getting mySQL data

[MaxD]

I'm trying to extract data from a mySQL record and use it to make a linked image. (the point of all this is to have dynamic content) The only problem is, the php manual only gives examples where the entire table is dumped. I just want one recordset, and just a few columns from that recordset.

echo"<a href=\"$ftr_link\"><img src=\"$ftr_img\" alt=\"$ftr_alt\"></a>";

where the columns of the database table are:
..., ftr_img, ftr_link, ftr_alt, ...

[bad76]

Originally posted by MaxD
I just want one recordset, and just a few columns from that recordset.

Hi,
this means you have to use a SELECT.

A look at the mysql manual tells us we can choice the fields list with a SELECT, and query for a single field. How to do this?

SELECT <list of field> FROM <name_of_table> WHERE <condition>

It's easy. If you want just a row, you can use a condition like:
<condition>: <keyfield>=<num>.

And to get only needed fields:
<list of field>: ftr_img, ftr_link, ftr_alt.

Your query now is:
SELECT ftr_img, ftr_link, ftr_alt
FROM <name_of_table>
WHERE <keyfield>=<num>

To use the resultset, you can access to each field with $row["<name_of_field>"], where $row is the name of an array.

I hope this helps you.

[MaxD]

I tried that, it makes sense, but the script won't execute.
this is what I have:

// How Many Features do we have in the database
$result = mysql_query("SELECT * FROM storefront_features", $db)

// Pick a random line
$random_line = rand(1,mysql_num_rows($result));
mysql_free_result($result);

// Get that random row
$result = mysql_query("SELECT * FROM storefront_features WHERE ftr_id = $random_line", $db)

echo"
<a href=\"$row["ftr_link"]\">
<img src=\"img/features/$row["ftr_big_img"]\" height=\"200\" width=\"400\" border=0 alt=\"$row["ftr_alt"]\">
</a>
";

the manual says that I shouldn't put semicolons after the query lines. But either way, the problem is somewhere else. I've tried taking out the mysql_free_result, I've tried taking out the second mysql_query line, taking out the echo'ed line...

[bad76]

You use mysql_error() for each mysql_query() to find the problem:

$result = mysql_query("SELECT * FROM storefront_features", $db) or die(mysql_error());

This is the best way to debug mysql statement.

[sundarn]

mysql_select_db(***); //for your database
$dd="your command for exc";
//then... "select from user";

mysql_query(**) // for run your query--> $dd
//then.... mysql_query($dd);

mysql_fetch_object or other function // for feed data

example :

$result=mysql_query($dd); // select * from user

while($aa=mysql_fetch_object($result)){
echo "$aa->user_name<br>";
echo "$aa->user_age<br>";
}

i think you are learning about mysql and php...... : )

[MaxD]

that yeilds no results, the script isn't executing. Nothing comes up that's inside the delimiters. There is some catastrophic fatal horrible world-ending error somewhere between those question marks. (humor)

any additional thoughts?

<?php

// How many features do we have in the database
$result = mysql_query("SELECT * FROM storefront_features", $db)
  or die(mysql_error());

// Pick a random line
$random_line = rand(1,mysql_num_rows($result));
mysql_free_result($result);

// Get that random row
$result = mysql_query("SELECT * FROM storefront_features WHERE ftr_id = $random_line", $db)
  or die(mysql_error());


echo"
<a href=\"$row["ftr_link"]\">
<img src=\"img/features/$row["ftr_big_img"]\" height=\"200\" width=\"400\" border=0 alt=\"$row["ftr_alt"]\">
</a>
";
?>

I have reson to believe its the $row['ftr_id'] thingies. When I try to call one the error occurs. Is there something I'm missing? can I call them from inside an echo like I did? or do I have to assign thier value to another string? (which I already tried and it didn't work). Am I writing it wrong? ()'s instead of []'s? heeeeelp!

[MaxD]

Nothing is working, I've tried several different approaches. Mostly from examples in the manual. Nothing is working. This is the code I'm using.

<?php

$db_server = localhost;
$db_server_username = root;
$db_server_password = "";
$db_database = store;

// mySQL database connection
$db = mysql_connect($db_server, $db_server_username, $db_server_password)
  or die("Could not connect: " . mysql_error());
  mysql_select_db($db_database,$db);

$result = mysql_query("SELECT * FROM storefront_features", $db)
  or die("Failure: " . mysql_error());

$random_line = rand(1,mysql_num_rows($result));
mysql_free_result($result);

$result = mysql_query("SELECT * FROM storefront_features WHERE ftr_id = '$random_line'", $db)
  or die("Failure: " . mysql_error());

echo"
  <a href=\"$row['ftr_link']\">
  <img src=\"$row['ftr_big_img']\" alt=\"$row['ftr_alt']\">
  </a>
";
?>

these are the column names in my table.

ftr_id
ftr_big_img
ftr_sml_img
ftr_link
ftr_alt
ftr_description

any suggestions?

[Weedpacket]

You do have one line missing; the one that takes the result and actually fetches a $row from it; e.g.,

$row=mysql_fetch_array($result);

between making the query and echoing the $row.

Incidentally, your method for finding the number of rows is very expensive. Instead of selecting every field of every record in the table, and using mysql_num_rows() to find out how many records there are, it would be much faster to use

$result=mysql_query("SELECT COUNT(*) AS total FROM storefront_features");
$row=mysql_fetch_array($result);
$count=$row['total'];

All this assumes that ftr_id runs strictly from 1 to (however many rows there are) of course.

For selecting certain columns from the table, select only those columns, instead of all of them.

SELECT ftr_link, ftr_big_img, etc.

instead of

SELECT *

That also makes for savings.