(Php, Mysql) Result Could Not Be Converted to String

Object of class mysqli_result could not be converted to string

The mysqli_query() method returns an object resource to your $result variable, not a string.

You need to loop it up and then access the records. You just can't directly use it as your $result variable.

while ($row = $result->fetch_assoc()) {
    echo $row['classtype']."<br>";
}

PHP/MYSQL Error : Object of class mysqli_result could not be converted to string

$var in your code is actually an object of type mysqli_result.

This is actually a handle to a resultset, and you have to unload the results from that handle using some sort of fetch.

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";

$conn = new mysqli($servername, $username, $password, $dbname);

// dont normally get ID's of zero so I chnaged this to 1
$result = $conn->query("SELECT Name FROM users WHERE ID=1");

// now get the result row
$row = $result->fetch_assoc();     // row will be an array

?>

Now in your HTML you can do this

<html>
<head>
    <title>Title</title>
</head>
<body>
    <?php include 'pehape.php' ?>
<span>Username:</span><?php echo $row['Name'];?>
</body>
</html>

Object of class mysqli_result could not be converted to number (PHP | MySQL)

I updated your SQL query to alias the aggregation for easy access.

$qty = $db->query("SELECT SUM(`qty`) as 'val_sum' FROM `transaction` WHERE `date` = '2021-04-01'");

You can cast a string to an int with "(int) $string_value"

if ($qty->num_rows > 0) {
   $row = $qty->fetch_assoc()
   $profit = (int) $row['val_sum'] * 10;
}

mySQL Object of class mysqli_result could not be converted to string problems

you should fetch the result.

http://php.net/manual/en/mysqli-result.fetch-array.php

That is,

$row = $result->fetch_array(MYSQLI_NUM);

var_dump($row);

You could see all the rows as array.

Your code will be like this.

<?php
// Create connection
$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

    // sql to create table
$url = 'http://google.com';
$ip = $_SERVER['REMOTE_ADDR'];

$sql = "
SELECT ID FROM urls
WHERE url LIKE ('http://google.com')
";

$result = $conn->query($sql);

$row = $result->fetch_array(MYSQLI_NUM);

echo $row[0] . " " . $row[1];

$conn->close();

How do you convert a mysqli_result into a string?

Since you limit the selection to one entry use

$row = mysqli_fetch_array($result);
echo $row['ImageURL'];

If you select more than one entry loop over the result.

while($row = mysqli_fetch_array($result)) {
      echo $row['ImageURL'];
}

Error converting SQL result to string

You use variable $result both to concatenate string
$result .= "<div class='post modelVariant'>".$value."</div>"; and for storing result from MySQL query $result = $query->get_result().