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Prepared Statements headache

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Posted (edited)

 I've had to rearrange a lot of code and I've been trying to put together a prepared statement in a registration form. I'm having a really hard time and being very new to PHP the issue is really confusing for me.

first, I have this function:

function process_database($post) {
  global $table;
  global $conn;

  //check database connection
  if ($conn->connect_error) {
    return false;

  } else {
    if($statement = $conn->prepare("INSERT INTO $table (username, email, password) VALUES ( ?, ?, ? )")){
      $username = $post['username'];
      $email = $post['email'];
      $password = $post['password'];

      $statement->bind_param("sss", $username, $email, $password);

      echo "Added: ".$username.", ".$email.", ".$password."<br>";

        printf("Connect Failed: %s\n", $conn->connect_error);
      } else {
        echo 'fuckin ay!!!';

    } else {
      return false;
  return true; 

The issue is very strange. I'll post the function call so it's clear:

//process database actions
  if (!process_database($data) ) {
    return array( 'status' => 0, 'message' => 'Unable to process database request' );

When I run the registration.py without process_database() everything is fine, so I'm confident in the error processing. Here's where it get weird -

when I process the form


echo "Added: ".$username.", ".$email.", ".$password."<br>";

is returned from the //DEBUGGING BLOCK but I also get back the error from the following if statement - "Connection Failed: ...."

BUT I also get back the registration successful message that only shows if the function returns true

In short, it's giving me 2 positive affirmations but also the Connection failed message and of course it's not adding anything to the database. I've been working this function all day, and I'm lost for answers. What's going on with this code? I can't see where I've gone wrong

Edited by TechnoDiver
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That is calling the execute() a second time. If username or email must be unique that could be generating a duplicate key error - check mysql error, don't just assume it's a connection error.

Better still, put this line directly before your mysqli_connect() line


the all errors will throw a notification without all those if()s

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