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Editorial Team
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Asked: 2026-06-03T10:47:30+00:00

please check the following code. $imagebaseurl = ‘support/content_editor/uploads/$name’; The $imagebaseurl is a variable that

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please check the following code. 

$imagebaseurl = 'support/content_editor/uploads/$name';

The $imagebaseurl is a variable that is containing a link to my image folder (uploads) and inside the folder I have some other folders which are named after my users name. for example: I have a user who’s name is john, so the the link should look like this-> support/content_editor/uploads/john.

The main idea is when any user is logged in and browses his image gallery I want to take him to his own gallery which basically is named after his name.

When he will visit the gallery the value of $name in the link will come from the user’s login name (from session). Now the problem is as you probably have already understood that the placement of $name in the above link is wrong and that is why it is not working. I am getting this whole URL> (support/content_editor/uploads/$name) instead of (support/content_editor/uploads/john)

Now could you please tell me how to use the $name in this $imagebaseurl = 'support/content_editor/uploads/$name';

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1 Answer

  1. Editorial Team
    $imagebaseurl = 'support/content_editor/uploads/' . $name;

    or

    $imagebaseurl = "support/content_editor/uploads/{$name}";

    Note that if you use double quotes, you can also write the above as:

    $imagebaseurl = "support/content_editor/uploads/$name";

    It’s good though to get in the habit of using {$...} in double quotes instead of only $..., for times where you need to insert the variable in a string where it’s not obvious to PHP which part is the variable and which part is the string.

    If you want the best performance, use string concatenation with single quotes.

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