CODEWITHSUNDEEP
phpmysqlsql
Josh N
Josh N

Reputation: 25

PHP - Populate a field depending on auto-populated select

I'm fairly new to php and have a question. I have an HTML form that has a SELECT auto-populated from an SQL table via PHP. The dropdown is populated with all users with the level of "Admin" or "Moderator". This is the code to connect:

$con = mysqli_connect("localhost", "root", "", "database") or die("Error " . mysqli_error($con));

And the dropdown itself:

<form name="htmlform" role="form" method="POST" action="result.php">
<select id="user" name="user" required>
    <option selected disabled>User</option>
    <?php
    $result = $con->query("SELECT username FROM users WHERE level='admin' OR level='moderator' ORDER BY level");
    while ($row = $result->fetch_assoc()) 
    {
      $username = $row['username'];
      echo '<option value="'.$username.'">'.$username.'</option>'; 
    }
    ?>
</select>

This works perfectly. The problem I'm having is that I am trying to reuse the data from this form (specifically $_POST['user']) on another page to auto-populate another field in a form. I need to see if the 'user' is an Admin or not and return $other as either "y" (Admin) or "n" (not Admin), which will then be added to another table.

Here's my code on the 2nd page (result.php):

$user=$_POST['user'];
$query = $con->query("SELECT level FROM users WHERE username=$user");
$variable=mysqli_query($con, $query);
if ($variable=="admin") {
    $other = 'y';
} else {
    $other='n';
}

At the moment all output for $other is "n" regardless of anything. So, obviously I have an error in the code, but don't know enough php to be able to spot or correct it. Please could someone help point out the error?

Upvotes: 1

Views: 364

Answers (4)

RiggsFolly
RiggsFolly

Reputation: 94662

text values have to be wrapped in quotes in a query

$query = $con->query("SELECT level FROM users WHERE username='$user'");

You also look like you were trying to execute that same query twice here:

$query = $con->query("SELECT level FROM users WHERE username=$user");
$variable=mysqli_query($con, $query);

this is not legal usage.

Also when you run this line

$variable=mysqli_query($con, $query);

$variable is not a value, but a mysqli_result object that will contain a resultset or FALSE if the query failed, but definitely not the content if the id column in your query.

However if you are using data got from the user, it is not safe to assume thay are not attempting a SQL Injection Attack

So you should use Prepared and Parameterised queries like this

$stmt = $con->prepare("SELECT level FROM users WHERE username=?");
$stmt->bind_param('s', $_POST['user']);
$stmt->execute();

I think you shoud start by reading the PHP manual for the mysqli extension

Upvotes: 2

Anthony Rutledge
Anthony Rutledge

Reputation: 7564

(Without getting into issues about best practices ...)

Your second code snippet's usage of the return value from mysql_query() is problematic.

The PHP Manual states:

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

Hence, $variable is a PHP resource and cannot ever be equal to a string. Use tripple === equals when possible. You still need to "fetch" the record from the result resource (you managed to to do this in the first code snippet).

Generally speaking ...

$result = mysqli_query($con, $query);
$record = result->fetch_assoc();  

//if(result->fetch_assoc()['level'] === 'admin') in PHP 5.4 and up.

//or

//if(mysqli_query($con, $query)->fetch_assoc()['level'] === 'admin') in PHP 5.4 and up.

if($record['level'] === 'admin')
{

}
else
{

}

Cheers!

Upvotes: 1

Juanan Guerrero
Juanan Guerrero

Reputation: 1

According with mysqli_query doc:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

So don't expect to get database value directly from call, you are comparing a mysql_result object (you made a SELECT) versus a constant string. You need to get your data from mysql_result object and then you can make comparison.

Upvotes: 0

Saliery
Saliery

Reputation: 119

This code should work for you:

$user = $_POST['user'];
$sql = "SELECT level FROM users WHERE username={$user}";

$variable = mysqli_query($con, $sql)->fetch_row();
if ($variable[0]=="admin") {
    $other = 'y';
} else {
    $other='n';
}

Upvotes: -1

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