Problem: trying to get property of non-object

Question

1.00/5 (1 vote)

See more:

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();
?>

Posted 30-May-18 4:17am

ZingXing

Updated 30-May-18 4:49am

Add a Solution

Comments

Richard MacCutchan 30-May-18 10:26am

You need to check which line throws the error, and which variable is null.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Jochen Arndt · Answer 1 · 2018-05-30T04:50:00

Always check for mysqli errors, especially during development because then you can get meaningful error messages:

PHP

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_errno) {
    printf("Connect failed: %s\n", $con->connect_error);
    exit();
}
if ($result = $conn->query($sql)) {
    // $result is an object and can be used to fetch row here
}
else {
    printf("Query failed: %s\n", $conn->error);
}

A possible error source is in your query string which has an unexpected comma between DeadLine and FROM. Such syntax errors result in a boolean FALSE be returned by the query() function instead of a mysqli_result object. And the boolean type is not an object type and has therefore no properties like num_rows.

See also PHP: mysqli::query - Manual[^]-