Reply to Re: [PHP] Array Select from database

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Posted by "A.J. Brown" on 09/28/05 17:27

There is a small bug in your code:

[snip]
$values = join(', ', $array);
$query = "SELECT * FROM client WHERE clientaccountmanager IN ($values)"
[/snip]

You'll need to surround "$values" with a single quote after joining:

[code]
$values = join(', ', $array);
$query = "SELECT * FROM client WHERE clientaccountmanager IN ('$values')"
[/code]

--

Sincerely,

A.J. Brown


"Robin Vickery" <robinv@gmail.com> wrote in message
news:5cb2dafd0509280555e650cc3@mail.gmail.com...
On 9/28/05, Frank Keessen <fkeessen@gmail.com> wrote:
> O.K. Again;
> I have an array with one or more values, which must be selected in the
> database
> Array ( [count] => 1 [0] => Array ( [clientaccountmanager] => Array (
> [count] => 2 [0] => 210 [1] => 149 )
> Now i this is my select..
> I've got the following Query="select * from client WHERE
> clientaccountmanager='$value of array1' OR '$2nd value of array 1'"
>
> How can i loop through the query with all the values out of the array as
> OR
> value..

Are you trying to generate the query from the array?

If so, you are better off using the SQL 'IN' construct.

SELECT * FROM client WHERE clientaccountmanager IN (value1, value2,
value3,...)

Then all you need to do is generate a comma separated list of values
from the array, which you can easily do with join().

$values = join(', ', $array);

Then insert it in your query:

$query = "SELECT * FROM client WHERE clientaccountmanager IN ($values)"

-robin

[Back to original message]


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