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Caveat: I'm not a PHP expert.
I think it's likely that your like array value is padded right with blanks and you'll have to remove them. Not sure of syntax but maybe DEBAC LIKE %TRIM(?).


On 9/26/2022 11:50 PM, Don Brown via MIDRANGE-L wrote:
I an trying to use a LIKE in my SQL to allow searching for the result.

From Run SQL Scripts this works exactly as expected

select * from mfsdta/mfsdebpf
where debcoy = 1 and debac like 'A%' and debact = 'A'
fetch first 10 rows only;

But from PHP running on the IBMi using the toolkitservice I have

$conn = getToolkitServiceForCurrentUser();
$sql = "select * from "
. " where DEBCOY = ? and DEBAC like ? and debact =
? "
. " order by DEBAC "
. " fetch first 10 rows only";

$values = array();
$values[0] = $company;
$values[1] = $search . "%";
$values[2] = $status;
$stmt = prepareAndExecute($conn, $sql, $values);

The $values array have these values

Array ( [0] => 1 [1] => A% [2] => A )

No errors are received but no records either.

If I change the where to use and DEBAC >= ? then I get records returned
so it has to be the way I am formatting the variable with the % sign.

As a separate question I can't find any log file that may hold the answer
- any suggestions ????



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