&quot;Array&quot;

"Array"

Henri

when I try to do this:

<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT * FROM SEA_news", $connect);
$news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
echo "$news";
?>

All I get is the word "Array"

anyone know what the problem could be?

thank you

hand

<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT * FROM SEA_news", $connect);
$news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
while($row = mysql_fetch_array($news)){
	echo "$row[news]<br>\n";

}
?>

Henri

It says this 😐

Warning: Supplied argument is not a valid MySQL result resource in /home/virtual/site191/fst/var/www/html/sea/news.php on line 18

This is the "full" code (so you can see what line 18 is):

<html><head><title></title>

<?php include("style.inc") ?>

</head>

<body bgcolor="#002660">
<center>
<font color="white">
<!-- news start --!>
<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT * FROM SEA_news", $connect);
$news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
while($row = mysql_fetch_array($news)){
    echo "$row[news]<br><br>";
}
?>
<!-- news end --!>
</font>
</center>

</body>
</html>

thanx

hand

Sorry

<html><head><title></title>

<?php include("style.inc") ?>

</head>

<body bgcolor="#002660">
<center>
<font color="white">
<!-- news start --!>
<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT * FROM SEA_news", $connect)
	or die("MySQL error: ".mysql_error());

while($row = mysql_fetch_array($sql)){
    echo "$row[news]<br><br>";
}
?>
<!-- news end --!>
</font>
</center>

</body>
</html>

Phantazm

Ok.. I suppose I'll take a stab at this one.

while($row = mysql_fetch_array($sql)){
    echo "$row[news]<br><br>";
}

The above code, in my experience will not work (as you obviously know). It's close though. Try this (change in YOUR code the 'news_field' to match your database. Make sure the field name IS contained with the single quotes.)

while($row = mysql_fetch_array($sql))
{
    echo "$row['news_field']<br><br>";
}

Give that a whirl.

And while I'm on the subject, if you want to retrieve multiple fields from the database (as I imagin you're doing by liiking at your SQL query -- "SELECT * FROM....") you can do this.

while($row = mysql_fetch_array($sql))
{
    $first_field = $row['db_field_1'];
    $second_field = $row['db_field_2'];

// and so on... then you can print thusly......
    echo "$first_field<BR><BR>";
// and so on...
}

Let us know if that works.

Phantazm

Henri

thanx, I got a feeling we're getting close, but I het this error :-/

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING' in /home/virtual/site191/fst/var/www/html/sea/news.php on line 19

my code:


<html><head><title></title>

<?php include("style.inc") ?>

</head>

<body bgcolor="#002660">
<center>
<font color="white">
<!-- news start --!>
<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT * FROM SEA_news", $connect);
$news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
while($row = mysql_fetch_array($sql)) {
echo "$row['content']<br><br>";
}
?>
<!-- news end --!>
</font>
</center>

</body>
</html>

Weedpacket

I guess this is line 19 (where's all this double-spacing coming from? Windows/Unix linebreak conversion mixups?):

echo "$row['content']<br><br>";

Try

echo "$row[content]<br><br>";

(that ought to work)

echo "{$row[content]}<br><br>";

echo "{$row['content']}<br><br>";

or even

echo $row['content'],"<br><br>";

For your original post:

$sql = mysql_query("SELECT * FROM SEA_news", $connect);

$news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());

echo "$news";

You never say which field of the $news array you want printed, which is why PHP just said "Array". It's generally good practice to say which fields you want in the SELECT statement instead of just * (it helps the database engine optimise the query and reduces the amount of data that has to get sent back to PHP and stored there); and if you use mysql_fetch_array() or mysql_fetch_assoc() instead of mysql_fetch_row(), you can refer to the field by name. Exempla gratia:

$sql = mysql_query("SELECT content FROM SEA_news", $connect);

$news = mysql_fetch_array($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
echo $news['content'];

Oh, and PS: HTML comments end with -->, not --!> 🙂

Phantazm

Took me a sec to get this one licked... something simple though 🙂

Try this.

  $sql = mysql_query("SELECT * FROM SEA_news", $connect);
  $news = mysql_fetch_row($sql) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
  while($row = mysql_fetch_array($sql))
  {
    $content = $row['content'];
    echo "$content<br><br>";
  }

This worked in my tests. I'm sure it's possible to do it in one statement, but after several attempts (failed I may add), the above code worked.

(It seems PHP didn't like the "$row['content']" -- in particular the double quotes it's wrapped in...!?!)

Henri

thanx so much guys, you've been great :-)