PHP Insert Record into MySQL DB (Multiple recorded are being added)

Question

I am having problems with this code:

<?php
$new_value = 'testing';

$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

mysql_query("INSERT INTO myDB (myField) VALUES ('$new_value')");

mysql_close($con);

?>

2 issues are happening:

1st - 2 records are being inserted instead of 1

2nd - $new_value is not changed and it's creating multiple instances with the same $new_value value when I only want 1.

Where in your code is $new_value supposed to change? Or do you mean that you insert the same value into the DB over and over again? That is correct because I don't see you specifying anything about unique values or nothing. — Bono
– Bono, Commented Mar 30, 2012 at 10:37
Correct I inserted the same value into the DB over and over again and only want one — Satch3000
– Satch3000, Commented Mar 30, 2012 at 10:38
Are you sure you are not executing this code segment twice(e.g. include/require)? It seems impossible for two records to be inserted instead of single one.... — Jovan Perovic
– Jovan Perovic, Commented Mar 30, 2012 at 10:39
Is this the whole code? or maybe just a snippet and this code is somewhere in a loop? — ab_dev86
– ab_dev86, Commented Mar 30, 2012 at 10:40
I'm sure....because I just commented the code above and nothing was inserted into the db — Satch3000
– Satch3000, Commented Mar 30, 2012 at 10:44

Bono · Accepted Answer · 2012-03-30 10:48:21Z

3

I think this is something similar to what you are after, after reading your comment:

INSERT IGNORE INTO tablename (fields)
VALUES(values);

Use INSERT IGNORE rather than INSERT. If a record doesn't duplicate an existing record, MySQL inserts it as usual. If the record is a duplicate, the IGNORE keyword tells MySQL to discard it silently without generating an error.

edit Or as Vytautas points out; if you want to replace the value use REPLACE in stead of INSERT

edited Mar 30, 2012 at 10:48

answered Mar 30, 2012 at 10:41

Bono

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1 Comment

Satch3000 Over a year ago

INSERT IGNORE is still adding it twice

AKOP · Accepted Answer · 2012-03-30 13:34:46Z

1

Why dont you try using transactions? It helps you to avoid some of insert/update errors:

 try
    {

        // Start TRANSACTION
        mysqli_autocommit($con, false);

        mysqli_query('INSERT INTO table SET field = "'.$fieldValue.'"');

        // If there is no error then COMMIT
        mysqli_commit($con);      

    }
    catch(Exception $e)
    {

        // If there is an error, then rollback any INSERTS or UPDATES in the TRY block
        mysqli_rollback($con);

        $theError = $e->getMessage();
    }

Its highly recommended that you use Try/Catch blocks in your code when working with databases.

answered Mar 30, 2012 at 13:34

AKOP

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Comments

Web Developer · Accepted Answer · 2012-03-30 10:52:40Z

0

2nd - $new_value is not changed and it's creating multiple instances with the same $new_value value when I only want 1.

Where is $new_value changed ? How is it changed ? Are you sure you are not pressing the browser refresh button many times ?

answered Mar 30, 2012 at 10:52

Web Developer

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1 Comment

Satch3000 Over a year ago

There will only be 1 instance of $new_value so, say $new_value = 'Value1'; ... The code is adding itself twice in the db table, I want to avoid duplicates

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