Access array elements using foreach

Question

I'm studying php and now I'm struggling in following: i have an array that contains other array like this:

$leftMenu = array(
    array('link'=>'Домой', 'href'=>'index.php'),
    array('link'=>'О нас', 'href'=>'about.php'),
    array('link'=>'Контакты', 'href'=>'contact.php'),
    array('link'=>'Таблицы умножения', 'href'=>'table.php'),
    array('link'=>'Калькулятор', 'href'=>'calc.php')                
);

What i need to do, is draw a menu with hyperlinks by html and this array, using foreach. That is what i tried to do:

foreach ($leftMenu as $key=>$value){
    foreach ($value as $html=>$link){
        echo "<a href=$html>$link </a>\n"; 
    }   
}

Obviously it doesn't work because i get invalid values in variable $link. What i want to, is pass only links to that variable, not text. How to achieve that?

possible duplicate of Create a dropdown list from a multi-dimensional array — Amal
– Amal, Commented Feb 24, 2014 at 12:27

Fabio · Accepted Answer · 2014-02-24 12:26:35Z

10

You don't need to loop twice in your array but once

foreach ($leftMenu as $value){
    echo '<a href="'.$value['href'].'">'.$value['link'].'</a>'."\n";   
}

Live working sample here

answered Feb 24, 2014 at 12:26

Fabio

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4 Comments

Evgeniy Kleban Over a year ago

Thank you Fabio. If i understand correct i can use $value like i use an array?

Amal Over a year ago

@EvgeniyKleban: $value is an array and you are using it as such. Add print_r($value); inside the loop to see the contents of the array on each loop iteration.

Fabio Over a year ago

@evgeniyKleban Amal is right you can use as array because it is an array

Seeker Over a year ago

@Fabio are you free atm?

OfirH · Accepted Answer · 2014-02-24 12:28:33Z

1

foreach ($leftMenu as $key=>$value){
        foreach ($value as $html=>$link){
                if ($html != 'link')    {
                    echo "<a href='{$html}'>{$link}</a>\n";
                }
        }   
    }

This should work.

answered Feb 24, 2014 at 12:28

OfirH

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Comments

Awlad Liton · Accepted Answer · 2014-02-24 12:37:54Z

1

Try like this:

foreach ($leftMenu as $a){
    $link = $a["link"];
    $href = $a["href"];
    echo "<a href=\"$href\">$link </a>\n";

}

demo : https://eval.in/105138

answered Feb 24, 2014 at 12:37

Awlad Liton

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