Why is this an invalid argument?
foreach(($row['inType']) as $inType){
echo $inType,'br';}
I am trying to output results from a data base that may have multiple
results for the same name
So trying to use an array and foreach that is the right track ...right?
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On Thu, 2009-07-16 at 15:41 -0400, Miller, Terion wrote:
Why is this an invalid argument?
foreach(($row['inType']) as $inType){
echo $inType,'br';}
I am trying to output results from a data base that may have multiple
results for the same name
So trying to use an array and
Miller, Terion wrote:
Why is this an invalid argument?
foreach(($row['inType']) as $inType){
echo $inType,'br';}
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?
Actually this ended up doing what I needed:
$result =
mysql_query($sql) or die(mysql_error());
$header = false;
foreach iterates over an array or a object (see Traversable ).
If you pass anything different he complains
?php
if( !is_scalar( $collection ) )
foreach( $collection as $element )
print_r( $element );
On Thu, Jul 16, 2009 at 4:53 PM, Kyle Smith kyle.sm...@inforonics.comwrote:
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