The prepared request is not executed in php database.

Question

$mysqli = new mysqli($dbhost, $dbuser, $dbpass, $dbname); if ($mysqli->connect_errno) { echo "Не удалось подключиться к MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; } $stmt = $mysqli->prepare("INSERT INTO users ( user_id, user_status, user_email, user_pass, user_name, user_surname, user_avatar) VALUES (NULL, '2', ?, ?, ?, ?, '/front/img/users/user.png');"); $stmt->bind_param($this->email, $this->pass, $this->name, $this->surname); /* выполнение подготовленного выражения */ if (!$stmt->execute()) {echo "ошибочка!";}; /* Закрытие соединения и выражения*/ $stmt->close();

the code above gives an error ..

PS All variables checked for the presence of values, they are not empty. Edited the code, now it does not even reach the error output

 $stmt = $mysqli->prepare("INSERT INTO users ( user_id, user_status, user_email, user_pass, user_name, user_surname, user_avatar) VALUES (?, ?, ?, ?, ?, ?, ?);"); $stmt->bind_param('NULL', '2', $this->email, $this->pass, $this->name, $this->surname, '/front/img/users/user.png');

1. The string 'NULL' has nothing to do with an empty NULL value.
2. Values cannot be entered directly into bind_param, but can only be passed as variables

Accepted Answer · 2016-10-18T11:13:13

mysqli_stmt::bind_param first parameter accepts a string that describes the types of parameters, followed by the parameters themselves. There are 4 different identifiers in total

i - integer (integer)
d - double (floating point number)
s - string (string)
b - BLOB

For example, for four string parameters, the call should be

 $stmt->bind_param('ssss', $this->email, $this->pass, $this->name, $this->surname);

Well, then, instead

if (!$stmt->execute()) {echo "ошибочка!";};

Write so

 if (!$stmt->execute()) echo $stmt->error;

There will be more food for thought

error "No data supplied for parameters in prepared statement"
@Nikolay Did you specify the first parameter bind_param necessary types of parameters?
there must be lines .. obtained in this way: $name = $_POST["name"];
@Nikolay in the response the first parameter is the value of ssss which means four string values.

The prepared request is not executed in php database.

1 answer 1

More articles: