select from mysql table using array - php

Question

i am trying display a the rows in my database table using the array of ids gotten from another table. i want it to display the first row which is $rowsfriend. and display the second row which is rows .......... but it only displays $rowsfriend
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ochat";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM friends where friend1='".($_POST[id])."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$rowsfriend = $row["friend2"];
echo $rows;
}
}
$sqll = "SELECT * FROM users WHERE id IN ($rowssfriend)";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0) {
while($roww = $resultt->fetch_assoc()) {
$rowsss = $row["username"];
echo $rowss;
}
}
else {
?>
<h1>Register</h1>
<form action="selectfriends.php" method="post">
id:<br />
<input type="text" name="id" value="" />
<br /><br />
<input type="submit" value="enter" />
</form>
<?php
}
?>

Answer 1

Instead of two queries, you can write this nested query.
$sqll = "SELECT * FROM USERS WHERE ID IN (SELECT friend2 FROM friends WHERE friend1='".$_POST[$id]."')";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0)
{
while($roww = $resultt->fetch_assoc())
{
$rowsss = $row["username"];
echo $rowss;
}
}
Hope this solve your problem .

Answer 2

Please try this version of code instead, if you might:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ochat";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT users.* FROM users WHERE users.id IN (SELECT friend2 FROM friends where friend1='".($_POST[id])."')";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$rows = $row["username"];
echo $rows;
}
}
else {
?>
<h1>Register</h1>
<form action="selectfriends.php" method="post">
id:<br />
<input type="text" name="id" value="" />
<br /><br />
<input type="submit" value="enter" />
</form>
<?php
}
?>
As far as I understood the code well, the problem was with the variable name typo as #Admieus wrote but also in the fact that in each iteration of the first loop variable $rowsfriend got overriden with a new value so after the end of the loop, $rowsfriend contained the last id from the result of the first query $sql.
The above version makes only one query using subquery in it to get directly usernames who are friends of friend1 given in $_POST[$id].
I hope it helps.

Answer 3

There are typos in the second loop.
You assign the value to $rowsss and try to echo from $rowss notice the difference.
Also, you assign the fetch_assoc result to $roww and then try to call it again with $row.
$sqll = "SELECT * FROM users WHERE id IN ($rowsfriend)";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0) {
while($roww = $resultt->fetch_assoc()) {
$rowsss = $roww["username"];
echo $rowsss;
}
}
Point of improvement is: check your variable names, make names that are easy to understand and hard to mix up.
For instance, the variable containing the sql query should not be named $sql and to make it worse a second query shoul not be named sqll. Instead use names that imply what you are doing.
$querySelectFriendsFrom = "SELECT * FROM users WHERE id IN ($friendId)";
Don't take this as a hard rule, it's more of a tip to prevent silly mistakes.
Update: there was also a type in the query referring to rowssfriend instead of rowsfriend. Fixed above.