Php 简单的MySQL调用来选择特定键处的整数不起作用?
我有一个PHP:Php 简单的MySQL调用来选择特定键处的整数不起作用?,php,mysql,sql,phpmyadmin,connection,Php,Mysql,Sql,Phpmyadmin,Connection,我有一个PHP: <?php $client_ip = $_SERVER['REMOTE_ADDR']; $connection = new mysqli("localhost", "MyNotSoSecretUsername", "MySuperSecretPassword", "MyNotSoSecretDatabaseName"); if ($connection->connect_error) { die("Connect
<?php
$client_ip = $_SERVER['REMOTE_ADDR'];
$connection = new mysqli("localhost", "MyNotSoSecretUsername", "MySuperSecretPassword", "MyNotSoSecretDatabaseName");
if ($connection->connect_error) {
die("Connection failed: " . $Connection->connect_error);
}
$check_emails_sent_query = "SELECT `emails` FROM `email-ips` WHERE `ip`='11.111.111.111'";
$check_emails_sent_result = $connection->query($check_emails_sent_query);
echo $check_emails_sent_result;
?>
失败时返回FALSE
。要成功地选择
,显示
,描述
或
EXPLAIN
querysmysqli\u query()
将返回一个mysqli\u结果
对象。对于
其他成功的查询mysqli\u query()
将返回TRUE
你可以通过使用
if ($result = $connection->query($check_emails_sent_query)) {
while ($row = $result->fetch_row()) {
printf ("%s (%s)\n", $row[0]);
}
}
Mysqliquery()。使用对象:
<?php
$client_ip = $_SERVER['REMOTE_ADDR'];
$connection = new mysqli("localhost", "MyNotSoSecretUsername", "MySuperSecretPassword", "MyNotSoSecretDatabaseName");
if ($connection->connect_error)
{
die("Connection failed: " . $Connection->connect_error);
}
$check_emails_sent_query = "SELECT `emails` FROM `email-ips` WHERE `ip`='11.111.111.111'";
if ($check_emails_sent_result = $connection->query($check_emails_sent_query))
{
while($obj = $check_emails_sent_result->fetch_object())
{
$echo $obj->emails;
}
}
?>
$check\u emails\u sent\u result
是一个对象。您希望打印什么?Mysqli->query()应该返回一个对象,而不是一个可以回显的变量。从这个对象中,您可以获得emailsA quick Google的价值,并且浏览一些PHP文档并没有得到任何关于mysqli对象的信息,只是了解了如何使用面向对象的mysqli。为了获得我的值,我应该向对象请求什么?从对我的问题的否决票来看,我应该停止使用不推荐的函数,并在不久前登上MySQLi列车。我找到了一个“速记”版本来获取我在另一个问题上寻找的单一值。。。很难相信MySQL的“改进”版本会让返回单个值变得如此痛苦。我弄错了吗?就我个人而言,我并不热衷于尽可能多地将其转换为速记(除了娱乐之外),因此明确使用额外的代码行对我来说是合适的。