Validate numbers before inserting into MySQL

[thaidomizil]

Hello,

 

i'm trying to server-side validate input from POST to insert it into mysql if the requirements for the insert are met.

 

i was using jquery before but this isn't reliable at all, i want the data to get inserted into mysql only if the length is 19 chars and only if it starts with 3227580 the next 12 digits can be any number.

 

So basically: IF chars=19 and data=3227580* insert into mysql -- else show error

 

Thank you for help.

[Buddski]

Try this:

if (strlen($_POST['your_field']) != 19 && substr($_POST['your_field'],0,7) != '3227580') { 
    // Throw Error // 
} else { 
    // Success// 
}

[thaidomizil]

Works, thanks !

[thaidomizil]

I don't want to open a new thread for this, your code works fine, i'm trying to put

if(isset($_POST['submit'])) 

infront, my final code looks like this:

 

if(isset($_POST['submit'])) {


if (strlen($_POST['test']) != 19 && substr($_POST['test'],0,7) != '1234567') { 
    echo "wrong code"; 
} else { 
    mysql_query("INSERT INTO table(code) VALUES('$_POST[test]')");
Print "Your table has been populated"; 
}}

 

now the script doesn't do anything, i guess i set the curly braces wrong or can i not put 2x if behind each other ? can you enlighten me on this please

[xyph]

Do you have a button/input named 'submit' ? Some clients won't send the submit button if you use {enter} to submit a form, from what I've read.

 

Don't forget to make sure the rest of the string is numeric as well. ctype_digit will help with that.

[thaidomizil]

Thanks xyph, i just noticed my button name wasn't set to submit only the id.

 

Also, i just saw that this isn't working:

if (strlen($_POST['test']) != 19

This is working fine though

&& substr($_POST['test'],0,7) != '1234567')

any idea?

[xyph]

<?php 

$str1 = '1234567890123456789';
$str2 = '12345678';

if( strlen($str1) != 19 )
echo '$str1 is not 19 numbers';
else
echo '$str1 is 19 numbers';

echo '<br>';

if( strlen($str2) != 19 )
echo '$str2 is not 19 numbers';
else
echo '$str2 is 19 numbers';

?>

 

Not sure why yours isn't working.