Question

我有连接数据库的功能。我也会检查是否经过，但是，当我故意失败时，它并没有打印出if(!$db)括号中的错误。

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;
    }
}

它打印出来自PHP的警告：Warning: mysqli::__construct(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: NO)我得到了。但为什么它仍然打印出来＆＃34;好的＆＃34;在else中，不应该调用printError吗？

Answer 1

这里是处理错误的mysqli的php文档示例

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 . "\n";

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

echo $mysqli->host_info . "\n";
?>

Answer 2

您的对象$ db不为null，因此这将始终打印'ok'。要检查数据库是否已连接，请在if语句中使用'$ db-＆gt; connect_error'。

Answer 3

你是通过说新的mysqli（...）来创建一个对象。您可以删除新部件，并可以像下面一样使用它。

<?php
$link = mysqli_connect('localhost', 'my_user', 'my_password', 'my_db');

if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') '
        . mysqli_connect_error());
}

echo 'Success... ' . mysqli_get_host_info($link) . "\n";

mysqli_close($link);
?>

连接到数据库不会打印出错误

3 个答案: