Posted by malcolm on 02/28/06 23:53

On Mon, 27 Feb 2006 19:49:28 -0500, J.O. Aho <user@example.net> wrote:

> malcolm wrote:
>> On Mon, 27 Feb 2006 17:29:51 -0500, J.O. Aho <user@example.net> wrote:
>>
>>> malcolm wrote:
>>>> Hello All,
>>>> On my rented space I have only one database. I have a few users
>>>> at my site
>>>> and any mysql tables I made for them have their initials as a
>>>> prefix i.e.
>>>> aa_phonelist, aa_addresses, aa_otherjunk
>>>> bb_phonelist, bb_addresses, bb_otherjunk
>>>> I want to make a drop down selection for the tables so that each can
>>>> see his/her tables
>>>> but ONLY theirs. What I really want is a query like 'Show tables
>>>> where prefix like aa_'
>>>> and send the results to the drop down. Obviously that won't work
>>>> but I have wracked my
>>>> small brain for quite some time now and can't seem to find something
>>>> that will work.
>>>> I have tried to use the show table results with things like in_array
>>>> but I just can't
>>>> make it work. This is on a logged-in page so I know the prefix but I
>>>> can't seem to limit
>>>> show tables at all and I can't find a way to strip out the other
>>>> prefixes from the resulting
>>>> array. Not zillions but quite a few tables and new ones all the time,
>>>> so this needs to be
>>>> dynamic.
>>>
>>> Look at the mysql_list_tables() function, if you aren't using mysql on
>>> the host, then it will be more tricky. If you use the function, then
>>> you will need to filter the result when you got it to PHP, while the
>>> alternative options could work with WHERE statements.
>>>
>>> http://www.php.net/manual/en/function.mysql-list-tables.php
>>>
>>>
>>> //Aho
>> Thanks Aho but I'm able to list the tables ok and do the drop down but
>> I can't figure how to filter the array returned by the 'show tables'
>> query.
>> That function does the same but I don't see any way there to filter the
>> result. I'm looking at array_clean now, thanks for reply.
>
> If you save the results from the "query" into an array, you can use
> array_filter()
>
> http://www.php.net/manual/en/function.array-filter.php
>
> and a function that returns true if the initials of a string is the same
> as for the user, something like this may work:
>
> //we assume that $initials contains the initials of the user
>
> function is_user($var) {
> global $initials;
> if(substr($var, 0, 2)==$initials) {
> return true;
> }
> return false;
> }
>
> //$table_array is a array with all the tables
> $user_tables=array_filter($table_array, "is_user"));
>
> //$user_tables should only contain tables with the same initials as the
> users
>
>
> or you can do a loop where you check which tables are okey to list and
> at the same time type out the options.
>
> for($i=0;$i<count($table_array);$i++) {
> if(substr($table_array[$i], 0, 2)==$initials) {
> echo "<option>".$table_array[$i]."</option>\n";
> }
> }
>
>
>
> //Aho


Thank you again Aho but I can't seem to get anything out of your code.
I have tried all the things I can think of but I get no output. No errors
and if I print_r the table_array I can see the array but if I run this code
no matter the initials I use, I get no output. Still working, thanks
again.

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