PHP Code not working (Insert to DB)

Question

I have a db connection and want to test inserting a variable value into a db field. It's not inserting anything but cannot see why...can anyone spot the problem please?

Updated - here's all the php code:

    //DB Conn

<?php
    $link = mysql_connect('localhost', 'root', '');
    if (!$link) {
    die('Could not connect: ' . mysql_error());
}
    mysql_select_db('mydb',$link);
?>

    ... then
    //Save.php

    <?php

    include('dbconn.php');

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

        $mytestvalue = 'hello';

        $query="insert into mytable (id, name) values ('null','".$mytestvalue ."')";

    ?>

Any ideas?

@DanGrossman means it should look like this: $query="insert into mytable (id, name) values (NULL,'".$mytestvalue ."')";. However if ID is a auto_increment primary key then there is no need to even pass in NULL you can just leave ID out of the insert query for the same effect. — Treffynnon
– Treffynnon, Commented Jul 30, 2011 at 9:01
You should do corrections in code you actually run, but It is better to leave the code in your question intact. People have pointed you miss a call to mysql_query($query, $link); and you put it in (and in the wrong place), now the answer looks irrelevant. — Majid Fouladpour
– Majid Fouladpour, Commented Jul 30, 2011 at 10:10

Treffynnon · Accepted Answer · 2011-07-30 08:54:09Z

2

You are missing the actual call to mysql_query() in your code that actually executes the SQL you have created.

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

Add the line above to the end of the code sample in your question.

Please see this example on the PHP manual for a fully worked example of an interaction with MySQL from PHP.

answered Jul 30, 2011 at 8:54

Treffynnon

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Comments

Nemanja · Accepted Answer · 2011-07-30 09:51:42Z

1

Mysql_query function should be after you define the actual query string.

So, you first define the $query and THEN execute the mysql_query function call.

Cheers!

answered Jul 30, 2011 at 9:51

Nemanja

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