Error in inserting a row to an mysql database table using php

Question

I try to insert a row to my DB table as follows:

$selected = mysqli_select_db($link, $db_name);
if(!$selected){
    $mail_Subject = 'Error in selecting DB: ' . mysqli_error($link);        
    mail($mail_To, $mail_Subject, $mail_Body, $mail_From);
    mysqli_close($link);
    exit;
}

// Add the line to the database!
if(!mysqli_query("INSERT INTO tokens (user_id, user_name, transaction_id, token_type, token_meta, date) VALUES ('-3', 'nisui2', 'blabla2', '111', '111 meta', '');")){
    $mail_Subject = 'Error in insertion: ' . mysqli_error($link);       
    mail($mail_To, $mail_Subject, $mail_Body, $mail_From);
    mysqli_close($link);
    exit;
}

From some reason I get an email with the subject "Error in insertion:", with no error details.

Can you spot what am I doing wrong?

Also, can you tell me a way to debug this? (how can I know what the problem was?)

You're using mysqli - but there's only one argument !?!? And you're mixing mysql_ and mysqli - that's not going to work — Strawberry
– Strawberry, Commented Jan 12, 2014 at 1:52
What happens when you enter that query directly into the mysql console? — Markus Malkusch
– Markus Malkusch, Commented Jan 12, 2014 at 4:36

lhadameck · Accepted Answer · 2014-01-12 01:52:52Z

2

You'r selecting the Database with "mysql_select_db" but you'r querying with msqli

try

   $selected = mysqli_select_db($db_name, $link);

answered Jan 12, 2014 at 1:52

lhadameck

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3 Comments

user1028741 Over a year ago

I changed it, but still I get the Error in insertion. Do you know how can I know what is the concrete error?

lhadameck Over a year ago

try "mysqli_error($link)", did you made the database connection with $link = mysqli_connect ?

user1028741 Over a year ago

yes, $link is the connection to db. I did change it, still doesn't work... (and doesn't give error info)

Markus Malkusch · Accepted Answer · 2014-01-12 04:48:42Z

1

Add $link as first parameter to your call:

mysqli_query($link, $query [, $resultmode])

mysqli_query($link, "INSERT …");

edited Jan 12, 2014 at 4:48

answered Jan 12, 2014 at 4:35

Markus Malkusch

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mysqli_query($link, $query [, $resultmode])

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