为什么 PHP echo 返回不完整的链接？

Question

So I have the following code:

echo '<li><a href="'.$Website.$Elements[2].'/'.$Row["povezava"].'" class="leftMenuArticlesInactive"></a><li>';

1. $Website contains: http://www.somesite.com/test/
2. $Elements[2] contains: sometext
3. $Row["povezava"] contains: sometext2

All variables have nothing but letters in them.

Now my problem. If I try to echo the following I get:

// I'm missing "sometext":
http://www.somesite.com/test/sometext2
// Should be like this:
http://www.somesite.com/test/sometext/sometext2

If I do print_r($Elements) I can see that $Elements[2] is "sometext".

Why can't I echo $Elements[2] along with a few other variables?

edit: added more code

$Website = 'http://' . $_SERVER["SERVER_NAME"] . '/' . $Elements[1] . '/';

function ShowMenuOfParent($Parent, $Spacing) {
    $Query = mysql_query("SELECT * FROM `izdelki_meni` WHERE `stars` = '$Parent';");
    if(mysql_num_rows($Query) != 0) {
        while($Row = mysql_fetch_array($Query)) {
            $s = '<li><a href="'.$Website.$Elements[2].'/'.$Row["povezava"].'" class="leftMenuArticlesInactive">';
            for($i = 0; $i < $Spacing;$i++) {
                $s .= '&nbsp;';
            }
            $s .= $Row["ime"].'</a></li>';
            echo $s;
            ShowMenuOfParent($Row["id"], $Spacing + 8);
        }
    }
}

$Query = mysql_query("SELECT * FROM `izdelki_meni` WHERE `stars` = 0;");
if(mysql_num_rows($Query) != 0) {
    echo '<ul>';
    while($Row = mysql_fetch_array($Query)) {
        echo '<li><a href="'.$Website.$Row["povezava"].'" class="leftMenuArticlesActive">'.$Row["ime"].'</a></li>';
        if($Elements[2] == $Row["povezava"]) {
            ShowMenuOfParent($Row["id"], 8);
        }
    }
    echo '</ul>';
}

Answer 1

Why can't I echo $Elements[2] along with a few other variables?

Actually, echo can do that, so only this part of your question remains unresolved:

Why can't I?

What you can and what you can not often depends on what you've learned so far to do. As learning is subjective, it's hard to answer that question by somebody else than you. But probably you can add more information what especially hinders you to understand your actual problem so others can better assist you.

---

The information you provide with your question is factually wrong. I can understand you have a problem and you would like to fix it, but what you actually do is that you make your problem bigger than it is by posting it here as an invitation for others to even do more guessing. I mean 10 persons are able to guess more than one person. So the way you ask is actually counter-productive and not helpful. Both for yourself as anybody else.

Instead of guessing you need to start to actually proof things step by step. Start with your own expectation:

Now my problem. If I try to echo the following I get:

// I'm missing "sometext":
http://www.somesite.com/test/sometext2
// Should be like this:
http://www.somesite.com/test/sometext/sometext2

And the code you've posted that is related to the output is:

echo '<li><a href="'.$Website.$Elements[2].'/'.$Row["povezava"].'" class="leftMenuArticlesInactive"></a><li>';

You write that $Website ends with /test/ , and additionally there is a slash in the output string ( .'/'. ), however the output itself does not even contain that extra string according to you:

http://www.somesite.com/test/sometext2

And that is already regardless what $Elements or $Row contains. Start thinking first, you need to analyze why that happens, and others can not help your with that, because we don't have your software and your data to debug it. With some time you will learn your own best ways to find the mistakes you do.

Take the time to setup yourself the xdebug debugger and step through your code then. It helps you to locate errors very quickly. And not only this time but also in the future.

Answer 2

You seem to be printing unescaped text inside an HTML attribute that's part of a URL. My educated guess is that you are inspecting the scripts's output as the browser renders it, eg by hovering the mouse and reading the tooltip. Instead, use your browser's "View Source" feature to see the exact output.

Additional tips:

1. var_dump() is by far the best built-in function for debugging purposes. All other functions either cast variables to strings or can hide some important information.

2. Abort the script with die() or exit() right after the offending line to ensure the output is not altered later.

Answer 3

PHP sometimes has a hard time with concatenation on multiple variables in a line (back to back). This depends on the server setup. You could try echoing using one of the following:

echo '<li><a href="'.$Website.''.$Elements[2].'/'.$Row["povezava"].'" class="leftMenuArticlesInactive"></a><li>';

echo '<li><a href="'.$Website;
echo $Elements[2].'/'.$Row["povezava"].'" class="leftMenuArticlesInactive"></a><li>';

echo '<li><a href="',$Website,$Elements[2],'/',$Row["povezava"],'" class="leftMenuArticlesInactive"></a><li>';