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POInput in database failure
 

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  1. POInput in database failure
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    <p>I'm having trouble with the following code:</p> <pre><code>$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES ('$hash', '$lastname', '$email', '$email')"; mysqli_query($MyConnection, $sql); if(!mysqli_query($MyConnection, $sql)) { echo 'We are sorry, there are some problems with saving your data. Please try again within a few minutes.'; } else { echo 'We have succesfully saved your data. An activation e-mail will now be send to the e-mail address that you have provided us.'; } </code></pre> <p>I get no direct errors as due to mistyping or misusing a function. I do get however the message of the if-statement in a failure, the "We are sorry(..)" text.</p> <p>There must be a problem with the execution of the mysqli_query($MyConnection, $sql) function. But I don't see where it is.</p> <p>P.S. I can't post images, because my reputation is below 10. (Which is quite weird to limit it to that point)</p> <p>As some of you have provided most / all of the code:</p> <pre><code>&lt;?php // Opens the connection of the MySQL Database $MyConnection = mysqli_connect('fdb6.biz.nf', '1446018_amp', '-') or die("Could not connect to the database, please try again"); mysqli_select_db($MyConnection,'Users'); mysqli_connect_errno(); // Website Url: $website = 'http://www.askmephilosophy.co.nf/'; // Information provided by the user $username = $_POST['username']; $password = $_POST['password']; // Will get encrypted. $lastname = $_POST['lastname']; $email = $_POST['email']; // A higher "cost" is more secure but consumes more processing power $cost = 5; // Create a random salt $salt = strtr(base64_encode(mcrypt_create_iv(16, MCRYPT_DEV_URANDOM)), '+', '.'); // Prefix information about the hash so PHP knows how to verify it later. // "$2a$" Means we're using the Blowfish algorithm. The following two digits are the cost parameter. $salt = sprintf("$2a$%02d$", $cost) . $salt; // Hash the password with the salt $hash = crypt($password, $salt); $sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES ('$hash', '$lastname', '$email', '$email')"; mysqli_query($MyConnection, $sql); var_dump(mysqli_error($MyConnection)); if(mysqli_query($MyConnection, $sql)) { echo 'We have succesfully saved your data. An activation e-mail will now be send to the e-mail address that you have provided us.'; } else { echo 'We are sorry, there are some problems with saving your data. Please try again within a few minutes.'; mysqli_error($MyConnection); } mysqli_close($MyConnection); ?&gt; </code></pre>
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    1. PO
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      1. PTAnswer
    1. This table or related slice is empty.
    1. PTQuestion
    1. USJesse Dijkstra
    1. USJesse Dijkstra
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    1. This table or related slice is empty.
    1. This table or related slice is empty.
    1. This table or related slice is empty.
    1. PO
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    2. PO
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    3. PO
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    1. This table or related slice is empty.
    1. COYou're executing the query twice. If you have some restraints on any of the columns, it will throw an error. You also declare four columns, but only send three values.
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      1. POInput in database failure
      1. USDavid Chen
    2. COTry placing the result in a variable and checking from that instead. `$query = mysqli_query($MyConnection, $sql);` then `if ($query === false) {`
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      1. POInput in database failure
      1. USDavid Chen
    3. COEven when commenting out / removing the first mysqli_query function, and remove the '!' mark in the if-statement, changing the display messages whether it failed or not, it still gives me the error.
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      1. POInput in database failure
      1. USJesse Dijkstra
 

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