CODEWITHSUNDEEP
phpmysqlarrayspdo
Isedit
Isedit

Reputation: 19

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

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!

Upvotes: 0

Views: 273

Answers (2)

Sean
Sean

Reputation: 12433

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));

Upvotes: 1

david strachan
david strachan

Reputation: 7228

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'];
   }
}
?>

Upvotes: 0

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