Getting error will executing the query (T_VARIABLE)

Question

I have script which downloading rss feed from the internet and store it in Microsoft SQL database. It's working perfectly, I'm able to download RSS feed and store it in MSSQL database.

But I'm getting error will executing the query:

PHP Parse error: syntax error, unexpected '$item' (T_VARIABLE) in C:\phptest\test7.php on line 37

This error query is supposed to check duplicate feed in the database, but it's not working.

php script

<?php


try {
$con = new PDO("sqlsrv:Server=hadi-tosh;Database=rssfeed", "rssuser", "rssuser");
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8");

$url = "http://www.albaldnews.com/rss.php?cat=24";
$rss = simplexml_load_file($url);

if($rss)
{
$items = $rss->channel->item;
foreach($items as $item)
{
$source = "اخبارالبلد";
$title = $item->title;
$link = $item->link;
$published_on = $item->pubDate;
$description = $item->description;
$category = $item->category;
$guid = $item->guid;
$enclosure = $item->enclosure[0]['url'];
$img2 = $item->enclosure[0]['url'];

$ch = curl_init ($enclosure);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_BINARYTRANSFER,1);
$rawdata=curl_exec ($ch);
curl_close ($ch);
$img = "$rawdata";

$query = $con->prepare('SELECT * from [feedtable] where title = '$item->title' AND link = '$item->link' AND published_on = '$item->pubDate' AND description = '$item->description' AND category = '$item->category' AND image = '$img' AND imagelink = '$img2' AND source = '$source'');

$query->execute();
$num_rows = $query->fetchColumn();

if($num_rows == 0)

{


      $stmt = $con->prepare('insert into [feedtable] (title,link,pubdata,description,category,image,imagelink,source) values (:title,:link,:pubdata,:description,:category,:image,:imagelink,:source)');
    $stmt->bindParam(':title',$item->title);
    $stmt->bindParam(':link',$item->link);
    $stmt->bindParam(':pubdata',$item->pubDate);
    $stmt->bindParam(':description',$item->description);
    $stmt->bindParam(':category',$item->category);
    $stmt->bindParam(':image',     $img, PDO::PARAM_LOB, 0, PDO::SQLSRV_ENCODING_BINARY);
    $stmt->bindParam(':imagelink',$img2);
    $stmt->bindParam(':source', $source );


$stmt->execute();    

echo "albaldnews feeds added\n";
}

else 
{
echo "albaldnews duplicate entry\n";

}



}
}
}

catch (PDOException $e) {

exit("Connection failed: " . $e->getMessage());
}

?>

possible duplicate of [Reference - What does this error mean in PHP?](http://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php) — Rizier123, Mar 03 '15 at 17:50
@Rizier123 i changed my code according to the article,but didnt work`id="'$id.'"';` — user3818265, Mar 03 '15 at 18:00

score 0 · Answer 1 · answered Mar 03 '15 at 18:03

0

You're not concatenating your query correctly.

You need to add . after every single quote in your $query = $con->prepare line. You also need to add escaped ' around every variable for SQL Server to be able to read your values correctly.

For example: SELECT * from [feedtable] where title = \'' . $item->title . '\' AND link = \''. $item->link. '\' AND ...

And so on and so forth for the rest of that line.

answered Mar 03 '15 at 18:03

Baher Ramzy

171
6

i did what you asked `$query = $con->prepare('SELECT * from [feedtable] where title = \''.$item->title.'\' AND link = \''.$item->link.'\' AND published_on = \''.$item->pubDate.'\' AND description = \''.$item->description.'\' AND category = \''.$item->category.'\' AND image = \''.$img.'\' AND imagelink = \''.$img2.'\' AND source = \''.$source.'\'');`but im gettimg error from sql `SQLSTATE[IMSSP]: An error occurred substituting the named parameters.` – user3818265 Mar 03 '15 at 18:18

Getting error will executing the query (T_VARIABLE)

1 Answers1