问题：试图获得非对象的属性 [英] Problem: trying to get property of non-object

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< br>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 = "";

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