Question

以下PHP代码有什么问题？它始终显示无法连接到数据库。

require "init.php";

$nottxt = $_POST["nottxt"];
$date = date("Y-m-d");
$time = date("H:i:s");
$sql = "INSERT INTO notice (NOTICE_DETAIL, TIME, DATE) VALUES ('$nottxt', '$time', '$date')";
if (!mysqli_query($con, $sql)) {
    echo "Unable to save the data to the database";
}

Answer 1

此代码有什么问题？

它不会将参数转义为查询，因此，例如，当$_POST['nottxt']包含撇号时，您可以轻松地中断查询。了解how to prevent SQL injection in PHP。

另外，正如@Daniel Dudas已经指出的那样，如果您的错误消息是“无法连接到数据库”，那么错误的原因不在您粘贴的代码中，而是在init.php某处。首先检查数据库主机，用户名和密码。

Answer 2

这是经过测试的代码。您可以使用此代码：

<?php
$connection = mysqli_connect("localhost", "root", "","db_name")or die("cannot connect....");

echo $nottxt=$_POST['nottxt'];
echo $date=date("Y-m-d");
echo $time=date("H:i:s");
$sql = "INSERT INTO `notice`(notice_detail,time,date) VALUES('$nottxt','$time','$date')";
$result=mysqli_query($connection,$sql);
if($result)
{
    echo "inserted successfully";
}

?>

Answer 3

            <?php
            $connection = mysqli_connect("localhost", "root", "","database_name");
            $nottxt=$_REQUEST['nottxt'];
            $date=date("Y-m-d");
            $time=date("H:i:s");
            $sqls = "INSERT INTO `notice`(notice_detail, time, date) VALUES('".$nottxt."','$time','$date')";
            $result=mysqli_query($connection,$sqls);
            if($result)
            {
                echo "inserted successfully";
            }

            ?>
please use this code...

Answer 4

试试这个....

var output = data.reduce(function(prev, cur) {
  if (!prev[cur.topic]) prev[cur.topic] = [];
  prev[cur.topic].push(cur);
  return prev;
}, {});

PHP使用mysqli将包含时间和日期的数据插入MySQL

4 个答案: