> What's in your array?
$array1[0] = 'firstname';
$array1[1] = 'lastname';
$array1[2] = 'tkt_title';
I got that fixed.
Thanks.
alex hogan
**
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Alex Hogan wrote:
I gotcha..,
I do have one question though. I ran your code and it only returns the
first letter for that element in the array. For instance fieldone
returns as f and tableone returns as t.
The output looks like this;
SELECT f,f,f FROM t,t.
alex
What's in your array?
--
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On 20 April 2004 16:44, Alex Hogan wrote:
> foreach ( $array1 as $flds ) {
> if ( $start ) {
> $query .= $flds;
> $start = false;
> } else {
> $query .= ", " . $flds;
>
> }
> }
$query .= implode(", ", $array1);
Cheers!
Mike
--
Apr 20 at 9:55am, Alex Hogan wrote:
> function myselect($array1, $array2){
> $query = "SELECT".foreach($array1 as $flds){ $flds[$i]; }.
> "FROM".foreach($array2 as $tbls){ $tbls[$n]; } ...
Alex, if you merely wish to place the values of an array into your
query, you might also try doing it th
[snip]
I do have one question though. I ran your code and it only returns the
first letter for that element in the array. For instance fieldone returns
as f and tableone returns as t. The output looks like this; SELECT f,f,f
FROM t,t.
[/snip]
Never mind I found it. It should be;
$query = "SELEC
[snip]
> $query = "SELECT ";
> $start = true;
> foreach ( $array1 as $flds ) {
> if ( $start ) {
> $query .= $flds[$i];
> $start = false;
> } else {
> $query .= ", " . $flds[$i];
> }
> }
> $query .= " FROM ";
> $start = true;
> foreach ( $
Alex Hogan wrote:
Hi All,
I am having a logic problem. (insert jokes here) I am trying to create a
function that will allow me to pass any number of fields and tables into a
blanket query.
I am getting an error; Unexpected foreach. I have tried placing the loop in
several different places and am
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