MySQL- Getting data in PHP

Serpent7

This is my script for a link table that is supposed to get a set of links from my databse and echo them out on the webpage. I know it will give all of the links in one straight line, I'm not worried about aesthetics at the moment.

My problem is that I get nothing out on the webpage whatsoever, and can't understand why. I know it's not an issue with connectiing to the database.

Here's my code:

<?php
	// Connecting, selecting uc database
	$link = mysql_connect('localhost', 'usera', 'xxxxx')
	   or die('Could not connect: ' . mysql_error());
	mysql_select_db('uconstruct') or die('Could not select database');

// Performing SQL query
$query = "SELECT `LinkStart` FROM `misc` LIMIT 0,1";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
$link_prefix = mysql_fetch_array($result);
$i = 0;
$j = 1;

while ($i < 30)
{
$query = "SELECT `name` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} ";
$result1 = mysql_query($query) or die('Query failed: ' . mysql_error());
$linkname = mysql_fetch_array($result1);
$query = "SELECT `URL` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} ";
$result2 = mysql_query($query) or die('Query failed: ' . mysql_error());
$linkurl = mysql_fetch_array($result2);
echo "<a href=\"${link_prefix}${linkurl}\">${linkname}</a>";
$i++;
$j++;
}
?>

The number is 30 in the while loop because I intend to set a 30-link maximum. That's sorted. 🙂

Any help will be appreciated, thankyou.

krotkruton

A trick I frequently use is to add in echo statements throughout my buggy code. This way I'll know exactly where it crashes. Try this code and see what outputs:

<?php 
    // Connecting, selecting uc database 
 echo "Before connecting<br>";
    $link = mysql_connect('localhost', 'usera', 'xxxxx') 
       or die('Could not connect: ' . mysql_error()); 
 echo "Connected, selecting database<br>";
    mysql_select_db('uconstruct') or die('Could not select database'); 
 echo "Selected database<br>";

// Performing SQL query 
$query = "SELECT `LinkStart` FROM `misc` LIMIT 0,1"; 
$result = mysql_query($query) or die('Query failed: ' . mysql_error()); 
 echo "After query<br>";
    $link_prefix = mysql_fetch_array($result); 
 echo "After array<br>";
    $i = 0; 
    $j = 1; 

while ($i < 30) 
{ 
 echo $i."<br>";
    $query = "SELECT `name` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} "; 
    $result1 = mysql_query($query) or die('Query failed: ' . mysql_error()); 
    $linkname = mysql_fetch_array($result1); 
    $query = "SELECT `URL` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} "; 
    $result2 = mysql_query($query) or die('Query failed: ' . mysql_error()); 
    $linkurl = mysql_fetch_array($result2); 
    echo "<a href=\"${link_prefix}${linkurl}\">${linkname}</a>"; 
    $i++; 
    $j++; 
    } 
?>

That will show you exactly where the code is crashing. For example, if it is your query that has the problem, then you will never see "After query" as an output.

Serpent7

Well, it appears that stupidly I hard coded the position = 1 bit in, whereas I meant to put $j. (I was tired. 😛)

But now I get a set of 12 links (the number I have in my database) all pointing to /ArrayArray and all with their names as Array: like this:

Array

It's confusing me beyond belief! :S

(Incase you couldn't tell by now, I'm new to MySQL. 🙂 )

Finally, thankyou for your help so far.

Houdini

Look closely right here

$link_prefix = mysql_fetch_ARRAY($result);

here too

$linkname = mysql_fetch_ARRAY($result1);

and lastly

$linkurl = mysql_fetch_ARRAY($result2);

NOTE I capped the ARRAY for emphasis in the code.
In all those code you set a variable to the result set of an sql query so
$link_prefix[$i] would give the values of the array and so on, try that out.

You have some unnecessary code also why con't you combine the

$query = "SELECT `name` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} ";
$result1 = mysql_query($query) or die('Query failed: ' . mysql_error());
$linkname = mysql_fetch_array($result1);
$query = "SELECT `URL` FROM `links` WHERE `position` = 1 LIMIT ${i},${j} ";

so you have this instead since they are in the same table.

$query="SELECT 'URL', 'name' FROM links";
$result= mysql_query($query)

and this should return your stuff in a table if you want the link to work add <a href=......>and </a> it to the code where needed.

echo "<table cellspacing= '2' border='2'";
	while ($row=mysql_fetch_array($result)){
	  echo"<tr><td>";
	  echo"{$row['URL']}{$row['name']}";
	  echo"</td></tr>";
	}
	echo "</table>";