Invalid query: Unknown column in 'field list'

[Tom_LR]

Ok..I got it to find the database and it return the values with an echo statement. How do I assign column names to variables? Thanks

[cbrknight]

You might also look at this...

$result=mysql_query($query);

Try
$result=mysql_query($query,$link);

so it knows what connection for the query...

[cbrknight]

$data=mysql_fetch_array($results);

$var1=$data[0];
$var2=$data[1];

etctra

you can also find out what your data offsets are by using this

echo('<pre>');
print_r($data);
echo('</pre>');

that will print out the first record of your array with all the offsets. the <pre> just makes it easier to read.

[Tom_LR]

cbrknight - I appreciate you taking the time to help me with this. Unfortunately I'm very new to all of this. I'm trying to put together a "web portal" for a client to access and look at their accounts. I'm not very familiar with the language yet.

When I used this: echo $row['guarantor']; It worked just find.

Now I want to use the print command so that I can format and put the information exactly where I want it on the page. Or can I use the echo command above within html formatting to get the same result?

Thanks again...very much appreciated.

[cbrknight]

you can use echo $row['guarantor'];
or set

$guarantor = $row['guarantor'];

and use echo $guarantor;

I believe you can print also...

but I never us print so couldnt tell you the pros and cons of echo vs. print

[cbrknight]

as to html I would do it something like this

<tr>
<td>Guarantor Name: <?php echo $guarantor; ?> </td>
</tr>

or you can

echo('
<tr>
<td>Guarantor Name: $guarantor </td>
</tr>
')
With the second method all the html is processed through php and you have to be careful with your quotes

[Tom_LR]

Thanks! Now that I see it, it's simple. Sometimes I make it harder than it should be. I'll give it a try and post my results.

[cbrknight]

I think I learn everthing in PHP the hard way... Good luck

[Tom_LR]

Ok...one more quick question...where do I put, for example, $guarantor = $row['guarantor']; in the code?

[cbrknight]

anywhere after the mysql_fetch_array.... and before you echo it out....

[Tom_LR]

That worked! You have no idea how much I appreciate your help. I've been fighting with this stuff for about 3 or 4 days. Slowly coming to understand though. Thanks Again.

[cbrknight]

No problem dont forget to Set link to RESOLVED