CODEWITHSUNDEEP
phpmysql
Neekoy
Neekoy

Reputation: 2533

Check if a PHP variable exists in a MySQL table

I'm trying to check if the content of the variable $host exists in the ID column of my database. Afterwards I want it to assign the corresponding value to $exists, if the entry is in the database or not.

For some reason this part of my script is always returning the same result, regardless if $host is contained in the database or not. Please help :)

$query = mysqli_query($conn, "SELECT 'ID' FROM 'servers' WHERE 'ID' = '$host'");
if (empty($query)) {
        $exists = "false";
}
else {
        $exists = "true";
}

Upvotes: 0

Views: 6305

Answers (2)

baao
baao

Reputation: 73241

This line 

$query = mysqli_query($conn, "SELECT 'ID' FROM 'servers' WHERE 'ID' = '$host'");

needs to be like this:

$query = mysqli_query($conn, "SELECT `ID` FROM `servers` WHERE `ID` = '$host'");

Right now, you are selecting ID as a string, so you need to put table and column names in `` and you put strings (or variables containing strings in ' ' )

and then do 

$count = $conn -> num_rows($query);
if ($count < 1 ) {
$exists = "false";
}
else
{
$exists = "true";
}

to actually check the number of rows containing $host 's value

Also, you should at least use 

$host = mysqli_real_escape_string($conn, $host);

before using a variable in a query to avoid mysql injection, but better use prepared statements. There are some links in the comments to your question which will help you with that.

Sidenote:

Having used or die(mysqli_error($conn)) to mysqli_query() would have signaled the error.

Upvotes: 3

user3522371
user3522371

Reputation: 

$query = mysqli_query($conn, "SELECT ID FROM servers WHERE ID = '".$host."'");
if (empty($query)) {
        $exists = "false";
}
else {
        $exists = "true";
}

Upvotes: 1

Related Questions