Check if a PHP variable exists in a MySQL table

Question

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";
}

A: You're using the wrong identifiers for your table and column. Remove the quotes. Plus, that's not the way to check if a record exists, you're just checking if the query returns true or not. — Funk Forty Niner
– Funk Forty Niner, Commented Oct 26, 2014 at 12:42
Well the sad part is that it returns the same result when I set $host to a value that's contained in the database, and when I set it to a different value. I am definitely going to check on the links that you sent. I'm writing the script for a bash alias, but it would definitely be a good idea to learn how to write securely. Thank you very much. — Neekoy
– Neekoy, Commented Oct 26, 2014 at 13:27

Funk Forty Niner · Accepted Answer · 2014-10-26 13:03:22Z

3

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.

edited Oct 26, 2014 at 13:03

Funk Forty Niner

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answered Oct 26, 2014 at 12:46

baao

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8 Comments

Funk Forty Niner Over a year ago

Needs to be like this; why? Plus, OP stands at coming back saying "it doesn't work".

Funk Forty Niner Over a year ago

I understand. Now, OP's question is "I'm trying to check if the content of the variable $host exists" - Whether OP truly wants to check if it returns TRUE or FALSE, is fine, but it doesn't really check if a row containing the variable being passed actually exists. Wait for OP to respond.

baao Over a year ago

Was just at writing the second part, if it's still wrong I don't see something right, @Fred-ii-

Neekoy Over a year ago

What you suggested is definitely working. There seems to be another issue somewhere around the code, however now it's correctly determining if $host is contained in the database or not. Thank you very much <3

baao Over a year ago

You're welcome @Neekoy , Please mark the answer as correct if it has helped you

|

score 1 · Accepted Answer · 2014-10-26 12:51:50Z

1

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

edited Oct 26, 2014 at 12:51

answered Oct 26, 2014 at 12:46

user3522371

4 Comments

Funk Forty Niner Over a year ago

I love "drop-in code" with no explanations. Plus, it will fail. WHERE ID = $host

user3522371 Over a year ago

@Fred-ii- What is unclear ? (and sorry for the quotes, i reedited the answer)

Funk Forty Niner Over a year ago

When an answer is given with no explanation as to why OP's code is failing, OP would have not learned the reason why it is failing and stands at doing the same mistake again.

user3522371 Over a year ago

@Fred-ii- I agree with you, but for this specific question, nothing much to explain, it is just a syntaxic error.

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