MySQL SELECT statement won't accept two php variables

Question

<?php
    function searchDatabase($key, $value)
    {
        $key = $key;
        $query = "SELECT * FROM user_data WHERE username='$value'";
        $result = mysqli_query(loadDatabase(), $query);
        $numRows = mysqli_num_rows($result);
        if ($numRows > 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }
?>

So I am using this code to search through my database to reveal a match on a key/value pair, but $key doesn't find the correct column in my database when passed into the query function. If I replace it with the word username, it matches fine. Is it a type issue? I am not explicitly stating its type so I can search other columns with the same function.

username: varchar(40)

Uh? $key = $key; must be a mistake? replace username by $key. — KIKO Software
– KIKO Software, Commented Feb 27, 2015 at 8:58
$query = "SELECT * FROM user_data WHERE username='$value'"; should be $query = "SELECT * FROM user_data WHERE username='".$value."'"; — Maximilian Ast
– Maximilian Ast, Commented Feb 27, 2015 at 9:00
Your code is vulnerable. Don't use mysqli without real-escape-string, because your site will attackable by SQL injection. — Joci93
– Joci93, Commented Feb 27, 2015 at 9:01

Carlos M. Meyer · Accepted Answer · 2015-02-27 09:00:38Z

2

<?php
function searchDatabase($key, $value)
    {
        $query = "SELECT * FROM user_data WHERE ".$key." = '".$value."'";
        $result = mysqli_query(loadDatabase(), $query);
        $numRows = mysqli_num_rows($result);
        return ($numRows > 0);
    }
?>

answered Feb 27, 2015 at 9:00

Carlos M. Meyer

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2 Comments

JohnMark Over a year ago

Thank you so much! This worked for me! So what exactly happened? Was it a type issue?

Carlos M. Meyer Over a year ago

You used username instead $key.

Deenadhayalan Manoharan · Accepted Answer · 2015-02-27 09:01:35Z

0

Try this

$query = "SELECT * FROM user_data WHERE ".$key."='$value'";

answered Feb 27, 2015 at 9:01

Deenadhayalan Manoharan

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Comments

KIKO Software · Accepted Answer · 2015-02-27 09:01:39Z

0

Like so:

function searchDatabase($key, $value)
{
  $query = "SELECT * FROM user_data WHERE $key='$value'";
  $result = mysqli_query(loadDatabase(), $query);
  $numRows = mysqli_num_rows($result);
  return ($numRows > 0);
}

answered Feb 27, 2015 at 9:01

KIKO Software

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Comments

Ataboy Josef · Accepted Answer · 2015-02-27 09:05:32Z

0

The problem is in the query. PHP variables are not properly echoed in the query.

<?php
function searchDatabase($key, $value)
    {
        $query = "SELECT * FROM user_data WHERE username='".$value."'";
        $result = mysqli_query(loadDatabase(), $query);
        $numRows = mysqli_num_rows($result);
        return ($numRows > 0);
    }
?>

answered Feb 27, 2015 at 9:05

Ataboy Josef

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