search mysql with php

[miles_rich]

This code should work, i know it will only output the last value, but i think this is wrong, because i use it when im selecting all fields, so i dont know if this code is different because im only selecting one field.

"$f_cust_id= $row[0];"

any ideas??

db_connect();
db_select ($db_info);

$query = "SELECT uid FROM aq_info WHERE cust_name LIKE '$p_cust_firstname' ORDER BY uid ASC";

$result = mysql_query( $query);
$f_cust_id= $row[0];
echo $f_cust_id;
db_close();

[bobthedog]

There seems to be something missing from your code here. You do the query, but never retrieve a row from the result set. You need a mysql_fetch_array() call before "$f_cust_id= $row[0];", otherwise $row is never getting set.

If you put this in, you should find you get the FIRST row returned only.

Hope this helps 🙂

[miles_rich]

I get this error now.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\wwwroot\Beaumont v 3.2\customer v1.3\customer.php on line 460

db_connect();
db_select ($db_info);

$query = "SELECT uid FROM aq_info WHERE cust_name LIKE '$p_cust_firstname' ORDER BY uid ASC";

$result = mysql_query( $query);
$row = mysql_fetch_array ( $result);

$f_cust_id= $row[0];

db_close();

[bobthedog]

Your query is messing up somewhere... try checking the status of the returned result from mysql_query, something like this -

$result = mysql_query( $query);

if (!$result)
{
    print(mysql_error());
    exit();
}

$row = mysql_fetch_array ( $result);

Then at least you can see the error coming back.

[mworld]

Assuming your query is valid, i would be doing it like this:

db_connect();
db_select ($db_info);

// optimised query hehehe

$query = "SELECT `aq_info`.`uid` FROM `aq_info` WHERE  `aq_info`.`cust_name`='" . $p_cust_firstname . "' ORDER BY  `aq_info`.`uid`";

$row = mysql_fetch_row(mysql_query($query));

$f_cust_id = $row['0'];

db_close();