Invalid Argument why?

Why is this an invalid argument?

foreach(($row['inType']) as $inType){

echo $inType,'
';}

I am trying to output results from a data base that may have multiple
results for the same name....

So trying to use an array and foreach that is the right track ...right?


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Re: Invalid Argument why?

On Thu, 2009-07-16 at 15:41 -0400, Miller, Terion wrote:
> Why is this an invalid argument?
>
> foreach(($row['inType']) as $inType){
>
> echo $inType,'
';}
>
> I am trying to output results from a data base that may have multiple
> results for the same name....
>
> So trying to use an array and foreach that is the right track ...right?
>
>
I imagine $row is the array, and ['inType'] is an element of the array.
This is not how you use a foreach. Can you show where you are getting
$row from?

Thanks
Ash
www.ashleysheridan.co.uk


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Re: Invalid Argument why?

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Miller, Terion wrote:
> Why is this an invalid argument?
>
> foreach(($row['inType']) as $inType){
>
> echo $inType,'
';}
>
> I am trying to output results from a data base that may have multiple
> results for the same name....
>
> So trying to use an array and foreach that is the right track ...right?
>
>
>
Looks like you meant to do something like this:

// Always better to be plural when you have an array.
$rows = whatever_your_rows_come_from();

foreach($rows as $row)
{
$inType = $row['inType'];
echo $inType . '
';
}


HTH,
Kyle

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Re: Invalid Argument why? (RESOLVED)

Actually this ended up doing what I needed:

$result =3D mysq=
l_query($sql) or die(mysql_error()); =
$header =3D false; =
while($row =3D mysql_fetch_assoc($result)){ =
if(!$header){ =
echo ($row['=
name']),'
'; =
echo ($row['address']),'
'; =
$header =3D true; =
} =
echo ($row['inDate']),'
'; =
echo ($row['inType']),'
';=
echo ($row=
['notes']),'
'; =
echo ($row['critical']),'
' ; =
echo ($row['cviolations']),'
'; =
} =
}


On 7/16/09 2:53 PM, "Kyle Smith" wrote:

Miller, Terion wrote:

Why is this an invalid argument?

foreach(($row['inType']) as $inType){

echo $inType,'
';}

I am trying to output results from a data base that may have multiple
results for the same name....

So trying to use an array and foreach that is the right track ...right?



Looks like you meant to do something like this:

// Always better to be plural when you have an array.
$rows =3D whatever_your_rows_come_from();

foreach($rows as $row)
{
$inType =3D $row['inType'];
echo $inType . '
';
}


HTH,
Kyle



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Re: Invalid Argument why?

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Content-Transfer-Encoding: 7bit

foreach iterates over an array or a object (see Traversable ).
If you pass anything different he complains


if( !is_scalar( $collection ) )
foreach( $collection as $element )
print_r( $element );

On Thu, Jul 16, 2009 at 4:53 PM, Kyle Smith wrote:

> Miller, Terion wrote:
>
>> Why is this an invalid argument?
>>
>> foreach(($row['inType']) as $inType){
>>
>> echo $inType,'
';}
>>
>> I am trying to output results from a data base that may have multiple
>> results for the same name....
>>
>> So trying to use an array and foreach that is the right track ...right?
>>
>>
>>
>>
> Looks like you meant to do something like this:
>
> // Always better to be plural when you have an array.
> $rows = whatever_your_rows_come_from();
>
> foreach($rows as $row)
> {
> $inType = $row['inType'];
> echo $inType . '
';
> }
>
>
> HTH,
> Kyle
>



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Martin Scotta

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