tell csv import script to ignore blank rows?

Posted on 16th Feb 2014 07:03 pm by admin

Hi i have the below script but i get an error if a row is blank ie a return in the csv file at the end i need to somehow tell it to skip the rows that have nothing on them:

Code: <?php
// define table name and csv file location and name

$supp_id = "15";
$pricecode = "6";
$csvupload = "../ccd/Catalogue_ZZMILGAT.csv";

// create DB connection
$host = "localhost";
$user = "user";
$pass = "pass";
$mydb = "db";
$table = "table";

$db = mssql_connect($host,$user,$pass);
mssql_select_db($mydb);

// delete all old data
$sql2 = "DELETE FROM ".$table." WHERE supp_id = ".$supp_id."";

mssql_query($sql2) or die("Failed to insert to db ");

// Define DB mapping (Fill in rest of the mapping in order of appearance in CSV)
$fields = array(
"manufacturer",
"ignore",
"manf_part_no",
"supp_part_no",
"description",
"rrp",
"cost_price",
"avail_qty");


// Open the CSV file
$handle = fopen($csvupload, "r") or die("Unable to open $csvupload for reading");

// Get each row's values to an array
$i = 1;
// define tab delimited or comma
while (($data = fgetcsv($handle, 512, ",")) !== FALSE) {
// First row's headers and not included, otherwise row field count must match wanted field count
if(count($data)==count($fields) && $i!=1){
$rows[] = $data;
}
elseif(count($data)!=count($fields)) {
die("Erroneus CSV file: incorrect field count on row $i");
}
$i++;
}

// Close file handle
fclose($handle);

// Create SQL
if(count($rows)==0) die("No rows found in user file");


// Remove last comma
foreach($rows as $i => $row){
$sql = "INSERT INTO ".$table." (";
foreach($fields as $field)
if($field!="ignore")
$sql .= "$field,";
$sql = substr($sql,0,-1).",supp_id, date_added, PriceCode) VALUES ";
$sql .= "(";
foreach($row as $j => $value){
if($fields[$j]!="ignore"){
if(empty($value))
$value = "0";
// Quote strings, try to remove existing quotes
elseif(!is_numeric($value))
$value = str_replace("'","",$value);
$value = str_replace(""","",$value);
$sql .= "'$value',";
}
}

$sql = substr($sql,0,-1).",'".$supp_id."', getdate(),'".$pricecode."')"; // Remove last comma
//echo "SQL became: $sql";

mssql_query($sql) or die("Failed to insert to db ");

}

?>

Your Answer:

Login to answer

Other forums

Help pulling in 'id' with "read more" link
I'm trying to create snippets of my articles in the cms I'm creating, and then have them redirect to

[Need Help] php timing issues
I don't know what is going wrong. I need some help with being able to set an image at 9:00am Colorad

mysql VARCHAR acting like INT
Hi, All.

I have a table that contains a varchar(10) column named weird_field. In this column

Reading waves
Hi,

I am trying to find a way of finding the highs and lows on a graph line, the line points

need help with mail()
hi i want to send an email with attaching pdf file using php. i have the following script but the pr

Calender Not Opening
The following code is not loading the javascript calender in another window. It's not doing anything

RTF fomatting to email content
Im trying to sen an email with content is picked up from a rtf-file (file_get_contents('*.rtf'). Mai

whats wrong with my code please help!!!
this is the error


Warning: mysql_close(): supplied argument is not a valid MySQL-Link res

WS-Security PL/SQL Forms
Oracle Database 10g Enterprise Edition Release 10.2.0.2.0 - Prod
PL/SQL Release 10.2.0.2.0 - P

login form can you find my error?
registrationform.php seems fine send data to registration.php
registration.php seems fine checks