PHP Array to string conversion in multiple selected dropdown

Question

I have this function for show selected value in multiple dropdown:

function _is_dropdown_( $name,array $selected=null,$table ,$class,$required,$type_name,$type, $size )
{
        $options = DB::fetch("SELECT name, id FROM " . $table . " WHERE ". $type_name ." = ? ORDER BY name",$type);
      //$options = array();
        /*** begin the select ***/
        $dropdown = '<select name="'.$name.'" id="'.$name.'" class="'.$class.'" required = "'.$required.'" size="'.$size.'" multiple>'."\n";

        /*** loop over the options ***/
        foreach($options as $key=>$option )
        {
                /*** assign a selected value ***/
                $select = in_array( $option, $selected ) ? ' selected' : null;

                /*** add each option to the dropdown ***/
                $dropdown .= '<option value="'.$key.'" '.$select.' >'.$option.'</option>'."\n";
        }

        /*** close the select ***/
        $dropdown .= '</select>'."\n";

        /*** and return the completed dropdown ***/
        return $dropdown;
}

for result:

$name = 'book_cover[]';
$selected = json_decode($DB_QUERY[0]['book_cover'] , true);
echo _is_dropdown_( $name,$selected ,NEWS_TYPE ,'contentgroup','required','book_type','4', '4' );

Now in result dropdown not show name of option:

<select name="book_cover[]" class="contentgroup" required = "required" size="4" multiple>
<option value="0"  >Array</option>
<option value="1"  >Array</option>
<option value="2"  >Array</option>
<option value="3"  >Array</option>
<option value="4"  >Array</option>
<option value="5"  >Array</option>
<option value="6"  >Array</option>
<option value="7"  >Array</option>
<option value="8"  >Array</option>
<option value="9"  >Array</option>
<option value="10"  >Array</option>
<option value="11"  >Array</option>
<option value="12"  >Array</option>
</select>

And show this error :

[2015-09-09 08:28:06] [E_NOTICE] [8] Array to string conversion in C:\xampp\htdocs\cms\class\functions.php:1032

EDIT: i print_r $options:

(
    [0] => Array
        (
            [name] => test
        )

    [1] => Array
        (
            [name] => test2
        )

    [2] => Array
        (
            [name] => test3
        )
)

EDIT 2: i change $options to manual array : $options = array( 'test', 'test1', 'test2' ); this worked fine. now i need to convert my db result to this.

how do fix this error ?!

Clearly $option is an array. Would need to see the result set of the query to further debug. — Chris
– Chris, Commented Sep 9, 2015 at 4:04

Pradeepta · Accepted Answer · 2015-09-09 04:50:20Z

1

Here your key are 0,1,2... and values are

$option = Array
(
    [name] => test
)

So instead of $option use $option["name"].

foreach($options as $key=>$option ) {

      $select = in_array( $option["name"], $selected ) ? ' selected' : null;
      $dropdown .= '<option value="'.$key.'" '.$select.' >'.$option["name"].'</option>'."\n";

 }

answered Sep 9, 2015 at 4:50

Pradeepta

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Comments

Padmanathan J · Accepted Answer · 2015-09-09 04:31:34Z

0

You have $options which are rows. So each $option is a row, an array with 2 elements (name, id).

edited Sep 9, 2015 at 4:31

Padmanathan J

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answered Sep 9, 2015 at 4:06

flcoder

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Comments

Vikas Umrao · Accepted Answer · 2015-09-09 04:31:42Z

0

Updated the function, Please use this:

 function _is_dropdown_( $name,array $selected=null,$table ,$class,$required,$type_name,$type, $size )
    {
            $options = DB::fetch("SELECT name, id FROM " . $table . " WHERE ". $type_name ." = ? ORDER BY name",$type);
          //$options = array();
            /*** begin the select ***/
            $dropdown = '<select name="'.$name.'" id="'.$name.'" class="'.$class.'" required = "'.$required.'" size="'.$size.'" multiple>'."\n";

            /*** loop over the options ***/
            foreach($options as $key=>$option )
            {
                    /*** assign a selected value ***/
                    $select = ($key==$selected ? ' selected' : null );

                    /*** add each option to the dropdown ***/
                    $dropdown .= '<option value="'.$key.'" '.$select.' >'.$option[$key]['name'].'</option>'."\n";
            }

            /*** close the select ***/
            $dropdown .= '</select>'."\n";

            /*** and return the completed dropdown ***/
            return $dropdown;
    }

edited Sep 9, 2015 at 4:31

answered Sep 9, 2015 at 4:06

Vikas Umrao

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2 Comments

Perspolis Over a year ago

<b>Parse error</b>: syntax error, unexpected 'if' (T_IF) in

Perspolis Over a year ago

not work. i change $options to manual array : $options = array( 'test', 'test1', 'test2' ); this worked fine. now i need to convert my db result to this.

Padmanathan J · Accepted Answer · 2015-09-09 04:43:54Z

0

check this

$dropdown .= '<option value="'.$key.'" '.$select.' >'.$option['name'].'</option>'."\n";

answered Sep 9, 2015 at 4:43

Padmanathan J

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