Retrieving and displaying data from database

[DarthJawns]

I'm working on the user area of my site and I'm trying to get it to display how much currency and the rank the user currently has. I know sql statement would be something like "select * from users where username = ?" but I'm a little lost on how I bind the session username so it knows what username to pull from the users table. 

 

[MJ12]

What language / framework are you using?

 

[DarthJawns]

What language / framework are you using?

Just procedural php

 

[MJ12]

You'd want to use a prepared statement
 

SELECT currency, rank FROM users WHERE name = ? LIMIT 1;


where ? is replaced by the username as a bound variable. An example of this is shown here: https://www.w3schools.com/php/php_mysql_prepared_statements.asp.

 

[DarthJawns]

You'd want to use a prepared statement
 

SELECT currency, rank FROM users WHERE name = ? LIMIT 1;


where ? is replaced by the username as a bound variable. An example of this is shown here: https://www.w3schools.com/php/php_mysql_prepared_statements.asp.

How would I echo the results then?

 

[MJ12]

You'd bind the output of the SQL statement to a variable and then echo it normally, I'd imagine. I haven't written PHP in many years, so you'd probably echo each column out as

echo $row['column_name'];


or something like that.

 

[Hare]

$query = mysqli_query($connect, "SELECT * FROM users WHERE username = '$username'");
$row = mysqli_fetch_array($query);
$usermoney = $row["money"];


This will get the user based on their username, and allow you to display their money. 

echo $usermoney;


If it worked, you'll be able to display the user's money like this.

This is all assuming that you have the user's name contained in the $username variable. If you don't have $username set, it won't work. You can set $username with a session or cookie. 

This also assumed you have your PHP/MYSQL database connection set up and contained in the $connect variable (or whatever you named yours).

 

[DarthJawns]

@Hare , this is what I currently have 

<?
ini_set('display_errors', 1);
require_once "header.php";

$sql = "SELECT * FROM users WHERE username = ?";

if($stmt = mysqli_prepare($link, $sql)){

mysqli_stmt_bind_param($stmt, 's', $_SESSION['username']);

if(mysqli_stmt_execute($stmt)){


$info = mysqli_fetch_array($stmt);

echo "Current Points:" . $info['points'];


} else {

echo "Can't find user";
}
}
mysqli_stmt_close($stmt);





?>


but get an error message of mysqli_fetch_array() expects parameter 1 to be mysqli_result