mySQL Error

[duyen]

I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/stliuq/public_html/guestbook.php on line 48

<?php
$con = mysql_connect("localhost","personremoved","passremoved");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("dbremoved", $con);

/*THIS IS LINE 48*/ $result = mysql_query("SELECT * FROM posts");

while($row = mysql_fetch_array($result))
  {
  echo '<div class="name">';
  echo $row['name'];
  echo '</div>';
  }

mysql_close($con);
?>

I double checked the DB and table names.

 

[TFlan]

echo mysql_error(); // after the query

Should tell you whats the problem.

But try this:

$con = mysql_connect("localhost","personremoved","passremoved");
if(!$con){
	die('Could not connect: '.mysql_error());
}

mysql_select_db("dbremoved", $con);

$result = mysql_query("SELECT `name` FROM `posts`");

while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
	echo '<div class="name">'
		. $row['name']
		. '</div>';
}

mysql_close($con);

 

[ertaboy356b]

in every mysql function you make.. put this code at the end

"or die(mysql_error())"

it will generate a mysql error report...

for example

$con = mysql_connect("localhost","personremoved","passremoved") or die(mysql_error());

mysql_select_db("dbremoved", $con) or die(mysql_error());

$result = mysql_query("SELECT `name` FROM `posts`") or die(mysql_error());

 

[TFlan]

don't use die. you want to see your page and how it's affected because of the error.

Use echo.

 

[JerseyFoo]

According to the error...

while($row = mysql_fetch_array($result))

... is line 48, so go with the error on this.

It is probably because it didn't return anything, so you could do this...

if ( mysql_num_rows($result) > 0 ){
  while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
  ...
  }
} else {
   echo '<div class="name"><i>Nothing to show</i></div>';
}

And as my OCD is kicking in, you may also use...

$q = "SELECT `name` FROM `posts` LIMIT 0,100"; // Limiting is usually nice
if ( $r = mysql_unbuffered_query($q) ){
  while ( $v = mysql_fetch_row($r) ){
    echo '<div class="name">' . $v[0] . '</div>';
  }
} else {
  // no results
}
mysql_free_result($r);

 

[duyen]

According to the error...

... is line 48, so go with the error on this.

It is probably because it didn't return anything, so you could do this...

if ( mysql_num_rows($result) > 0 ){
  while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
  ...
  }
} else {
   echo '<div class="name"><i>Nothing to show</i></div>';
}

And as my OCD is kicking in, you may also use...

$q = "SELECT `name` FROM `posts` LIMIT 0,100"; // Limiting is usually nice
if ( $r = mysql_unbuffered_query($q) ){
  while ( $v = mysql_fetch_row($r) ){
    echo '<div class="name">' . $v[0] . '</div>';
  }
} else {
  // no results
}
mysql_free_result($r);

The table is not empty.

 

[JerseyFoo]

You need to give us a bit more information then.

 

[duyen]

You need to give us a bit more information then.

...Like?

 

[TFlan]

anything that you think is relevant.

<?php
$con = mysql_connect("localhost","personremoved","passremoved");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("dbremoved", $con);

$result = mysql_query("SELECT * FROM posts");
if($result==FALSE){ echo mysql_error(); echo "<br>".$result."<br>"; } // Test if result is even there.
while($row = mysql_fetch_array($result))
  {
  echo '<div class="name">';
  echo $row['name'];
  echo '</div>';
  }

mysql_close($con);
?>

Use that, give us the result of it.

 

[duyen]

It says:

No database selected

Then the error in the original post.

 

[TFlan]

Then thats your error.

No DB is connected.

to confim this use this:

<?php
$con = mysql_connect("localhost","personremoved","passremoved");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

$db = mysql_select_db("dbremoved", $con);
if (!$db)
  {
  die('Could not connect: ' . mysql_error());
  }

$result = mysql_query("SELECT * FROM posts");
if($result==FALSE){ echo mysql_error(); echo "<br>".$result."<br>"; } // Test if result is even there.
while($row = mysql_fetch_array($result))
  {
  echo '<div class="name">';
  echo $row['name'];
  echo '</div>';
  }

mysql_close($con);
?>

And if that gets confirmed, then we have no control over it. Check your spelling and make sure the DB is in that MySQL connection.