PHP PDO resulting in "Array" value rather than actual value

Question

I have created a form which queries a MySQL database and displays results. Now I want to be able to filter my results. I have:

<input type="checkbox" name="multi_cif[]" value="<?php echo $g1 ?>"><font size="1" face="Arial, Helvetica, sans-serif">

In the output which is displaying all results via foreach and the variable $g1 is a MySQL query value (address). I want to be able to click these check boxes next to the results so when the user clicks the button labeled "Filter" only the results checked are displayed.

So far my code is as follows:

<?PHP
if (isset($_POST['Submit2']))
{
$stmt = $dbh->prepare('SELECT * FROM CIF WHERE address LIKE ?');
$stmt->execute(array("%$_POST[multi_cif]%"));
//$stmt->execute(array(":term" => "%" . $_POST["multi_cif"] . "%"));
//$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
/*while ($results = $stmt->fetch())  //WILL UNCOMENT AND ADD OR LIKE AFTER SINGLE QUERY WORKING
{
    echo $results['address'];
    echo $results['alternativeid'];
}*/
print_r($_POST);
$results = $stmt->fetch();
echo $results['address'];
echo $results['alternativeid'];
}
?>

The commented out stuff is other things I have tried. I am very close to my results. The results of the following code ends in:

 [multi_cif] => Array ( [0] => test.13ann.com [1] => testfortestltd444557.com.tw ) ) coralarray.ruhttp://mirror3.malwaredomains.com/files/domains.txt

So clearly "Array" is being passed as a value instead of the desired address assigned to multi[]. Could someone please explain why this is and help me fix it? I am new to PDO as of yesterday but have chosen to use it and re-work my other statements to implement prepared statements instead of dynamically building them. Thanks in advance!

Edited: I took some of Brad's advice but kept the statement without the additional "OR address LIKE ?" as right now I am only clicking one checkbox but still getting "Array" instead of "test.13.ann.com" Once I figure out why "Array" vs. value I will add the additional OR --thanks Brad for pointing out!

verify that it is not in your database that way... this is likely a problem with the code that inserts the record — Orangepill
– Orangepill, Commented Jul 19, 2013 at 14:57
You're passing an array in to a single LIKE comparison? Shouldn't this be ... address LIKE ? OR address LIKE ? then ->execute($_POST['multi_cif'])? (and, of course, add the % to each value within $_POST['multi_cif']) — Brad Christie
– Brad Christie, Commented Jul 19, 2013 at 15:01
Actually, since the input name is multi[] and the php is checking for multi_cif, it's probably checking '%%' — aynber
– aynber, Commented Jul 19, 2013 at 15:07
@aynber That is just a typo, I pasted the wrong section of code. <div align="left"><h2><input type="checkbox" name="multi_cif[]" value="<?php echo $g1 ?>"><font size="1" face="Arial, Helvetica, sans-serif"> is the correct with "multi_cif". I have 3 tables total that are queried and once I figure this out I plan to do this for the other two. — Isedit
– Isedit, Commented Jul 19, 2013 at 15:26

Sean · Accepted Answer · 2013-07-19 18:30:48Z

1

Try accessing the first array value, rather than the outer array -

$stmt->execute(array("%{$_POST['multi_cif'][0]}%"));

to do it dynamically you could try

// create n number of placeholders based off number of $_POST['multi_cif']
$place_holders = implode(' OR', array_fill(0, count($_POST['multi_cif']), ' address LIKE ?'));

// create n number of values based off number of $_POST['multi_cif']
$values = '"%'. implode('%","%', $_POST['multi_cif']).'%"';
// explode the values into an array
$values = explode(',',$values);

$stmt = $dbh->prepare("SELECT * FROM CIF WHERE $place_holders");
$stmt->execute(array($values));

edited Jul 19, 2013 at 18:30

answered Jul 19, 2013 at 18:18

Sean

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4 Comments

Isedit Over a year ago

Thank you very much this worked. I created it dynamically but using:

$cif_num = count($multi_cif);         while ($x < $cif_num)         {                 $stmt = $dbh->prepare('SELECT * FROM CIF WHERE address= ?');                 $stmt->execute(array("%$multi_cif[$x]%"));                  while ($results = $stmt->fetch())                 {                         echo $results['address'];                         echo $results['alternativeid'];                 }                 $x++;          }

Isedit Over a year ago

I am now changing to change the like to an = but for some reason having issues with the results. In MySQL commandline my query SELECT * FROM CIF WHERE address = 'albface.com' works but not in PDO. This is just a side task to help me learn PDO. Anyone have any ideas why = won't work?

Sean Over a year ago

$stmt = $dbh->prepare('SELECT * FROM CIF WHERE address = ?'); $stmt->execute(array($multi_cif[$x])) should work. (remove the "%/%" in your current code)

Isedit Over a year ago

ahhh right you are, again! I forgot to remove "%/%"...it's been a long week of coding. Thanks a bunch!

david strachan · Accepted Answer · 2013-07-19 18:16:19Z

0

You were nearly there. You didn't concatenate your parameter correctly.

Try

<?php
if (isset($_POST['Submit2']))
{
   $term = "%" . $_POST["multi_cif"] . "%";
   $stmt = $dbh->prepare('SELECT * FROM CIF WHERE address LIKE ?');
   $stmt->execute(array($term));
   $rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
   while ($rows = $stmt->fetch())  
   {
       echo $results['address'];
       echo $results['alternativeid'];
   }
}
?>

answered Jul 19, 2013 at 18:16

david strachan

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