Help to fix the error in the request:

Warning: mysqli_error () expects exactly 1 parameter, 0 given in D: \ OpenServer \ OSPanel \ domains \ train2 \ second.php on line 29
ERROR QUERRY

index.php

<form action="second.php" method="POST"> <fieldset> <label for="first_name">first name</label> <input type="text" name="first_name"> <br> <label for="first_name">last_name</label> <input type="text" name="last_name"> <br> <label for="first_name">bio</label> <input type="text" name="bio"> <br> <label for="first_name">email</label> <input type="text" name="email"> <br> <label for="first_name">facebook_url</label> <input type="text" name="facebook_url"> <br> <label for="first_name">twitter_url</label> <input type="text" name="twitter_url"> <br> </fieldset> <fieldset> <input type="submit" name="ok"> </fieldset> </form>

second.php

 <?php require "db_connection.php"; $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; $bio = $_POST['bio']; $email = $_POST['email']; $facebook_url = $_POST['facebook_url']; $twitter_url = $_POST['twitter_url']; $show_tb = mysqli_query($connection, "SHOW TABLES"); $sel_tb = mysqli_query($connection, "USE TABLES my"); if ($sel_tb) { mysqli_error(); }else{ echo "<br> okkkkkkkkkkk <br>"; } $qq = mysqli_query($connection,"SELECT * FROM my"); if (!$qq) { echo "noooooo" . mysqli_query(); }else{ echo "select good"; } $query = mysqli_query($connection, "INSERT INTO my(first_name, last_name, bio, email, facebook_url, twitter_url) VALUES ($first_name, $last_name, $bio, $email, $facebook_url, $twitter_url)"); if (!$query) { die("ERROR QUERRY" . mysqli_error()); } while ($row = mysqli_fetch_row($query)) { echo "You'e use {$row[0]}"; } mysqli_close($connection); ?>

There is a connection to the database. Only I can not make a request to the database by entering information in the input.

  • Use prepared statements and there will be no such problems - ArchDemon

1 answer 1

Warning: mysqli_error () expects exactly 1 parameter, 0 given

There is a missing parameter in the function call. You did not specify a connection identifier

Procedural style

string mysqli_error ( mysqli $link )

Returns the error message of the last call to the MySQLi function, which may succeed or fail.

List of parameters:

link Procedural style only: Connection ID obtained using mysqli_connect () or mysqli_init ()

If you use a procedural style, then you need to specify

 mysqli_error($connection);

If object-oriented, then:

 $connection->mysqli_error();

By the way

 echo "noooooo" . mysqli_query();

will cause the same error