[SOLVED] Unexpected Output using "while ($row = mysql_fetch_array($result, MYSQL_ASSOC))"

[FirstBorn]

Hi,

 

Thank you for reading.

 

I have a php script that I'm using that has

been used before on a different site, but now,

when attempting to use the following code,

it is not displaying the data that I expect for

it to.

<?php

	// connecttodb($servername,$dbname,$dbuser,$dbpassword)
	$query="SELECT * FROM directory WHERE active='1' 
	AND location='L' AND type='paid' ORDER BY sortorder ASC";
	$result = mysql_query ($query) or die(mysql_error());

	$num=mysql_num_rows($result);

	// mysql_close();


	$id=mysql_result($result,$i,"id");
	$title=mysql_result($result,$i,"title");
	$url=mysql_result($result,$i,"url");
	$sortorder=mysql_result($result,$i,"sortorder");
	$type=mysql_result($result,$i,"type");
	$location=mysql_result($result,$i,"location");
	$class=mysql_result($result,$i,"class");
	$active=mysql_result($result,$i,"active");

	while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
		echo "<b>$class<br><a href=\"$url\">$title</a><hr>";
	}
	mysql_free_result($result);

	mysql_close();

?> 

 

The above displays 1 record when there should be 2.

 

The code below displays 12 rows of the same record

when there should be 12 different records displayed:

<p><h3>Available</h3><hr>
<?php

	connecttodb($servername,$dbname,$dbuser,$dbpassword);

	$query="SELECT * FROM directory WHERE active='1'
	AND location='L' AND type='order' ORDER BY sortorder ASC";
	// $result=mysql_query($query);
	$result = mysql_query ($query) or die(mysql_error());

	$num=mysql_num_rows($result);


	$id=mysql_result($result,$i,"id");
	$title=mysql_result($result,$i,"title");
	$url=mysql_result($result,$i,"url");
	$sortorder=mysql_result($result,$i,"sortorder");
	$type=mysql_result($result,$i,"type");
	$location=mysql_result($result,$i,"location");
	$class=mysql_result($result,$i,"class");
	$active=mysql_result($result,$i,"active");

	while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
		echo "<b>$class<br><a href=\"$url\">$title</a><br>";
	}

	mysql_close();

?> 

 

So, I'm guessing that either:

a.  The "while" statement is in the wrong place

(of which, I don't think so because this is how it is used on a different site)

OR

b.  There is nothing to count and then display the next record, but, again,

(I don't think so because this is how it is used on a different site)

 

So, what am I missing here?

 

Any Ideas?

 

Thank you in advance for your help.

 

Thanks,

FirstBorn

 

MySQL Ver:

5.0.51a-community

 

phpMyAdmin:

2.11.9.1

MySQL client version: 4.1.22

 

[rhodesa]

the code should be:

 

   <?php

      // connecttodb($servername,$dbname,$dbuser,$dbpassword)
      $query="SELECT * FROM directory WHERE active='1' 
      AND location='L' AND type='paid' ORDER BY sortorder ASC";
      $result = mysql_query ($query) or die(mysql_error());
      
      while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
         echo "<b>{$row['class']}<br><a href=\"{$row['url']}\">{$row['title']}</a><hr>";
      }
      mysql_free_result($result);

      mysql_close();

   ?> 

[FirstBorn]

the code should be:

 

   <?php

      // connecttodb($servername,$dbname,$dbuser,$dbpassword)
      $query="SELECT * FROM directory WHERE active='1' 
      AND location='L' AND type='paid' ORDER BY sortorder ASC";
      $result = mysql_query ($query) or die(mysql_error());
      
      while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
         echo "<b>{$row['class']}<br><a href=\"{$row['url']}\">{$row['title']}</a><hr>";
      }
      mysql_free_result($result);

      mysql_close();

   ?> 

 

Dear Aaron,

 

Thank you VERY MUCH!!!

 

That worked!

 

How do I get this Thread marked as 'resolved?'

 

Thanks,

Christopher

 

[rhodesa]

there should be a solved button somewhere

[FirstBorn]

there should be a solved button somewhere

 

Hi Aaron,

 

Found it, Thanks...

 

Was on the bottom of the page and nearly out of sight...

 

Have an Excellent New Year!

 

Thanks,

Christopher