Why does the connection to the database fail?

Question

I study OOP in php . I wrote a simple class for connecting to the database and creating a table. I do not cut why it does not work, I understand that there is no connection to the database in the $link= $this->mysql_connect_djo(); but I do not understand why. Poke my nose knowing, please! Here is the code

 <?php $host = "localhost"; $user = "test"; $password = "123456"; $database = "class2"; $query = "CREATE TABLE notepad_info ( id INT PRIMARY KEY AUTO_INCREMENT, fio VARCHAR(60) NOT NULL, notepad_text TEXT )"; class MYSQL_RABOTA { //подключаемся к базе protected function mysql_connect_djo() { global $host, $user, $password, $database; $link = mysqli_connect($host, $user, $password, $database); if (!$link) { echo "Не возможно подключиться к базе" . " " . mysqli_connect_error(); exit(); } } //Создание необходимой таблицы function create_table() { global $query; $link= $this->mysql_connect_djo(); if (mysqli_query($link, $query)) { echo "Запрос отправлен"; } else { echo "Ошибка отправки запроса "; } mysqli_close($link); } } $mysql_djo = new MYSQL_RABOTA; $mysql_djo->create_table(); ?>

For the student - always write exactly how something does not work: errors, exceptions, white screen, smoke from the system, etc.
While it is difficult to understand why somewhere something somehow does not work)
It is not necessary to use global in the class, describe the logins / passwords in the constructor, and put the parameters to create the table in the create_table method, put $ link in $ this-> link;
as a result, the call should be something like this: $ mysql_djo = new MYSQL_RABOTA ("test", "123456", "class2");
$ mysql_djo-> create_table ("notepad_info", "id INT PRIMARY KEY AUTO_INCREMENT", "fio VARCHAR (60) NOT NULL", "notepad_text TEXT")
well, use the mysqli object interface to not switch back and forth.

Accepted Answer · 2013-06-10T11:53:56

You do not pass the parameters! In general, I would change the class so:

  class MYSQL_RABOTA { private host, user,password, database, link; //Добавлена строка; function __construct($host, $user, $password, $database) //Добавлена функция; { $this->host = $host; $this->user = $user; $this->password = $password; $this->database = $database; } protected function mysql_connect_djo() { //global $host, $user, $password, $database; Строка удалена; $this->link = mysqli_connect($this->host, $this->user, $this->password, $this->database); //Строка отредактирована; if (!$this->link) { echo "Не возможно подключиться к базе" . " " . mysqli_connect_error(); exit(); } return ($this->link); //Строка добавлена; } //Создание необходимой таблицы function create_table($query) { //Строка отредактирована; //global $query; Строка удалена; $this->mysql_connect_djo(); //Строка отредактирована; if (mysqli_query($this->link, $query)) { //Строка отредактирована; echo "Запрос отправлен"; } else { echo "Ошибка отправки запроса "; } mysqli_close($this->link); //Строка отредактирована; } } $mysql_djo = new MYSQL_RABOTA($host, $user, $password, $database); //Строка отредактирована; $mysql_djo->create_table($query); ?>

It seems so. By the way, since you are already writing object-oriented, it makes sense to replace the procedural inserts

 mysqli_query($this->link, $query) на $this->link->query($query); mysqli_close($this->link) на $this->link->close();

And you can also:

 if (mysqli_query($this->link, $query)) { echo "Запрос отправлен"; } else { echo "Ошибка отправки запроса "; } //Заменить на: mysqli_query($this->link, $query) or exit("Ошибка отправки запроса"); echo "Запрос отправлен";

Answer 2 · 2013-06-10T11:52:43

One:
Your connection function returns nothing. And it would be nice, at least to return the link.

Two:
start learning OOP with global - a bad omen. =) Especially for the purpose for which you use.

It is better:

 class my_class{ const db_name="db1"; const db_pass="db2";

Or even in the configuration file.

Why does the connection to the database fail?

2 answers 2

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