PHP for each loop error

Question

I have a loop to go through an array and each checkbox that is selected I want to put the value (also brought in dynamically) into my QuestionSelected table. I get this error "Warning: Invalid argument supplied for foreach()" and I cannot get the results into my table. Here is the code I am trying:

//Declare the QuestionID as a array
$QuestionID = array();


while($row = mysqli_fetch_array($run,MYSQLI_ASSOC)){
echo '<div id="QuestionSelection"><input id="chkQuestion" type="checkbox" value=" '.$row['QuestionID'].'" name="question_'.$row['QuestionID'].'">' .$row['Question']. '</p></div><br/><br/>';

//Assign the QuestionID from the table to the var
$QuestionID[] = $row['QuestionID'];

}


if($_POST['submitted']) { 


$ids_list = '';

foreach($_POST["QuestionID"] as $id)
{
$ids_list .= (strlen($ids_list) > 0 ? ',' : '').mysql_real_escape_string($id);
}

$sql2 = "INSERT INTO tbl_QuestionSelected (`QuestionID`) VALUES (".$ids_list.")";

}//End of IF 'submitted

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — user895378
– user895378, Commented Jul 12, 2012 at 16:05
If I do a print_r(QuestionID) it returns 3 results so it looks as though its populating correctly? Wouldn't this be the right way to see? — Keith Montgomery
– Keith Montgomery, Commented Jul 12, 2012 at 16:06
do print_r($_POST['QuestionID']); as this is what you are foreach-ing — Waygood
– Waygood, Commented Jul 12, 2012 at 16:08

jeroen · Accepted Answer · 2012-07-12 16:06:22Z

1

The problem is that in your form, you are not using an array, but building the name values like:

name="question_'.$row['QuestionID'].'"

So $_POST["QuestionID"] does not exist and is not an array.

You can change your name building to:

name="QuestionID[]"

So that it is an array and you can even use the actual id as a key:

name="QuestionID[' . $row['QuestionID'] . ']"

answered Jul 12, 2012 at 16:06

jeroen

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2 Comments

Keith Montgomery Over a year ago

I was under the assumption that my attempt is building name values and value values both (as ridiculous as that sounds) I wanted each checkbox to have a new name and then the value be my questionID from my table

Keith Montgomery Over a year ago

I just did print_r($QuestionID); and got the result: Array ( [0] => 1 [1] => 2 [2] => 3 ) before I try this foreach loop. So my array values are in $QuestionID based on my understanding. So my issue is with the foreach and something I have done there. I just can't figure out what. Sorry if I'm missing something in your explaination

Keith Montgomery · Accepted Answer · 2012-07-12 18:11:44Z

0

This resolves the issue and works:

$ids_list = '';

foreach($_POST["QuestionID"] as $key=>$value) {
{
$ids_list .= (strlen($ids_list) > 0 ? ',' : '').mysql_real_escape_string($value);
}

$sql2 = "INSERT INTO tbl_QuestionSelected (`QuestionID`) VALUES (".$ids_list.")";

//Run the query 
$run2 = @mysqli_query ($conn,$sql2);


}

answered Jul 12, 2012 at 18:11

Keith Montgomery

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