How to remove "warning Invalid argument supplied for foreach()" in php?

Question

I want to get distinct values from database in php but i got this warning

Invalid argument supplied for foreach()

How i remove this warning? here is my code:

$startdate1=date('Y-m-d h:i:s',strtotime($_POST['registration_opens_date']));
$enddate1=date('Y-m-d h:i:s',strtotime($_POST['registration_ends_date']));
$startdate1  = strtotime($startdate1 ); // Convert date to a UNIX timestamp  
$enddate1  = strtotime($enddate1 ); // Convert date to a UNIX timestamp  
$dates=array();
// Loop from the start date to end date and output all dates in between  
for ($i = $startdate1; $i <= $enddate1 ; $i += 86400) {
    $dates[$i]=date("m-d-Y", $i);
}
$strQuery="select Distinct DATE_FORMAT(transactions.transaction_date,'%c-%d-%Y') as transaction_date,sum(amount)as Amount from transactions where transaction_date BETWEEN '".$startdate1."' AND '".$enddate1."' group by  DATE_FORMAT(transactions.transaction_date,'%c-%d-%Y')";
$result = $GLOBALS ['mysqli']->query ($strQuery) or die ($GLOBALS ['mysqli']->error . __LINE__);
while($rs=$result->fetch_assoc ()){
    $res[]=$rs;
}
$strXML = "<chart caption='Reports of transactions' showValues='0' xAxisName='Date' yAxisName='Amount' useRoundEdges='1' palette='3'>";
for ($i = $startdate1; $i <= $enddate1 ; $i += 86400) {
    foreach($res as $r){
        if($r['transaction_date']==$dates[$i]){
            $substrXML  .= "<set label='" .$r['transaction_date'] ."' value='" .$r['Amount'] ."' />";
            break;
        }
        else {
            $substrXML = "<set label='".$dates[$i]."' value='0'/>";
        }
    }
    $strXML .=$substrXML;
}

The error means what you pass to `foreach` is **not** an array. In this case, when the loop iterates for the first time, the array isn't defined yet, and that error message will be thrown. First of all, initialize an empty array using `$res = array();` before the loop. To make sure the argument is ***always*** an array, you can use [`is_array()`](http://php.net/is_array). — Amal Murali, Feb 01 '14 at 15:52

score 1 · Accepted Answer · answered Feb 01 '14 at 15:50

1

Immediately before the foreach add this line:

if(!isset($res) && ! is_array($res)) continue;

answered Feb 01 '14 at 15:50

krowe

2,129
17
19

score 0 · Answer 2 · answered Feb 01 '14 at 15:50

0

You are not initializing your variable and it's empty.

Add this before your while loop:

$res = array();

answered Feb 01 '14 at 15:50

Hardy

5,590
2
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27

score 0 · Answer 3 · answered Feb 01 '14 at 15:50

$res is never defined. It is only ever assigned values inside your while() loop from your SQL query. In case the query returns a blank set of data, $res will be undefined and thus PHP will throw an error.

Try defining $res as a blank array at the start of your code:

$res = array();

pirs · Answer 4 · 2017-04-25T09:48:54.447

0

Ultimate foreach trick

if(!empty($rows)) foreach( is_array($rows) ? $rows : array($rows) as $row) {
    // no error possible
}

edited Apr 25 '17 at 09:48

answered Feb 01 '14 at 16:41

pirs

2,410
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How to remove "warning Invalid argument supplied for foreach()" in php?

4 Answers4