注册脚本 - mysqli_num_rows（）期望参数1为mysqli_result [duplicate]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

What I am trying to do:

Trying to create a register php script that works. Basically I need new records in my table.

Problem:

I am learning php and trying to find out how to solve, after loads of fixing, this last error message which I absolutely dont understand.

What I tried:

I set up the database and wrote the script. After that I encountered 13 errors from which I only fixed 12. The remaining one really gives me a hard time. I know it has something to do with mysqli_result but no idea how to implement it. I did try this:

$amount = mysqli_num_rows(mysqli_result, $result);

Error message:

mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given on line 22

<?php

    $connection = mysqli_connect("localhost", "root", "", "project")
    or die("cant connect");

$name = $_POST["name"];
$email = $_POST["email"];
$age = $_POST["age"];
$about = $_POST["about"];
$occupation = $_POST["occupation"];
$special_talent = $_POST["special_talent"];
$password = $_POST["password"];
$repeat_password = $_POST["repeat_password"];

if($password != $repeat_password OR $name == "" OR $password == "")
    {
    echo "Error";
    exit;
    }

$result = mysqli_query($connection, "SELECT id FROM members WHERE name, email LIKE '$name', '$email'");
$amount = mysqli_num_rows($result);

if($amount == 0)
    {
    $record = "INSERT INTO members (name, email, age, about, occupation, special_talent, password, repeat_password) VALUES ('$name', '$email', '$age', '$about', '$occupation',  '$special_talent', '$password', '$repeat_password')";
    $register = mysqli_query($connection, $record);

    if($register == true)
        {
        echo "Congratulations $name, you are now a Member!</a>";
        }
    else
        {
        echo "Unfortunately the registration failed. Try again please.</a>";
        }
    }

else
    {
    echo "Username already in use";
    }

?> 

</div>

Answer 1

$result is returning false. Your query has an error in it.

Change from

SELECT id FROM members WHERE name, email LIKE '$name', '$email'

To

SELECT id FROM members WHERE name LIKE '$name' AND email LIKE '$email'