Fetch row and explode - PHP mySQL

Question

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?

If you want to display the list of tags then don't explode the list. — VMai, Sep 06 '14 at 10:00
but there are tags more then one, and i want them to have separate links. — David, Sep 06 '14 at 10:02

score 0 · Answer 1 · answered Sep 06 '14 at 10:06

0

$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 ',').

answered Sep 06 '14 at 10:06

Peter van der Wal

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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 06 '14 at 10:09
If you already have the comma separated value in variable than what is the need of implode? – Ram Sharma Sep 06 '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 06 '14 at 10:17
Does your db also include the spaces after the comma? In that case just replace `{$postCatTag}` with `{$postCat}`. Otherwise use `..." . implode(', ', $postCatTag) . "...` – Peter van der Wal Sep 06 '14 at 10:26

Abhishek Shetty · Answer 2 · 2014-09-06T10:29:01.810

0

$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>";  

}

edited Sep 06 '14 at 10:29

answered Sep 06 '14 at 10:17

Abhishek Shetty

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1
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and how to insert this? – David Sep 06 '14 at 10:22
replace `
kategoria: {$postCatTag}

Abhishek Shetty

Sep 06 '14 at 10:25

There was a small error in my code. I re-edited it in my answer. Its `$i – Abhishek Shetty Sep 06 '14 at 10:30

just replace? like this: `echo "

{$postTitle}

avtori: {$postAuthor}
TariRi: {$postDate}

kategoria: $postCatTag[$i]"; }";` it shows nothing/ replaced (") with (') and result was: `

TariRi: 2014-08-22 13:00:00

for(=0;kategoria: '; }`

– David Sep 06 '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 06 '14 at 10:49

score 0 · Answer 3 · answered Sep 06 '14 at 12:39

0

    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.

answered Sep 06 '14 at 12:39

Tristup

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Fetch row and explode - PHP mySQL

3 Answers3

{$postTitle}