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up vote 0 down vote favorite

Here is my code:

$postsql = "SELECT * FROM post WHERE id='{$_GET['id']}'";
$posts = mysqli_query($connect,$postsql) or die("Error: ".mysqli_error($connect));
$post = mysqli_fetch_array($posts);

$postAuthor = $post['author'];
$postDate = $post['date'];

$postCat = mysqli_fetch_row($post['cat']);// I've tags here separated by commas.
$postCatTag = explode(",",$postCat);

$postText = $post['text'];

echo "<img class='view_newsimg' src='{$postImg}'>
<h3 class='lath'>{$postTitle}</h3>
<ul class='det'><li class='adc'>avtori: {$postAuthor}</li>
<li class='adc'>TariRi: {$postDate}</li>
<li class='adc'>kategoria: <a href='#'>{$postCatTag}</a></li>"; // tags are shown here
echo <<<TEXT
<p class="news">{$postText}</p>
TEXT;

As a result for tag section in browser it displays: kategoria: Array instead of displaying tags from table. Why? How can get desired result?

share|improve this question
    
If you want to display the list of tags then don't explode the list. –  VMai Sep 6 '14 at 10:00
    
but there are tags more then one, and i want them to have separate links. –  David Sep 6 '14 at 10:02
    
You should use a loop then. –  VMai Sep 6 '14 at 10:04
    
what kind of out put you need for tags? –  Ram Sharma Sep 6 '14 at 10:06
    
i need separated links shown side by side –  David Sep 6 '14 at 10:11

3 Answers 3

up vote 0 down vote

$postCatTag is an array. Array to string conversions always results in Array. You could use the implode-function to impode an array with a given seperator:

implode(', ', $postCatTag);

Or you can just use $postCat since that is already a string (tags seperated by ',').

share|improve this answer
    
he already have comma separated value in $postCat –  Ram Sharma Sep 6 '14 at 10:08
    
I know, see last line of my answer. The implode-part was for completeness, and in case he wanted a custom separator. –  Peter van der Wal Sep 6 '14 at 10:09
    
If you already have the comma separated value in variable than what is the need of implode? –  Ram Sharma Sep 6 '14 at 10:14
    
Maybe because tag1, tag2, tag3 looks better than tag1,tag2,tag3. I didn't recommend to implode again with the same separator as used with explode –  Peter van der Wal Sep 6 '14 at 10:17
    
i.stack.imgur.com/jpO5d.png need this –  David Sep 6 '14 at 10:24
up vote 0 down vote

$postCatTag is an array. Arrays cannot be displayed by echo.

Use loop for this:

for($i=0;$i<count($postCatTag),$i++){

  echo "<li class='adc'>kategoria: <a href='#'>$postCatTag[$i]</a></li>";  

}
share|improve this answer
    
and how to insert this? –  David Sep 6 '14 at 10:22
    
replace <li class='adc'>kategoria: <a href='#'>{$postCatTag}</a></li>"; with my code Check and let me know. –  Abhishek Shetty Sep 6 '14 at 10:25
    
There was a small error in my code. I re-edited it in my answer. Its $i<count($postCatTag) not $i<=count($postCatTag) –  Abhishek Shetty Sep 6 '14 at 10:30
    
just replace? like this: echo "<img class='view_newsimg' src='{$postImg}'><h3 class='lath'>{$postTitle}</h3><ul class='det'><li class='adc'>avtori: {$postAuthor}</li><li class='adc'>TariRi: {$postDate}</li> for($i=0;$i<count($postCatTag),$i++){ echo "<li class='adc'>kategoria: <a href='#'>$postCatTag[$i]</a></li>"; }"; it shows nothing/ replaced (") with (') and result was: <li class='adc'>TariRi: 2014-08-22 13:00:00</li> for(=0;<count(Array),++){ echo '<li class='adc'>kategoria: <a href='#'></a></li>'; } –  David Sep 6 '14 at 10:41
    
Thats because your array $postCatTag is empty. Dont replace (") with ('). Compiler reads and convert everything inside (") but when it comes accross (') it simply prints it. write print_r($postCatTag) somewhere and see the result. It will display content inside array. Will help you to debug the code. –  Abhishek Shetty Sep 6 '14 at 10:49
up vote 0 down vote 
    echo "<li class='adc'>kategoria:'; 
    for($i=0;$i<count($postCatTag),$i++)
    {

              echo "<a href='#'>$postCatTag[$i]</a>";  

    }
    echo "</li>";

Hope it will work for you.

share|improve this answer

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