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"Undefined variable: payout_item_1" so it's getting the variable name correctly but I must have the format wrong. 

for ($x = 1; $x <= 5; $x++) {
    echo "<input name = 'payout_item_" . $x . "' type = 'text' value = '" . $row[${"payout_item_" . $x}] . "' style = 'width : 150px;' ";
}
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  • please check you have passed payout_item from server side or not.. Commented Apr 6, 2017 at 3:22
  • What does a var_dump($row) show? Commented Apr 6, 2017 at 3:34
  • Can you supply more code in your question to help people help you? Commented Apr 6, 2017 at 6:54

1 Answer 1

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I'm making a couple assumptions

  1. You have a table with columns "payout_item_1" through "payout_item_5"
  2. You do not have variables called $payout_item_1 through $payout_item_5 in which the actual column names are stored.

Currently your code is building variable variables:

This statement builds a variable name with payout_item_1 (in the first iteration). Effectively $payout_item_1.

${"payout_item_" . $x}

The code is then looking for a value in that variable to use as the column header name. Effectively, it's expecting somewhere further up for there to be something akin to

$payout_item_1 = "column1";

Which, as the error suggests, it cannot find. If my assumption in 1. was correct, all you need to do is reformat to

$row["payout_item_" . $x]

and you will be referencing the column payout_item_1 (through 5) from your $row object. Written literally:

$row["payout_item_1"]
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1 Comment

Yup, that did the trick. I had tried many variations of that but must have missed the magic combination. Lack of sleep probably didn't help either. Thanks! You rock!

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