[PHP Header help]

$aid = $_POST['artcid'];

$uid = $_POST['userid'];

header('location:../artikeldetail.php?msg=86&articleid='.$aid'&userid='.$uid);

 

it returns Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING

 

the correct way to do this?

 

 

header('location:../artikeldetail.php?msg=86&articleid='.$aid

 

returns http://localhost/articles/artikeldetail.php?msg=86&articleid=24

 

i want to show user id too

[joining 3 tables mysql]

 

SELECT * FROM user_table as u JOIN content_article as a ON u.user_id = a.user_id JOIN article_category as c ON a.category_article_id=c.category_article_id WHERE . . . .

 

it works somehow, thank you

[joining 3 tables mysql]

please help

 

i have assignment to selecting tables by joining them..

 

how do i use join, or inner join?

 

is there another option selecting 3 tables by using foreign key?

 

Here's the table

 

 

this one is user table

 

 

here is article category

 

 

and this is content of article

 

 

 

 

i was willing to choosing 

 

 

article_news table = 

 

article_id, category_article_id, user_type_id, user_id, judul(title), artikel, photo, tag, tanggal

 

 

tcategory_article = 

 

category_article_id, category_article

 

 

tuser = 

 

user_type_id, firstname, lastname

 

 

 

thank you

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

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

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

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

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

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

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

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?

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

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..