MySqLi not working for some reason

Question

Actually I am trying to use MySqli but its is not working for me even it is showing that database is connected but query is not working even not giving any error. I tried using the code bellow:

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);


    $connect = mysqli_connect("localhost",'dbu', 'password');
    if($connect){
    echo 'connected';   
    }
    else {
    echo 'Failed:'.mysqli_error();  
    }
    mysqli_select_db('dbname',$connect);

    if($query = mysqli_query("SELECT * FROM `users`")){
    $num = mysqli_num_rows($query);
    while($fetch = mysqli_fetch_array($query)){
    echo $id = $fetch['id'];
    }
    }
    else {
    echo 'error:'.mysqli_error();   
    }

What do you mean it's not working. You just said query works and database is connected — Rotimi, Oct 04 '17 at 09:44
Sorry I mean database connected but query is not working.. That was typo I fixed in the question too. — SimulationCode, Oct 04 '17 at 09:47
Try mysqli_connect( "127.0.0.1", "dbu", "password", "dbname"); and remove mysqli_select_db() — shashi, Oct 04 '17 at 09:51
If you do a `die(var_dump($num))` right after `$num = mysqli_num_rows($query);` what do you get? — apokryfos, Oct 04 '17 at 10:04
Possible duplicate of [How to get MySQLi error information in different environments? / mysqli\_fetch\_assoc() expects parameter 1 to be mysqli\_result](https://stackoverflow.com/questions/22662488/how-to-get-mysqli-error-information-in-different-environments-mysqli-fetch-as) — Dharman, Jul 05 '19 at 21:12

Rotimi · Answer 1 · 2017-10-04T09:52:05.973

You have an issue with how you select the database. First parameter is the link while second is the name

mysqli_select_db($connect, 'db_name');

Alternatively you could pass the database as the fourth parameter as

$connect = mysqli_connect("localhost",'dbu', 'password','db_name');

Also the same issue is with the query

mysqli_query($connect, $query);//the first parameter is the link while second is the query itself 

$query = mysqli_query($connect, "SELECT * FROM `users`");

Where $query is the query itself. Also do not forget to use prepared statements when you have where clauses for where statements and also when you have insert values as well as delete

score 0 · Answer 2 · answered Oct 04 '17 at 09:46

please check mysqli_query syntax before start to write code https://www.w3schools.com/PhP/func_mysqli_query.asp

Your select query having syntax problem, Please change like below,

$query = mysqli_query($connect, "SELECT * FROM `users`") //connection paramter missing

May this will help

score 0 · Accepted Answer · edited Jul 04 '19 at 18:43

0

Turn error on you have errors in mysqli_* functions. change your code to

$connect = mysqli_connect("localhost", 'dbu', 'password');
if ($connect) {
    echo 'connected';
} else {
    echo 'Failed:'.mysqli_connect_error();
}
mysqli_select_db($connect, 'dbname'); // <-------change this

if ($query = mysqli_query($connect, "SELECT * FROM `users`")) { //<---------------change this
    $num = mysqli_num_rows($query);
    while ($fetch = mysqli_fetch_array($query)) {
        echo $id = $fetch['id'];
    }
} else {
    echo 'error:'.mysqli_error($connect);
}

edited Jul 04 '19 at 18:43

Dharman

30,962
25
85
135

answered Oct 04 '17 at 09:47

B. Desai

16,414
5
26
47

Thank you so much. It fixed my problem. – SimulationCode Oct 04 '17 at 09:54
every time we need to send connection id. if you used $mysqli = new mysqli('localhost','dbu','password','dbname'); as object then does not required to add every time please check my answer as well – Abhijit Oct 04 '17 at 10:10
1

@Abhijit its own choice either use `procedural way ` or `object-oriented way` – B. Desai Oct 04 '17 at 10:11
@B. Desai : object-oriented way is most preferred way. php support oop then why use procedural way – Abhijit Oct 04 '17 at 10:14

Abhijit · Answer 4 · 2017-10-04T10:12:21.177

    ini_set('display_errors', 1);
    ini_set('display_startup_errors', 1);
    error_reporting(E_ALL);
    //Open a new connection to the MySQL server

    $mysqli = new mysqli('localhost','dbu','password','dbname');

    //Output any connection error
    if ($mysqli->connect_error) {
        die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
    }

    //MySqli Select Query
    $results = $mysqli->query("SELECT * FROM users");

    while($row = $results->fetch_assoc()) {
        echo $row["id"];
    }  


    // Frees the memory associated with a result
    $results->free();

    // close connection 
    $mysqli->close();

It's Work for me please check

@SimulationCode : please try this. – Abhijit Oct 04 '17 at 11:06 — Abhijit, Oct 04 '17 at 11:06

score 0 · Answer 5 · answered Oct 04 '17 at 09:58

0

Try using this.

$con = mysqli_connect("localhost","my_user","my_password","my_db");

// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

answered Oct 04 '17 at 09:58

Himanshu Ray

38
1
12

score 0 · Answer 6 · answered Oct 04 '17 at 09:59

Hi Order of your method pparameters are wrong Try this

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

        $connect = mysqli_connect("localhost",'dbu', 'password');
        if($connect){
            echo 'connected';
        }
        else {
            echo 'Failed:'.mysqli_error($connect);
        }
        mysqli_select_db($connect,'dbname');

        if($query = mysqli_query($connect,"SELECT * FROM `users`")){
            $num = mysqli_num_rows($query);
            while($fetch = mysqli_fetch_array($query)){
                echo $id = $fetch['id'];
            }
        }
        else {
            echo 'error:'.mysqli_error($connect);
        }

MySqLi not working for some reason

6 Answers6