in_array() expects parameter 2 to be array, null given error

Question

I have a form that displays a checkbox for each engineer in a table. When the form is submitted if the engineer's checkbox isn't ticked they are removed from the job ($diary_id).

The code as it is at the moment works but a page briefly displays this error before going back to the next page.

The code on the form...

  $sql = "SELECT * FROM users WHERE type = 'engineer' ORDER BY first_name";
  $result = mysql_query($sql) or die(mysql_error());
  while ($row = mysql_fetch_array($result)){ ?>

      <div style="width:70px; float:left">
      <input type="hidden" name="engineer_on_page[]" value="<? echo $row['id'];  ?>"/>
          <input type="checkbox" name="engineer[]" <?
          $eng_id= $row['id'];

              $result2 = mysql_query("SELECT * FROM calls_engineers WHERE call_ce = '$diary_id' AND engineer_ce = '$eng_id'");
              if((mysql_num_rows($result2)) > 0)
              { echo 'checked="checked"';
             ?> value="<? echo $row['id']; ?>" />
          <? echo '   '.$row['first_name']; }
             else {
                 echo "value=".$row['id']." />";
          echo '   '.$row['first_name'];
             } 
              ?>
      </div>

  <? } ?>

The mysql to remove the unselected engineers

foreach($_POST['engineer_on_page'] as $engineer_id) {
            $sql = "SELECT * FROM calls_engineers WHERE call_ce = '$diary_id' AND engineer_ce = '$engineer_id'";
            $result = mysql_query($sql)or die(mysql_error());
                if(!in_array($engineer_id, $_POST['engineer'])){
                    if(mysql_num_rows($result) > 0){
                        $sql = "DELETE FROM calls_engineers WHERE engineer_ce = '$engineer_id' AND call_ce = '$diary_id'";
                        $result = mysql_query($sql)or die(mysql_error());
                    }           
                }

            }

I have tried to initialise $_POST['engineer'] as an array ($_POST['engineer'] = array();) before the foreach loop which stops the error, but it also stops the page working, no error messages, it just doesn't work any more.

I am aware that the page needs to be updated to mysqli but I would like to resolve this issue before I do.

Thanks

replace if(!in_array($engineer_id, $_POST['engineer'])){ with if(isset($_POST['engineer']) && is_array($_POST['engineer']) && !in_array($engineer_id, $_POST['engineer'])){ — Ejaz
– Ejaz, Commented May 2, 2013 at 20:54
stops the page from working you mean you get some error? or it doesn't execute the delete statement? — Ejaz
– Ejaz, Commented May 2, 2013 at 21:13
you should post your complete generated HTML for the <form> that's making POST request to the script. — Ejaz
– Ejaz, Commented May 2, 2013 at 21:35

jon__o · Accepted Answer · 2013-05-02 21:00:54Z

2

If only 1 engineer is checked, the value might be a string, not an array.

Then, I would add an is_array() before calling the in_array():

if(is_array($_POST['engineer']) && !in_array($engineer_id, $_POST['engineer'])){
    // code          
}

There are actually a lot of ways that you could structure that, but you should get the idea.

answered May 2, 2013 at 21:00

jon__o

1,51913 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

tatty27 Over a year ago

Hi, If I add that it stops the page from working, the checked engineer does not get removed from the job

jon__o Over a year ago

Before your foreach, try adding this: $engineer = is_array($_POST['engineer']) ? $_POST['engineer'] : array($_POST['engineer']); Then, in your original code, change the $_POST['engineer'] on that 4th line to the variable $engineer. Or else, check your logic and SQL results.

Collectives™ on Stack Overflow

in_array() expects parameter 2 to be array, null given error

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related