PHP foreach statement produces no output

Question

the following produces no output, when really it should show a list including Milk, Cheese and Yoghurt. It is probably something really simple, I just can't see it.

 <?php 

 $FoodList=array();
 $newArray =array();
 echo "<p>";


 $Dairy= array(
    'a' => 'Milk',
    'b' => 'Cheese',
    'c' => 'Yoghurt',
 );
 $Fruit = array(
    'd' => 'Apples',
    'e' => 'Oranges',
    'f' => 'Grapefruit',
 );
 $GlutenFree = array(
    'g' => 'GF Cookies',
    'h' => 'GF Pancakes',
    'i' => 'GF Bread',
 );
 $Sweets = array(
    'j' => 'Musk Sticks',
    'k' => 'Caramels',
    'l' => 'Chocolate',
 );


 if ($_POST['running'] == 'yes')
 {
    $newArray = array_merge($FoodList, $Dairy);

    foreach ($newArray as $key => $value)
    {
       echo $value;
    }
 }


 echo "<p>";
 ?>

This may because the FoodList Array does not have anything in it, so I will look into that, but I have a strong feeling it is related to something else.

What is in $newArray? Use print_r( $newArray ) to see what's in it. — Andy Lester
– Andy Lester, Commented Sep 7, 2013 at 3:48
@AndyLester $newArray is an empty array. Also I've tried your code and it works perfectly. Maybe the $_POST['running'] !== 'yes'? — Class
– Class, Commented Sep 7, 2013 at 3:50
Actually, your handler should contain $Running=$_POST['Running']; then you should be checking if it's true meaning if it has been checked if ($_POST['Running'] == true) and your input should be <input type="checkbox" name="Running" /> so when it's checked, it will return MilkCheeseYoghurt and nothing if it isn't checked. — Funk Forty Niner
– Funk Forty Niner, Commented Sep 7, 2013 at 4:04

Anthony Nandaa · Accepted Answer · 2013-09-07 03:51:50Z

1

Your "bug" must be coming from the line, the array merge is fine:

 if ($_POST['running'] == 'yes')

answered Sep 7, 2013 at 3:51

Anthony Nandaa

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4 Comments

Andrew Smith Over a year ago

The HTML page states: Running <input type="checkbox" name="Running" value="Yes" />

Andrew Smith Over a year ago

I have changed 'running' to 'Running' but it still produces no output

Class Over a year ago

'running' !== 'Running'

Funk Forty Niner Over a year ago

@AndrewSmith In the future, post all relevant code in your question. One of the most important factors (50%) are form elements which were not posted in your question.

khan · Accepted Answer · 2013-09-07 08:26:32Z

0

foreach($Dairy as $key => $value) {

echo $value; }

answered Sep 7, 2013 at 8:26

khan

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