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Add random password into my SQL database

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Skippie1

Technical User
May 7, 2012
29
0
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ZA
Good day,

I am new to PHP and I am trying to add a random password for my clients when they are verified. The script I am using is as follows:

<?php
function genpwd($cnt)
{
$pwd = str_shuffle('abcefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890@#%$*');

return substr($pwd,0,$cnt);
$password=genpwd(6);
}
echo genpwd(6);

$conn = mysql_connect('localhost','database','password') or die('Iam dying');
mysql_select_db('database_table',$conn);
$string = "insert into `Table` (`password`) values ('$password')";
mysql_query($string,$conn);
mysql_close();
?>

What is happening is that a database entry is created but there is nothing in the password field. Any suggestions??
 
Hi

[tt]function genpwd($cnt)
{
$pwd = str_shuffle('abcefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890@#%$*');

return substr($pwd,0,$cnt);
$password=genpwd(6);
}[/tt]
[ul]
[li]There is no life after an unconditional [tt]return[/tt] - the assignment to $password is never executed. ( Fortunately. Otherwise the interpreter would run into stack overflow error due tu the infinite recursive call.)[/li]
[li]Assigning to the local variable $password inside a function will not be visible to the code outside the function.[/li]
[/ul]


Feherke.
feherke.github.io
 
Sorry, like I said I am new to PHP. How do I fix it so that I can add the random generated password to my database?
 
Also, you probably want to echo the generated $password instead of calling the function again, which would echo another (different) shuffled/random password.
Code:
$password=genpwd(6);
echo $password;
 
Thanks so much, it is working now. Such a small thing....
 
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