MYSQL Insert the same value again and again

Question

I'm having problem with my query, it inserts the same the value again and again even though the value of the variable is different.

Here is the first page. it contains a randomizer for the creation of accounts.

 function getRandomString($length = 10) {
        $validCharacters = "1234567890";
        $validCharNumber = strlen($validCharacters);

        $result = "";

        for ($i = 0; $i < $length; $i++) {
            $index = mt_rand(0, $validCharNumber - 1);
            $result .= $validCharacters[$index];
        }

        return $result;
    }

    $idNo=getRandomString();


    function getRandomPassword($length = 6) {
        $validCharacters = "qwertyuiopasdfghjklzxcvbnm1234567890";
        $validCharNumber = strlen($validCharacters);

        $result = "";

        for ($i = 0; $i < $length; $i++) {
            $index = mt_rand(0, $validCharNumber - 1);
            $result .= $validCharacters[$index];
        }

        return $result;
    }

    $password=getRandomPassword();

The values will then be submitted to the next page which is below. The submitted values are correct because i print them out first to check. It always inserts the value "2147483647" for the ID_No even though this is not the randomize value.

   $insert = "INSERT INTO person (ID_No, P_Word, E_Status)
        VALUES('$userName', '$userPass', 'Applicant')";

  $result = mysql_query($insert);

How do you fix this? Is it a bug?

It shouldn't do, but there isn't enough code in the question to reproduce the problem. — Quentin, Jan 25 '15 at 11:01
**Warning**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). — Quentin, Jan 25 '15 at 11:01
post the code where you are assigning new values to the variables. — Junius L, Jan 25 '15 at 11:02
hmm... saw you additional code. You might need to show, where the values $username are getting picked up, such as POST array. May be there is a line after it where a constant gets assigned or something. — Whirl Mind, Jan 25 '15 at 11:12

score 0 · Answer 1 · answered Jan 25 '15 at 11:47

We can only guess given the code posted.

I don't see where $index ends up being $userName.
The sql posted shows userName being inserted as a character string, quotes around it.

ID_No is likely an integer type and thus the character string fills it up.

This should just blow up on insert so you know that this is wrong. Best practice is to use sql_mode = 'STRICT_ALL_TABLES' in the MySQL configuration file.

score 0 · Answer 2 · answered Jan 25 '15 at 11:49

0

I think you should check data type for id_no column..You must give double data type to that column

answered Jan 25 '15 at 11:49

Kedar B

774
5
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47

score -2 · Answer 3 · answered Jan 25 '15 at 11:05

-2

I think what you have in mind is this : You must use the string concatenation operator (here, in PHP, it is the DOT) if you are thinking of $ symbols as variable names.

$insert = "INSERT INTO person (ID_No, P_Word, E_Status) " . 
        " VALUES( " . 
         "'" . $userName . "'" . " , " . 
         "'" . $userPass. "'" . " , " . 
         "'" . "Applicant. "'" . " ) " ;

$result = mysql_query($insert);

answered Jan 25 '15 at 11:05

Whirl Mind

884
1
9
18

1

He can use variable inside a string without concatenation if he encloses them with double quotes, which he does. – Ali Jan 25 '15 at 11:07
1

He's using double quotes there's no need for concatenation. – Junius L Jan 25 '15 at 11:08

MYSQL Insert the same value again and again

3 Answers3