mysqli extension error in php script

Question

I'm not good in PHP, I have problem to process with my sign form to using the php script, the server had returned me a message below:

Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in /home/tz005/public_html/test2/email-activation-script.php on line 11

Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/tz005/public_html/test2/email-activation-script.php on line 19
Failed to sent email activation code

Can anyone check what is the problem on my script.Here is my script:

<?php
$username=$_POST['username'];
$password=$_POST['password'];
$email=$_POST['email'];
$postcode=$_POST['postcode'];
$db_host = "server";
$db_name = "table";
$db_use = "user";
$db_pass = "pass";
$link = mysqli_connect($db_host, $db_use, $db_pass);
mysqli_select_db($db_name, $link);
$chars = array("1","2","3","4","5","6","7","8","9");
$length = 5;
$textstr = "";
for ($i=0; $i<$length; $i++) {
   $textstr .= $chars[rand(0, count($chars)-1)];
}
$query = "INSERT INTO email_activation (username, password, email, postcode, activation) VALUES ('$username','$password','$email','$postcode','$textstr')";
$result = mysqli_query($query);
if ($result) {
echo "Sign up successfully,please check your email for activation code.<br>";
$mail_to="$email";
$mail_subject="Email Activation";
$mail_body = "This is the email to activate your account.\n";
$mail_body.="Your activation code is: $textstr \n";
$mail_body.="Click the following link to activate your account.\n";
$mail_body.="<a href='activation-form.php?username=$username&activation=$textstr'>Click here</a>";
$sent = mail($mail_to,$mail_subject,$mail_body, "From: freecycle@greenwich.co.uk");
echo "$mail_to<br><b>$mail_subject</b><br><br>$mail_body";
}else{
echo "Failed to sent email activation code";
}
?>

Rakesh Sharma · Answer 1 · 2014-05-20T12:01:46.453

0

You can pass db name in connection string so need to select db with function try

$link = mysqli_connect($db_host, $db_use, $db_pass, $db_name);
$result = mysqli_query($link, $query);

For more :- http://www.php.net/manual/en/book.mysqli.php

edited May 20 '14 at 12:01

answered May 20 '14 at 11:56

Rakesh Sharma

13,680
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score 0 · Answer 2 · answered May 20 '14 at 11:56

You have the arguments the wrong way around for the mysqli_select_db() call. The first argument should be the link, the second being the string database name. So change to:

mysqli_select_db($link, $db_name);

Secondly, In the query call, you need to pass the link, not just the query:

$result = mysqli_query($link, $query);

References:

score 0 · Answer 3 · answered May 20 '14 at 11:57

0

Mysqli doesn't work the way the Mysql functions work, You need to define the connection in a variable to make use of the function.

However you could just change your functions to mysql (if installed) and it'll work.

Basicly:

mysqli_select_db

Becomes

mysql_select_db

Etcetera, this will probably be fine for your project. If you get more at ease with PHP look at the classes to handle your Database, I guess you'll find those yourself.

answered May 20 '14 at 11:57

Mathijs Segers

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This isn't good advice, see this: http://stackoverflow.com/questions/8891443/when-should-i-use-mysqli-instead-of-mysql and http://stackoverflow.com/questions/548986/mysql-vs-mysqli-in-php – Joe May 20 '14 at 11:59
True I guess. He is using the mysqli functions as mysql functions though. – Mathijs Segers May 20 '14 at 12:01

score 0 · Accepted Answer · answered May 20 '14 at 12:05

0

change

mysqli_select_db($db_name, $link);

to

mysqli_select_db($link, $db_name);

also change

$result = mysqli_query($query);

to

$result = mysqli_query($link, $query);

answered May 20 '14 at 12:05

fortune

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mysqli extension error in php script

4 Answers4