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Php 从SQL表中获取奇异值_Php_Mysql_Sql_Database - Fatal编程技术网

Php 从SQL表中获取奇异值

Php 从SQL表中获取奇异值,php,mysql,sql,database,Php,Mysql,Sql,Database,我试图从我的一个表中获取一个值,但是它似乎没有检索到它,因为当我回显变量时,没有显示任何内容 这是我的密码 <?php $conn = new mysqli('localhost','root','root','neadatabase'); //Creating connection to database //Connection Error Checking if ($conn->connect_error) { die("Connection Failed:

我试图从我的一个表中获取一个值,但是它似乎没有检索到它,因为当我回显变量时,没有显示任何内容

这是我的密码

<?php
$conn = new mysqli('localhost','root','root','neadatabase'); //Creating 
connection to database



//Connection Error Checking
if ($conn->connect_error) 
{
    die("Connection Failed: " . $conn->connect_error);
}

echo ("Connection Successful");

//selecting bookings
$sqlSelectBookings = "
    SELECT ActivID, DateRequired, TimeRequired
    FROM activbooking
    WHERE ActivID = '$_POST[ActivID]' AND DateRequired = 
   '$_POST[DateRequired]' AND '$_POST[DateRequired]' AND 'TimeRequired = ' 
   $_POST[TimeRequired]'
";

$sqlSelectBookings = $conn->que($sqlSelectBookings);

//getting member limit
$sqlGetMemberLimit = "
    SELECT memberLimit
    FROM activity
    WHERE activID = '$_POST[ActivID]'
";

$sqlGetMemberLimit = ($conn->query($sqlGetMemberLimit)-
>fetchObject(memberLimit));

echo $sqlGetMemberLimit;

您的代码易受SQL注入攻击。您应该使用或准备中所述的语句。除了@Alex所说的之外,您是否在查询之前检查了“ActivID”的值?它是从哪里来的?它是怎么来的?好吧,我已经添加了我的代码的其余部分-我不完全确定你所说的@oldPadawan1是什么意思。首先,$_POST[ActivID]来自哪里?任何带有method=post的表单?然后,在页面->活动ID是活动表中的一个字段,用户在使用方法=post表单之前在页面上输入该字段。我选择的活动记录包含用户输入的任何内容的activID。在活动表中有另一个名为MemberLimit的字段。我试图检索这个变量,但是用户不知道这个值是@OldPadawan