本文介绍了问题:试图获得非对象的属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 我在html中的表单中的输入类型是datetime-local并保存在mysql的表中。 在我的mysql表中,我创建了一个名为deadline的列,其类型为datetime。 /> 我尝试显示截止日期之后的行。 但是,它显示错误试图获取非对象的属性。 我不明白为什么。 我尝试了什么: <?php $ servername =localhost; $ username =root; $ password =; $ dbname =MyDB; //创建连接 $ conn = new mysqli($ servername,$ username,$ password,$ dbname); //检查连接 $ sql =SELECT Ass,Deadline,FROM Board WHERE Deadline> = NOW(); $ result = $ conn->查询($ sql); if($ result-> num_rows> 0) { //每行输出数据 while($ row = $ result-> fetch_assoc()){ echo作业:。 $行[驴。 - 最后期限: 。 $行[截止日期。 ; } }其他{ echo0 results; } $ conn-> close(); ?> My input type in a form in html is datetime-local and keep in a table in mysql.In my table in mysql, i created a column named deadline whose type is datetime.I try to display a row which deadline is after now. But, it shows the error "Trying to get property of non-object". I don't understand why.What I have tried:<?php$servername = "localhost";$username = "root";$password = "";$dbname = "MyDB";// Create connection$conn = new mysqli($servername, $username, $password, $dbname);// Check connection$sql = "SELECT Ass, Deadline, FROM Board WHERE Deadline >= NOW()";$result = $conn->query($sql);if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo " Assignment: ". $row["Ass"]. " - Deadline: ". $row["Deadline"]. ""; }} else { echo "0 results";}$conn->close();?>推荐答案 servername =localhost; servername = "localhost"; username =root; username = "root"; password =; password = ""; 这篇关于问题:试图获得非对象的属性的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!
10-19 07:52