql_query($sql,$bd);
$data=mysql_fetch_assoc($result);
while ($array = mysql_fetch_array($result)) {
// equate the value to a variable to use outside
// this while loop
$acc_balance = $array['com_balance'];
$comm = $array{$_SESSION['refcode']}; //#3: I am not sure what you're doing here?
$commision = $array['id'];
}
$remainder = $acc_balance + $val;
$update_query = mysql_query("UPDATE users SET com_balance = '". mysql_real_escape_string($remainder) ."'
WHERE id=refcode"); // #4: where are you defining refcode? if variable, the query needs to be WHERE id='$refcode'");
if ($update_query) {
print "Update successful";
Firstly, change:
$sql="SELECT * FROM users";
as you do not need to loop through an entire table just to get the details of the user logged in. Something like this should do the trick:
$yourID=3;// Assume this is sanitized data from a cookie or a login script.
$sql="SELECT * FROM users where userID=$yourID";
Secondly, when you verify the login of the user who referred the next user, just add in a query like this:
$update_query = mysql_query("
UPDATE users SET
com_balance = (select * from (
select com_balance from users where id=$yourID))");
This will update the new user with the balance from the referring user (which seems to be what you want). You will need to use the double subquery to get past the annoying mysql bug/feature where it cannot update a table from a subquery on the same table unless you encapsulate it in a second subquery.