How to assign php variable in select statement where condition [duplicate] - php

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What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
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I tried various ways to get the id from database but I can't get that Please help me!
Below is my query :
$username = $_POST['username'];
$query = "SELECT id FROM wp_users WHERE user_login ='".$username."'";
$result=mysqli_query($conn, $query);
if(mysqli_num_rows($result) == 1)
{
while($row = mysqli_fetch_array($result))
{
$id = $row['id'];
}
}
echo $id;
Also I tried this:
$query = "SELECT id FROM wp_users WHERE user_login ='$username'";

Answer 1

You should, at the very least, be using prepared statements in your queries if you're passing them user-supplied data.
use prepared statements ? and bind_param
use bind_result to bind the column to a variable. This variable is now bound by reference which means it will be updated on every iteration of the loop.
it is important to realize that you want to access the $id variable inside the loop as you're iterating over the dataset. If you use it outside/below the loop you are only working with the final row of data because it is being overwritten on every iteration.
turn on error reporting
Finally, I left your loop in place but usually, you'd only have a single user for a given username so you could use mysqli_fetch_assoc - Single Result from Database by using mySQLi
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$username = $_POST['username'];
$query = "SELECT id FROM wp_users WHERE user_login = ?";
$stmt = $conn->prepare($query);
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->bind_result($id);
while ($stmt->fetch()) {
echo "ID: $id\n";
}
$stmt->close();