"Array" is displaying in dropdown instead of values fetched from database

Question

code snippet from html_form_class

<?php
$frmStr = $frm->addSelectList(
                      'city',
                      $city,
                      true,
                      '',
                      '--- Select City ---',
                      array(
                          'class' => 'dropdown-style5',
                          'id' => 'city')); 
echo $frmStr; ?>

code snippet from seachcar.php

   $city = $db->select('City','City_Name');
foreach($city as $row)
{
 $row;
}

"Array" is displaying in dropdown instead of values fetched from database Please Advice!

function addSelectList($name, $option_list, $bVal = true, $selected_value = NULL,
            $header = NULL, $attr_ar = array() ) {
        $str = "<select name=\"$name\"";
        if ($attr_ar) {
            $str .= $this->addAttributes( $attr_ar );
        }
        $str .= ">\n";
        if ( isset($header) ) {
            $str .= "  <option value=\"\">$header</option>\n";
        }
        foreach ( $option_list as $val => $text ) {
            $str .= $bVal? "  <option value=\"$val\"": "  <option";
            if ( isset($selected_value) && ( $selected_value === $val || $selected_value === $text) ) {
                $str .= $this->xhtml? ' selected="selected"': ' selected';
            }
            $str .= ">$text</option>\n";
        }
        $str .= "</select>";
        return $str;
    }

html output of addSelectList function is

  <select name="city" class="dropdown-style5" id="city">
  <option value="">--- Select City ---</option>
  <option value="0">Array</option>
  <option value="1">Array</option>
  <option value="2">Array</option>
  <option value="3">Array</option>

Can we see all your code? I can't tell a damn thing without seeing the whole addSelectList() function. If you're using a framework or something, what is it? — searsaw
– searsaw, Commented Jul 23, 2013 at 22:45
Dude, really? Edit your post and put that stuff in there so its readable. — searsaw
– searsaw, Commented Jul 23, 2013 at 22:55

seymar · Accepted Answer · 2013-07-23 23:40:36Z

2

You need to rebuild the array of cities:

$city = $db->select('City','City_Name');
$city_rebuild = array();
foreach($city as $row) {
    $city_rebuild[] = $row['City_Name'];
}
$city = $city_rebuild;

edited Jul 23, 2013 at 23:40

answered Jul 23, 2013 at 23:20

seymar

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2 Comments

Hassan Over a year ago

blank output as well as its my generic class function to hard code it wont work for me at other places

seymar Over a year ago

Okay I took that in consideration. Edited my answer.

Fanda · Accepted Answer · 2013-07-23 22:52:20Z

0

You do echo of array. Something is wrong in your abstraction objects. You must iterate on array to show up its values.

answered Jul 23, 2013 at 22:52

Fanda

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Comments

Dylan · Accepted Answer · 2013-07-23 23:24:30Z

0

function addSelectList creates a "dropdown" (actually a select element)

You need to remove the html from the function output.

Edit 1 I was confused as to what you were going for. In your foreach($option_list... you need to know what keys are available in the $option_list array and what you want to appear in the select dropdown.

edited Jul 23, 2013 at 23:24

answered Jul 23, 2013 at 23:04

Dylan

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6 Comments

Dylan Over a year ago

@Mubbashir... I misunderstood, but I know know what you're trying to do. First we need to know what the values available in the array are. Do something like print_r($option_list); and let me know what it says.

Hassan Over a year ago

values are 'Lahore', 'Karachi', 'Multan','Hyderabad'

Dylan Over a year ago

That's not what print_r returns. We need to be able to see the structure of the array

Hassan Over a year ago

I've solved my problem thankyou so very much for anticipation :)

seymar Over a year ago

Please post your solution so we can benefit from it. Thats the idea of this community.

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