SQL Query of checking existing entry in database is not working

Question

<?php
          error_reporting(0);
          $link = mysqli_connect("localhost", "root", "", "checksql"); 

          if($link === false){
            die("ERROR: Could not connect. " . mysqli_connect_error());
          }

          $myemailaddress=$_POST['useremail']; 
          $mypassword=$_POST['userpassword'];

          $sql = mysqli_query("SELECT * FROM register WHERE Email = '$myemailaddress' ");
          $count = mysqli_num_rows($sql);
          echo $count;

          if($count > 0){
            echo "success";
          } else{
            echo "failed";
          }
       ?>

I am trying to check whether an email exists in the database or not. I searched different thread on stackoverflow and tried to correct it but failed. Even the echo of $count isn't showing it's value. Is there any other way to check it?

`error_reporting(0);` disables error reporting. Remove that line and post the error message, please. — PhilMasteG, Aug 23 '15 at 22:08
Looks like you want to check whether an email has "already been used" for registration and deny using it again. Instead of using a race-condition prone SELECT-then-INSERT combo, do that with a unique contraint on the Email field in your table definition and then check for the specific error code ER_DUP_KEY. see https://dev.mysql.com/doc/refman/5.5/en/error-messages-server.html and https://dev.mysql.com/doc/refman/5.0/en/constraint-primary-key.html — VolkerK, Aug 23 '15 at 22:14
You need to pass the connection into the query function when using the procedural approach. `$sql = mysqli_query($link, QUERY);` http://php.net/manual/en/mysqli.query.php Also you shouldn't pass user input direct to your query this opens you to SQL injections. — chris85, Aug 23 '15 at 22:16
if `$count` is zero ( 0 ) then echo'ing it out will generally show nothing as zero is considered false. Are you expecting a value greater than zero? — Professor Abronsius, Aug 23 '15 at 22:17
@chris85 darn it, saw your comment soon as I posted my answer. — Funk Forty Niner, Aug 23 '15 at 22:17
so, where are we with this question? there are quite a few unknowns which I have made points about in my answer below. Go over it carefully and check for errors while making sure there are no spaces in your db's data and your inputs. — Funk Forty Niner, Aug 23 '15 at 23:11
Your code must check the error result of the query. Get into this habit and you will avoid issues like this in future. — crafter, Aug 24 '15 at 00:59

score 5 · Accepted Answer · edited May 23 '17 at 11:52

You didn't pass db connection to your query

$sql = mysqli_query($link, "SELECT ...
                    ^^^^^^

Btw, your code is open to SQL injection.

Use a prepared statement

https://en.wikipedia.org/wiki/Prepared_statement

More on SQL injection:

Also make sure your POST arrays are not failing you.

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Displaying errors should only be done in staging, and never production.

error_reporting(0); doesn't help you, it turns it off.

Add or die(mysqli_error($link)) to mysqli_query() to check for errors.

http://php.net/manual/en/mysqli.error.php
Your form should be using a POST method with name attributes for both your POSTs. That is unclear and wasn't posted in your question; call it an insight.

If you are using both your form and PHP/MySQL inside the same file, then that will trigger undefined index notices on initial page load.

Use !empty() for them.

Reference(s):

At first I really want to thank you so much for helping me this much. Actually I am just learning php html css and in a beginner stage. So I didn't know so many informations but thanks to you I can study about them now. Problem solved. — , Aug 23 '15 at 23:18
@ArnabDas You're very much welcome Arnab, I'm glad to have been of help and that my answer helped solved it, *cheers* — Funk Forty Niner, Aug 23 '15 at 23:26

SQL Query of checking existing entry in database is not working

1 Answers1