Question

I have the below code:

class config_model
{
    public $host;
    public $root;
    public $root_password;
    public $db;

    public function __construct() {
        $this->host = "localhost";
        $this->root = "root";
        $this->root_password = "";
        $this->db = "Data";
    }

    public function init() {
        $mysqli = new mysqli($this->host, $this->root, $this->root_password, $this->db);
        if ($mysqli->connect_error) {
            die('Connect Error (' . $mysqli->connect_errno . ') '
                . $mysqli->connect_error);
        }
        return $mysqli;
    }
}

$config_model = new config_model();
echo $config_model->init();

When I check the script I see this error:

"Catchable fatal error: Object of class mysqli could not be converted to string".

The erorr is here: "echo $config_model->init();"

How do I handle this error?

Answer 1

Your error is this:

"Catchable fatal error: Object of class mysqli could not be converted to string".

now, read the error: it says Object of class mysqli not be converted to string. Actually read it. Now, I hope you know what an object is, and I also hope you know what a string is. So the Object is somewhere being converted to a string and the object can't handle that conversion.

Where is this happening? Reading your code line by line and you even give the line the error occurs on:

echo $config_model->init();

you are echoing out an object because that is what is being returned by the ->init() method call, init is giving you back an object type. You are then immediately telling PHP to output this object as a string type. This is causing the issue.

Solution

PHP has a __ToString() magic method which you can add to your object so that what you call it as a string (which you shouldn't, but...) that the object will run this magic method and output something of your designation.

A simpler solution is also not to try and output objects as strings, instead using routines such as print_r or var_dump if you really need to (but as stated, you shouldn't be doing this at all in a perfect world).

Answer 2

the code $config_model->init() returns object of Config_model class, so you can not 'echo' it, because echo is used for string, If you wan to test your configuration you may use var_dump().

for Ex:

$config_model = new config_model();
var_dump($config_model->init());

Answer 3

mysqli_report(MYSQLI_REPORT_STRICT);

try {
     $connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db') ;
} catch (Exception $e ) {
     echo "Service unavailable";
     echo "message: " . $e->message;   // not in live code obviously...
     exit;
}

Answer 4

$ config_model = new config_model（）;

$ config_model是对象，无法回显它，只需使用var_dump（）。

Mysqli connection issue fatal error

4 个答案:

Solution