doubipiao1611 2018-05-20 11:44
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连接数据库不会打印出错误

I have a function to connect to a DB. I also check if that went through, however, when I fail it intentionally, it doesn't print out the errors in the if(!$db) brackets.

function connectToDb(){
    //Connect to a database
    $db = new mysqli("localhost", "root", "", "vm_ski");
    if(!$db){
        echo "error: ConnectToDB failed";
        printError("Could not connect to db: ".$db->error);
    }
    else{
        echo "OK";
        return $db;
    }
}

It prints out the warning from PHP: Warning: mysqli::__construct(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: NO) which I get. But why does it still print out "OK" in the else, shouldn't it call printError?

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  • duanba3707 2018-05-20 11:57
    关注

    here is php document example of mysqli to deal with error

    http://php.net/manual/en/mysqli.quickstart.connections.php

    <?php
    $mysqli = new mysqli("localhost", "user", "password", "database");
    if ($mysqli->connect_errno) {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }
    echo $mysqli->host_info . "
    ";
    
    $mysqli = new mysqli("127.0.0.1", "user", "password", "database", 3306);
    if ($mysqli->connect_errno) {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }
    
    echo $mysqli->host_info . "
    ";
    ?>
    
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