无法从数据库phpMyAdmin检索数据

Question

I'm a beginner in learning how to set up database & PHP script and follow example to do that, then when I run login.php script I can't retrieve data from the database , I really feel that is a very simple question for others but I tried to solve it But didn't succeed, so can someone take look on my code then Corrects it?

here is my php script :

init.php :

<?php
    $db_name = "webapp";
    $mysql_username = "root";
    $mysql_password = "";
    $server_name = "localhost";
    $con=mysqli_connect($server_name, $mysql_username, $mysql_password, $db_name);

    if (!$con)  {
        echo "Connection Error ......." . mysqli_connect_error();
    } else {
        echo "<h3>Database connection Success .....</h3>";
    }
?>

login.php :

<?php
    require "init.php";
    $user_name = "YASER";
    $user_phone = "123456";

    $sql_query = "select name from user_info where user_name like'$user_name'and
        user_phone like'$user_phone';";

    $result = mysqli_query($con,$sql_query);
    if (mysqli_num_rows($result)>0)
    {
        $row = mysqli_fetch_assoc($result);
        $name = $row["name"];
        echo "<h3> Hello And Wellcome" . $name . "</h3>";
    }  else  {
        echo " No Info Is Available .......";
    }
?>

Answer 1

1st : First check that query is executing or failing

if(!$result){ echo mysqli_error($con); }

2nd : use = instead of like

$sql_query = "select name from user_info 
              where user_name='$user_name' and
              user_phone='$user_phone'";

3rd : You need to give proper spacing In query

like'$user_name'and
   ^^^        ^^^ 

To

like '$user_name' and

Answer 2

You have error in your query.

try this to find error

$result = mysqli_query($con,$sql_query) or die(mysqli_error($con));

Your query should be like as follow...

'SELECT name FROM user_info WHERE user_name LIKE "'.$user_name.'" AND user_phone LIKE "'.$user_phone.'"';