Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

[ardhix]

i have found warning message =  Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given inC:\xampp\htdocs\articles\artikel.php on line 7

 

 

  

 

<?php
   include 'koneksi.php';
$articleid = $_GET['articleid'];
$sql_query = 'SELECT ta.*, tca.category_article FROM article_news AS ta INNER JOIN tcategory_article AS tca ON ta.category_article_id = tca.category_article_id WHERE article_id = '.$articleid;
$query = mysql_query($sql_query);
$row = mysql_fetch_array($query); //--> this is line #7 


$msg = $_GET['msg'];
if(!empty($msg)){
if($msg = 100){
echo '<h5 style="text-align:center; color:#f00; font-weight: normal">.</h5>';
}
} 
?>


  <!--articles-->
  <a name="TemplateInfo"></a>
  <input type="hidden" name="prodid" value="<?php echo $row['article_id'];?>" size="1"/>
<h1><?php echo $row['judul'];?></h1>
<p><a href=""><img src="images/upload/<?php echo $row['photo'];?>" width="100" height="120" class="float-left"></a>
                <p><?php echo $row['artikel'];?></a>.
                </p>
<!--end of articles-->
                
<p><?php echo $row['category_article'];?></p>
<p class="post-footer"> 
<a href="index.html" class="readmore">Read more</a>
<a href="index.html" class="comments">Comments (7)</a>
<span class="date"><?php echo $row['tanggal'];?></span> 
</p>
 

 

please help me

 

i'm sorry newbie here..

Edited March 27, 2013 by ardhix

[haku]

Please wrap your code in code tags. You can do this using the <> button in the editor. As well, always point out which line is referred to in the error (in this case, which line is #53?).

[ardhix]

i've change 

 

Please wrap your code in code tags. You can do this using the <> button in the editor. As well, always point out which line is referred to in the error (in this case, which line is #53?).

 

i've change with editor, is that correct?

[haku]

Your query is failing for some reason. Change this line:

 

 

$query = mysql_query($sql_query);

 

To this:

$query = mysql_query($sql_query) or die(mysql_error());

 

Run the script again, and let us know the output.

[ardhix]

Your query is failing for some reason. Change this line:

 

 

$query = mysql_query($sql_query);

 

To this:

 

$query = mysql_query($sql_query) or die(mysql_error());

 

Run the script again, and let us know the output.

 

 

now it shows message :

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

[haku]

What does this show:

 

 

die($sql_query = 'SELECT ta.*, tca.category_article FROM article_news AS ta INNER JOIN tcategory_article AS tca ON ta.category_article_id = tca.category_article_id WHERE article_id = '.$articleid);

 

I'm thinking $articleid is empty.

Edited March 27, 2013 by haku

[ardhix]

What does this show:

 

 

die($sql_query = 'SELECT ta.*, tca.category_article FROM article_news AS ta INNER JOIN tcategory_article AS tca ON ta.category_article_id = tca.category_article_id WHERE article_id = '.$articleid);

 

I'm thinking $articleid is empty.

 

Solved now.. by removing $articleid, the data shows up, thank you very much

[ardhix]

there's no value in $articleid, because no data passed to it (it used for detail of article)