PHP - Select Pieces From Checkboxess

This script is supposed to run in the command line.

Like the title says, I need to:

Import some columns (input the names with an argument) from a .CSV file Split the column into blocks of 500 lines Translate this pieces Put them back into the original file

I already have some code but it is unfinished and not working:
<?php

//name of the .CSV file
$filename = $argv[1];

//name of the column to be translated
$header_name = $argv[2];

//open file
$file_handle = fopen($filename, "r");

$i=0;
$j=0;
$counter = 0;
$translated = '';
$string = '';

while (!feof($file_handle) ) {

$counter++;

$line_of_text = fgetcsv($file_handle, 1024);

//find the column id that contains the header name.
// headers are always on the the first line
if ( $counter == 1 ){
     while ($line_of_text[$i] != $header_name) $i++;
      $header_id = $i;
                              }

//create the blocks
$string .= $line_of_text[$i];

if ($counter % 500 == 0)
   {
  $gt = new GoogleTranslateWrapper();
 $translated .= $gt->translate($string, "en");
    }


}

//Write the translate column in the same .CSV file

fclose($file_handle);


?>

Any help in rewriting this to work? 

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

I have 2 queries that I want to join together to make one row  

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>

Hi all,
Im trying to get the requested genre and compare it with the <SELECT> list to add the word SELECTED on the option.
So if the requested genre is the same as the option name them make that the SELECTED option.

Cheers.

Here's a bit of code I made that does not work
Code: [Select]
<?
function selected(){
if ($_REQUEST['genre'] = $name){
echo"SELECTED";
}}
?>

Gen
<SELECT class="sort" align="right" onChange="window.location.href=this.options[this.selectedIndex].value;">
<option value="<?=$_SERVER['PHP_SELF'];?>" <?$name = ''; selected(); ?>>Any</option>
<option value="<?=$_SERVER['PHP_SELF'];?>?genre=Action" <?$name = 'Action'; selected(); ?>>Action</option>
<option value="<?=$_SERVER['PHP_SELF'];?>?genre=Adventure" <?$name = 'Adventure'; selected(); ?>>Adventure</option>
<option value="<?=$_SERVER['PHP_SELF'];?>?genre=Animation" <?$name = 'Animation'; selected(); ?>>Animation</option>
</SELECT>