Simple code not working [closed] - php

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Hello guys I want to show the database on my website using this code which is given below but it's giving me annoying error again and again. I have tried everything but nothing is working it give me same error notice
Here is the error Notice
Notice: Undefined index: name in C:\xampp\htdocs\test3\index.php on line 15
Here is the PHP Code
<?php
$connect = mysql_connect("localhost","root","123");
if(!$connect) {
die("Failed to Connect: " . mysql_error());
}
if (!mysql_select_db ("login")){
die("Failed to Select DB: ". mysql_error());
}
$results = mysql_query ("Select * from users ");
while($row = mysql_fetch_array($results)){
echo $row['name'];
}
?>
I have also tried to replace mysql_fetch_array($results) with this mysqli_fetch_assoc($result) and it's also not working please run this code yourself and then give me that code. Thanks

Answer 1

you can use var_dump() to check the result
while($row = mysql_fetch_array($results)){
var_dump($row);
}
check the output, whether field name exists

Answer 2

Your database "login", table "users" doesn't have a column "name". That could be because of a typo with CaSe sEnsitivitY or other just missing.

Answer 3

<?php
$host="localhost"; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table name
// Connect to server and select databse.
mysql_connect("localhost", "root", "abcd")or die("cannot connect");
mysql_select_db("testDB")or die("cannot select DB");
$sql="SELECT * FROM login WHERE userid='$myusername' and password='$mypassword'";
$result=mysql_query($sql);
//you can then mysql_fetch_array or mysql_fetch_ob
...
?>
this will surely help you

Answer 4

You can also simply do it this way to avoid confusion on the if's
mysql_connect("localhost","root","123") or die("Failed to Connect");
mysql_select_db ("login") or die ("cannot connect to db");
$results = mysql_query ("Select * from users ");
while($row = mysql_fetch_assoc($results))
{
echo $row['name'];
}
Since the error refers to the name, you might want to check your users table if the column name exists.