Question

I’m trying to grasp OOP and I decided to build a site that accesses a sql database.
It’s been working out great so far I have several tables, but I’ve run into a snag

I access the database and create an object array with

$Dog_array = $sth->fetchAll(PDO::FETCH_CLASS, 'Dog');

Simplified but my class looks like this.

class Dog {
     private $name;

     function get_name {
     return $this->name;
    } 

list that build my table

function edit_table($Dog) {

echo "<table><tr><form action='update.php' method ='post'>";
echo "<input type='hidden' name='id' value='" . $id ."'>";

echo "<td>". $Dog->get_id() ."</td>";
?>
<td><input type='text' name='rname' value="<?php echo $Dog->get_rname(); ?>"></td>
<td><input type='text' name='cname' value="<?php echo $Dog->get_cname(); ?>"></td>
<td><input type='text' name='dob' value="<?php echo $Dog->get_dob(); ?>"></td>
<td><input type='radio' name='gender' value='male' <?PHP if($Dog->get_gender() == 'male'){ echo "checked=\"checked\""; } ?> /> Male
<input type='radio' name='gender' value='female' <?PHP if($Dog->get_gender() == 'female'){ echo "checked=\"checked\""; } ?> />Female</td>
<td><input type='text' name='sire' value="<?PHP echo "builddropdown" ?> "></td>
<td><input type='text' name='dam' value="<?PHP drop_menu() ?>"></td>
<?php

if ($Dog->get_available()) {
    $checked = "checked=\"checked\"";
} else {
    $checked = NULL;
echo "<td><input type=\"checkbox\"" . $checked . "name=\"available\" value=\"TRUE\"/></td>";

if ($Dog->get_display()) {
    $checked = "checked=\"checked\"";
} else {
    $checked = NULL;
}
echo "<td><input type=\"checkbox\"" . $checked . "name=\"display\" value=\"TRUE\"/></td></tr></table>";
?>
<input type='submit' value='change image'/></form>
<FORM METHOD="LINK" ACTION="upload.php">
<INPUT TYPE="submit" VALUE="Change Image Thumb">
</FORM>
<input type='submit' value='change image'/></form>
<FORM METHOD="LINK" ACTION="upload.php">
<INPUT TYPE="submit" VALUE="Change Image Thumb">
</FORM>
    <?php
    }
?>
<input type='submit' value='update'/></form>
<FORM METHOD="LINK" ACTION="index.php">
<INPUT TYPE="submit" VALUE="Back">
</FORM>
<?php 
}

?>

My drop down

function drop_menu() {

?>
<form name ="dropdown">
<select ="alldogs">
<?php
        foreach ($Dog_array as $Dogs) { 
        ?>
        <option value="<?php echo $Dogs->get_id() . "\">" . $Dogs->get_rname();   ?></option>
<?php } ?>
</select>
</form>
<? }

My drop down menu creates a list of all of the dogs.

I get the error Invalid argument supplied for foreach

I get why I get the error I just don’t know how to clean this up so I don’t have to pass my Object array into my edit_table function so I can pass it into drop down function
.
I thought about creating a child class but the menu is built from an array of the objects not one object so I don’t know how to make this work either. New to forum posting, so feel free to recommend some corrections in how I ask for help.

Answer 1

It’s a problem with the arrage $Dog_array scope.
You need to pass the $Dog_array to the drop_menu function.

<?php
function drop_menu($Dog_array) {

?>
<form name ="dropdown">
<select ="alldogs">
<?php
foreach ($Dog_array as $Dogs) { 
?>
<option value="<?php echo $Dogs->get_id() . "\">" . $Dogs->get_rname();   ?>></option>
<?php } ?>
</select>
</form>
<? }

And whenever you need to call the function, pass the array normally.

Also you can pass by reference

<?php
function drop_menu(&$Dog_array) {?>

In this case any manipulation to the array inside the function will impact the original array directly.