syntax of a url using a variable in the url

Question

I have the following script in php:

<?php

$my_array_of_vars['v'];

    $row['comment'] = "<img src='http://img.youtube.com/vi/xxxxxx/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";

?>

All I want is to replace the "xxxxxx" with the result in variable $my_array_of_vars['v']; I have tried a lot of things but always getting wrong result. Any idea how to do this please?

This is what I have tried:

<?php
    $row['comment'] = "<img src='http://img.youtube.com/vi/$my_array_of_vars['v']/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";
?>

Did you try just regular concantenation with closing the quotes and periods. — adeneo
– adeneo, Commented Feb 1, 2014 at 14:59
What are you trying to do in the end? For many small problems like this one, lots of different ways exist and giving the context of your question helps finding the best appropriate answer. — Jivan
– Jivan, Commented Feb 1, 2014 at 15:02

kero · Accepted Answer · 2014-02-01 15:00:07Z

2

You could use those fancy brackets {} like this:

$row['comment'] = "<img src='http://img.youtube.com/vi/{$my_array_of_vars['v']}/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";

or properly concatenate the variable to the string:

$row['comment'] = "<img src='http://img.youtube.com/vi/".$my_array_of_vars['v']."/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";

answered Feb 1, 2014 at 15:00

kero

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1 Comment

Emissary Over a year ago

Also quotes around array keys are not required with "fancy" brackets.

user2509485 · Accepted Answer · 2014-02-01 15:03:09Z

0

you can use like this:

     $row['comment'] = "<img src='http://img.youtube.com/vi/".$my_array_of_vars['v']."/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";

answered Feb 1, 2014 at 15:03

user2509485

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Comments

Nadir M. Aliyev · Accepted Answer · 2014-02-01 15:05:51Z

0

You must use urlencode(). Here is best way:

$row['comment'] = "<img src='http://img.youtube.com/vi/".urlencode($my_array_of_vars['v'])."/hqdefault.jpg' title='YouTube Video' alt='YouTube Video' /> ";

answered Feb 1, 2014 at 15:05

Nadir M. Aliyev

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