I am building a friends list of logged in user verified by session id. I have code that query's, results, and loops perfectly. BUT, it is posting an integer (maybe the id#?), (as this is my unique identifying condition of the select query) AND what I want is to output the (users) table rows of "firstname" and "lastname" instead of the user-id!
Sounds simple enough, yet, I lack the knowledge to make it work! So, here I am inquiry from the world of coders for help! Any assistance would be greatly appreciated!
Here is my code:
// Get Friend Array
$sql = "SELECT * FROM users
INNER JOIN friends
ON users.id = friends.id
WHERE (friends.user1='$u' OR friends.user2='$u' AND friends.accepted='1')";
$query = mysqli_query($db_conx, $sql);
$numrows = mysqli_num_rows($query);
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
array_push($my_friends, $row["user2"]);
array_push($my_friends, $row["user1"]);
}
//remove your id from array
$my_friends = array_diff($my_friends, array($u));
//reset the key values
$my_friends = array_values($my_friends);
mysqli_free_result($query);
// Loop through $my_friends array and build results
foreach($my_friends as $friends => $v2);
$friends .= ' ';
}
Here is my html code:
<ul data-role="listview" data-autodividers="true">
<?php
echo "<li>$friends</li>";
?>
</ul>
It is the last line of code
$friends .= ' ';that I'm trying to workout. How the syntax of that line should be to add ($row["users.firstname"] and $row["users.lastname"]) and somehow not list the user "$id #" that is joining the two tables of (users) and (friends) in the select query!
It may be that I need a different array altogether for what I want, and If you know of the right way to do this, please inform me of how to do it...Thank you all!
What I understood that you are missing User's id in row. This issue is because php array can not have two keys with same name.
Change you query little bit like below
$sql = "SELECT *, friends.id as friendsID FROM users
INNER JOIN friends
ON users.id = friends.id
WHERE (friends.user1='$u' OR friends.user2='$u' AND friends.accepted='1')";
Now you will have id' as User Id andfriendsID` as Friend's ID
Related
I have several id's in a table called "leaderboards" that belong to different users. They're named as:"id_user" and they're not in order. What I want to do is printing divs in a leaderbord which should contain some info that I get from those id_user's.
The only problem I have about it is that after a research on stackoverflow and other websites, I still couldn't find how to select those id_user's in descending order AND be able to take one by one to get the info from that user and then continue with the next id_user, and so on.
I don't know how to select the specific row of each id_user in descending order to do the other codes that I already know how to do.
I hope it's not a duplicate of any other previosly asked question on this website (I really did a research and I couldn't find any specific answer to this question, for the sql part and the php part all together).
Thank you so so much beforehand.
An INNER JOIN between your tables will achieve what you intend.
SELECT *
FROM users
JOIN leaderboards WHERE users.id = leaderboards.id_user
ORDER BY users.id DESC
In each returned row, you will get the columns from both your users and leaderboards tables, so loop over the result and echo the information from the user you need.
$query = 'SELECT...';
$res = mysqli_query($query);
while ($row = mysqli_fetch_assoc($res)) {
echo '<div>'.$row['id'].' - '.$row['username'].' - '.$row['image'].'</div>';
}
You could do with a good read up on both PHP and MySql but I'll give you a clue.
EDIT
$query = "SELECT * FROM `the_name_of_your_table` ORDER BY `user_id` DESC;";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
print $row["user_id"] . " - " . $row["username"] . "<BR>";
}
/* free result set */
mysqli_free_result($result);
}
I'm building a followers list for a user. I have two tables the first shows the relationships between users and the second table holds every users profile info. In the following code, I first select from the user relationships table to get a variable {$userid1} which is the value of all the user ids that follow the current user. When I echo out {$userid1} I get all the ids of the users who follow the current user but it is one giant connected string. I want to take the variable {$userid1} and use it to pull every one of those follower's user data from the profile table. I want every user's data to show up from the profile table that follows the current user. The code works however only the newest follower's profile data is pulled from the profile table. I was thinking putting the variable {$userid1} into an array and using foreach, but I'm not sure how the syntax would be. Anybody know how it could work? The problem is a variable can only hold one value at a time.
Output of the first query are the iduser numbers from the users following the current user.
e.g. when echoed echo $userid1 . " "; the results are the ids look like this: 45 56 67
I added a space between numbers in the echo statement.
$sql = mysql_query("SELECT * FROM followrelations Where iduser2='$uid' "); //followers from relations table
while($row = mysql_fetch_array($sql)){
$userid1 = $row['iduser1'];
echo $userid1 . " ";
}
$sql = mysql_query("SELECT * FROM profile where iduser=$userid1 ORDER BY username ");//get followers infor from profile using variable
while($row = mysql_fetch_array($sql)){
$iduserf = $row['iduser']; //userid2 requires different var name so program does not get mixed up
$username = $row['username'];
$bio = $row['bio'];
$avatar = $row['avatar'];
echo "
<div style='width:500px;height:100px;padding:20px 20px;float:left;border: solid black 1px;'>
<a href='profile.php?uid=$iduserf'><img src=$avatar height=50px width=50px /></a></br>
<a href='profile.php?uid=$iduserf'>$username </a></br>
</div>" ;
}
You might be looking for subqueries:
SELECT username
FROM user_table
WHERE id IN( SELECT user_id WHERE linked_user_id = 123 )
This is simplefied to be a better example. This will select the username of all users from the user_table, where the ID exist in the subquery.
In turn, the subquery selects all user_id where the linked user has id=123.
The subquery is quite similar to making an array in PHP and use that info for the next query.
Small note: Be carefull with subqueries. They're perfect in my example above, but eg not all servers support a limit in the subquery. This might effect performance. The more complecated functions don't work in a subquery. Try to keep those simple.
You can use something like the following to execute this in a single query:
$id = mysql_real_escape_string($uid);
$sql = <<<SQL
SELECT
p.*
FROM profile as p
INNER JOIN followrelations as fl
ON fl.iduser1 = p.userid
WHERE fl.iduser2 = $id
ORDER BY username
SQL;
$stmt = mysql_query($sql);
while($row = mysql_fetch_array($stmt)){
// Rest of the logic here
}
This will return all the fields in your profile table. Note that the mysql_ extension is deprecated and unsafe - you should not be using it in new code. Take a look at How to replace MySQL functions with PDO? and How can I prevent SQL injection in PHP? for some good practices regarding this.
Found my mistake, the second $sql should be another name (I named it $sql1), and the second half should all be within the first while.
$sql = mysql_query("SELECT * FROM followrelations Where iduser2='$uid' "); //followers session uid
while($row = mysql_fetch_array($sql)){
$userid1 = $row['iduser1'];
$sql1 = mysql_query("SELECT * FROM profile where iduser='$userid1' ORDER BY username ");//get followers infor from profile using variable
while($row = mysql_fetch_array($sql1)){
$iduserf = $row['iduser']; //userid2 requires different var name so program does not get mixed up
$username = $row['username'];
$bio = $row['bio'];
$avatar = $row['avatar'];
echo "
<div style='width:500px;height:100px;padding:20px 20px;float:left;border: solid black 1px;'>
<a href='profile.php?uid=$iduserf'><img src=$avatar height=50px width=50px /></a></br>
<a href='profile.php?uid=$iduserf'>$username </a></br>
</div>" ;
}
}
I am trying to print all members reports (complains) to the admin in this form:
sender name:...
reported user name:...
First I wanted to select each sender name directly from complaint table(using for loop), then find the matched reported user name and print all the reports in this way.
Note that I am not saving the sender name and reported user name at the same table so I need to do mysql join in order to find reported user name.
<?php
$raw_results = mysql_query("SELECT * FROM complaint ");
$ln = mysql_num_rows($raw_results);
if($ln > 0){
$results = array();
$results = mysql_fetch_array($raw_results);
for($i=0;$i<$ln;$i++){
$SenderName = mysql_query("SELECT m_name FROM member WHERE (`m_id`LIKE '".$results[1]."') ");
$fetchA = mysql_fetch_array( $SenderName, MYSQL_ASSOC);
$ReportedUserId = mysql_query("SELECT r.d_id FROM ride r, complaint c WHERE r.r_id= c.r_id");
$fetchB = mysql_fetch_array($ReportedUserId , MYSQL_ASSOC);
//Afer finding ID I use it to select name
$dname = mysql_query("SELECT m_name FROM member WHERE (`m_id` LIKE '%".$fetchB['d_id']."%') ");
$fetchDname = mysql_fetch_array($dname, MYSQL_ASSOC);?>
<ul data-role="listview" data-divider-theme="d" data-inset="false">
<li data-role="list-divider" role="heading">
<?php echo 'From:'; ?>
<?php echo $fetchA['m_name']; ?></li>
<li data-theme="c" >
<?php echo 'Reported user Name: '; ?>
<?php echo $fetchDname['m_name']; ?>
</li>
</ul>
The output that I get from this code is the first row of the table complaint printed multiple ($ln) times like :
Nada
Ali
Nada
Ali
....until $ln
Why It doesn't go to the next line although I am using loop?
I tried the queries in database and it works so the problem is not from db.
I am ready to write more details if needed :)
My Schema:
Member(m_id, m_name, ...etc)
Ride(r_id, p_id ....etc) *p_id refers to m_id
Complaint(c_id, d_id ..etc)*d_id refers to m_id
the idea is member reports member in a ride
You are not iterating over your original result set. Move $results = mysql_fetch_array($raw_results); inside of your for loop and you should start to see your desired output.
I would like to say, however, that you might want to look into offloading some of your logic to your database by using JOIN statements. It is difficult to say without seeing your underlying database schema, but it is generally not a good idea to have multiple database queries when one would be capable of retrieving the data that you need.
Lastly, the mysql_fetch_array statements are deprecated, so you should also look into the use of either the PHP PDO or the mysqli extension.
Awww I did find the query guys Thank You
That is it
$name = mysql_query("SELECT m.m_name
FROM member m
INNER JOIN ride r
on m.m_id = r.d_id
INNER JOIN complaint c
on r.r_id = c.r_id
");
Now I need to know how to go throght it using the loop
Change $results[1] to $results[$i] in the first line of your loop.
I am trying to select everything from two different mysql tables. I imploded an array called friends so that I can select everything about the user's friends from both tables. Originally I wanted to perform one query on one table and then in a while loop a query on the other. But that didn't work. If you know I way I can nest while loops then please be sure to comment/answer this question.
Here's my code:
$friends = implode("','", $friends);
//implode the friends array for sql query
$sql = mysqli_query($con, "SELECT * FROM users, text_post WHERE username IN ('$friends')") or die(mysqli_error($con));
while ($row = mysqli_fetch_assoc($sql)) {
echo $row['last_name']. ' ' . $row['body'];
echo '<br><br>';
}
Note: $row['last_name'] is from the users table and $row['body'] is from the text_post table. I am receiving this error whenever I run my code Column 'username' in where clause is ambiguous
Please help me.
UPDATE:
I changed my query to: $sql = mysqli_query($con, "SELECT * FROM users, text_post WHERE users.username IN ('$friends')") or die(mysqli_error($con)); but now it echo's every match twice:
parker hello
parker hello
simms what is up
simms what is up
simms it's raining
simms it's raining
jorge potato
jorge potato
Why is it doing that?
The only thing that is the same in both tables is the username.
Thanks for your question. The issue here is the error that you are receiving of:
Column 'username' in where clause is ambiguous.
What this means is that the statement that you are running is too obscure.
What you need to do to avoid this is to explicitly define the tables and columns that you are attempting to access and thus be implicit in your statements.
As a rule of thumb try and always define your columns in the following format:
{table_name}.{column_name}
With that being said the following should work:
$friends = implode("','", $friends);
/* implode the friends array for sql query. */
$sql = mysqli_query($con, "SELECT * FROM users, text_post WHERE users.username IN ('$friends') AND text_post.username IN ('$friends')") or die(mysqli_error($con));
while ($row = mysqli_fetch_assoc($sql)) {
echo $row['last_name']. ' ' . $row['body'];
}
Also I would also recommend that you use PHP's PDO Object as appose to the mysqli_query method.
I trust this helps.
both tables have a username. specify it using table.username
apparently both tables have the same column, and you are not prefixing the username column with either table name.
Maybe what you want is a union?
Use table prefix users in where block i.e. users.username
$sql = mysqli_query($con, "SELECT * FROM users, text_post WHERE users.username IN ('$friends')") or die(mysqli_error($con));
Try this:
$friends = implode("','", $friends);
//implode the friends array for sql query
$sql = mysqli_query($con, "SELECT users.last_name,text_post.body FROM users, text_post WHERE username IN ('$friends')") or die(mysqli_error($con));
while ($row = mysqli_fetch_assoc($sql)) {
echo $row['last_name']. ' ' . $row['body'];
}
Your logic would work were it not for username being present in both tables. Because it is, you need to alias your table names so that you can uniquely identify which column you're trying to use in the where clause:
SELECT * from users AS u, text_post as tp
WHERE u.username IN(....)
or shorter:
SELECT * from users u, text_post tp
WHERE u.username IN(....)
As a side note, it's bad practice to SELECT *, you should always explicitly select the columns you want, otherwise your code may crash if you add a column to a table later and forget to update your for loop to include/ignore any changed columns.
Let's put an easy example with two tables:
USERS (Id, Name, City)
PLAYERS (Id_Player, Number, Team)
And I have to do a query with a subselect in a loop, where the subselect is always the same, so I would like to divide it into two queries and put the subselect outside the loop.
I explain. What works but it is not optimize:
for($i=0;$i<something;$i++)
{
$res2=mysql_query("SELECT Team from PLAYERS WHERE Number=$i
AND Id_Player IN (SELECT Id FROM USERS WHERE City='London')");
}
What I would like to do but it doesn't work:
$res1=mysql_query("SELECT Id from USERS where City='London'");
for($i=0;$i<something;$i++)
{
$res2=mysql_query("SELECT Team from PLAYERS WHERE Number=$i
AND Id_Player IN **$res1**");
}
Thanks!
Something like this should work.
<?
$sql = "SELECT Team from PLAYERS
JOIN USERS on (Id_player=Id)
WHERE Number BETWEEN $minID AND $maxID
AND City='London'
GROUP BY Team";
$results=mysql_query($sql) or die(mysql_error());
// $results contain all the teams from London
// Use like normal..
echo "<ul>\n";
while($team = mysql_fetch_array($results)){
echo "\t<li>{$team['Team']}</li>\n";
}
echo "</ul>";
Placing SQL quires in loops can be very slow and take up a lot of resources, have a look at using JOIN in you SQL. It's not that difficult and once you've got the hang of it you can write some really fast powerful SQL.
Here is a good tutorial worth having a look at about the different types of JOINs:
http://www.keithjbrown.co.uk/vworks/mysql/mysql_p5.php
SELECT PLAYERS.*, USERS.City FROM PLAYERS, USERS WHERE USERS.City='London' AND PLAYERS.Number = $i
Not the best way to do it; maybe a LEFT JOIN, but it should work. Might have the syntax wrong though.
James
EDIT
WARNING: This is not the most ideal solution. Please give me a more specific query and I can sort out a join query for you.
Taking your comment into account, let's take a look at another example. This will use PHP to make a list we can use with the MySQL IN keyword.
First, make your query:
$res1 = mysql_query("SELECT Id from USERS where City='London'");
Then, loop through your query and put each Id field one after another in a comma seperated list:
$player_ids = "";
while($row = mysql_fetch_array($res1))
{
$player_ids .= $row['Id'] . ",";
}
$player_ids = rtrim($player_ids, ",");
You should now have a list of IDs like this:
12, 14, 6, 4, 3, 15, ...
Now to put it into your second query:
for($i = 0; $i<something; $i++)
{
$res2 = mysql_query("SELECT Team from PLAYERS WHERE Number=$i
AND Id_Player IN $player_ids");
}
The example given here can be improved for it's specific purpose, however I'm trying to keep it as open as possible.
If you want to make a list of strings, not IDs or other numbers, modify the first while loop, replacing the line inside it with
$player_ids .= "'" . $row['Id'] . "',";
If you could give me your actual query you use, I can come up with something better; as I said above, this is a more generic way of doing things, not necessarily the best.
Running query in a loop is not a great idea. Much better would be to get whole table, and then iterate through table in loop.
So query would be something like that:
"SELECT Team from PLAYERS WHERE Number BETWEEN($id, $something)
AND Id_Player IN (SELECT Id FROM USERS WHERE City='London')"
$res1=mysql_query("SELECT Id from USERS where City='London'");
for($i=0;$i<something;$i++)
{
$res2=mysql_query("SELECT Team from PLAYERS WHERE Number=$i
AND Id_Player IN **$res1**");
}
Would work, but mysql_query() returns a RESULT HANDLE. It does not return the id value. Any select query, no matter how many, or few, rows it returns, returns a result statement, not a value. You first have to fetch the row using one of the mysql_fetch...() calls, which returns that row, from which you can then extract the id value. so...
$stmt = mysql_query("select ID ...");
if ($stmt === FALSE) {
die(msyql_error());
}
if ($stmt->num_rows > 0) {
$ids = array();
while($row = mysql_fetch_assoc($stmt)) {
$ids[] = $row['id']
}
$ids = implode(',', $ids)
$stmt = mysql_query("select TEAM from ... where Id_player IN ($ids)");
.... more fetching/processing here ...
}