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Mysql Results in variable


paegn

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I'm going to start by apologizing....

I'm sure this has been asked and answered before, but I've been searching all day and I'm not having any luck.

 

I've set up a Mysql database for product specs. There are about 70 fields at the moment, but this will likely increase.

Basically, I need to be able to select 2 records, and display them side by side in 2 columns.

 

I'm sending:

 

$a = 1;

$b = 72;

 

$result=mysql_query(SELECT * FROM `fixtures` WHERE id = ".$a." OR id = ".$b);

$row = mysql_fetch_array($result);

 

I've had a look at the result using phpmyadmin, and the query is returning 2 recors, as it should.

When I try get the data for the second record, I'm getting stumped...

 

If I use

 

mysql_num_rows($result);

 

I'm seeing 2 records, but I can't see it using

 

var_dump($row);

 

$row[1]['id']

 

isn't working, either.

What am I doing wrong?

 

I'd also like to be able to access the field names as well, so I can create a function that will be reusable throughout the whole table.

 

In order to keep everything laid out correctly, I'm using a containing <div> then floating 3 <div>'s inside it. the first one being the field name in readable english, the second 2 being the data from the records. Ideally, I'd like to be able to list 3 or 4 products as well.

 

I'm sure it's something simple that I'm missing, probably to do with the arrave variable, but as I said, I've been trying to sort this out all day, and my brain is fried now.

 

thanks in advance.

 

martin

 

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https://forums.phpfreaks.com/topic/203343-mysql-results-in-variable/
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The function mysql_query returns a pointer to the results set. To get all the results you need to use a fetch function in a loop;

<?php
$q = "SELECT * FROM `fixtures` WHERE id = ".$a." OR id = ".$b;
$result=mysql_query($q) or die("Problem with the query: $q<br>" . mysql_error());
while ($row = mysql_fetch_assoc($result)) {
    echo '<pre>' . print_r($row) . '</pre>';
}
?>

 

Ken

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