函数返回一个不起作用的对象

Question

I have some code which works:

$user = 'xxx';
$pass = 'xxx';
$db='vive';
$host ='localhost';
$name = 'chris';

function test($user, $pass, $db, $host){
$mysqli = new mysqli($host, $user, $pass, $db);
if ($mysqli->connect_error) {
die('Connect Error (' . $mysqli->connect_errno . ') '
        . $mysqli->connect_error);
}
else{
    echo 'Checkpoint 1.0 <br>';
        return $mysqli;
}
}
$mysqli2 = test($user, $pass, $db, $host);
$sql = "SELECT * FROM `vive_user` WHERE `username` LIKE"."'$name'";
$result = $mysqli2->query($sql);
$num_results = $result->num_rows;
if ($result->num_rows>0)  ...

I am able to connect to the database and pull the information that I want out of the database everytime. I wanted to clean up my code a little (all the details are not shown), so i made a new function register() out of the last part of the code. In this new function, i want to call on function test() to return me a database connection object which I can then use to perform queries:

<?php

$user = 'root';
$pass = 'root';
$db='vive';
$host ='localhost';
$name = 'chris';

function test($user, $pass, $db, $host){
$mysqli = new mysqli($host, $user, $pass, $db);
if ($mysqli->connect_error) {
    die('Connect Error (' . $mysqli->connect_errno . ') '
            . $mysqli->connect_error);
}
    else{
        echo 'Checkpoint 1.0 <br>';
            return $mysqli;
    }
}

function register($name){
    echo'test';

    global $user;
    global $pass;
    global $db;
    global $host;

    $mysqli2 = test($user, $pass, $db, $host);
    $sql = "SELECT * FROM `vive_user` WHERE `username` LIKE"."'$name'";
    $result = $mysqli2->query($sql);
    $num_results = $result->num_rows;
    if ($result->num_rows>0)...
}

For some reason the function register() will never give me any values from the database. I am unable to get anything for $result. Any help is appreciated, I have been dancing around the problem for a few days now. Note that in my actual code I have these two functions in different php files.

Answer 1

The LIKE statement there in the query is missing %...% wrapper.

change the register to :

function register($name){

  global $user;
 global $pass;
 global $db;
 global $host;


  $name = "%".$name."%"; // see here..
    echo'test';
    $mysqli2 = test($user, $pass, $db, $host);
    $sql = "SELECT * FROM `vive_user` WHERE `username` LIKE"."'$name'";
    $result = $mysqli2->query($sql);
    $num_results = $result->num_rows;
    if ($result->num_rows>0)...
}

and make sure the method is called