cs1h Posted September 7, 2007 Share Posted September 7, 2007 Hi, I'm getting the following error message, Parse error: syntax error, unexpected $end in D:\Inetpub\vhost\myroho.com\httpdocs\asearching.php on line 34 The php script is as follows, <?php // Connects to your Database mysql_connect("localhost", "adder", "clifford") or die(mysql_error()) ; mysql_select_db("real") or die(mysql_error()) ; $country = isset($_POST['menuFilesDMA']) ? preg_replace('/[^a-zA-Z0-9\ ]/', '', $_POST['menuFilesDMA']) : ''; $type = isset($_POST['Type']) ? preg_replace('/[^a-zA-Z0-9\ ]/', '', $_POST['Type']) : ''; $Abstract = isset($_POST['keyword']) ? preg_replace('/[^a-zA-Z0-9\ ]/', '', $_POST['keyword']) : ''; $where = array(); if ($country) { $where[] = "country='$country'"; } if ($type) { $where[] = "type='$type'"; } if ($Abstract ) { $where[] = "Abstract LIKE '%$Abstract%'"; } if ($where) { $sql = "SELECT * FROM items WHERE " . implode(' AND ', $where); // query and display results while($info = mysql_fetch_array( $data )) { //Outputs the image and other data Echo "<b>country:</b> ".$info['country'] . "<br> "; Echo "<b>type:</b> ".$info['type'] . " <br>"; Echo "<b>Abstract:</b> ".$info['Abstract'] . " <hr>"; } ?> I am new to php and cannot see what my error might be, Any help would be much appriciated, Thanks Colin Quote Link to comment Share on other sites More sharing options...
jitesh Posted September 7, 2007 Share Posted September 7, 2007 if ($where) { $sql = "SELECT * FROM items WHERE " . implode(' AND ', $where); // query and display results } Quote Link to comment Share on other sites More sharing options...
beboo002 Posted September 7, 2007 Share Posted September 7, 2007 $sql = "SELECT * FROM items WHERE " . implode(' AND ', $where); u hav error in this line i dont know what u want this query?? ??? Quote Link to comment Share on other sites More sharing options...
ToonMariner Posted September 7, 2007 Share Posted September 7, 2007 jitesh got it - missing a closing brace after the snippet of code he posted. there is nothing wrong with that code Beboo002.... Quote Link to comment Share on other sites More sharing options...
xyn Posted September 7, 2007 Share Posted September 7, 2007 $where = array(); if ($country) { $where[] = "country='$country'"; } if ($type) { $where[] = "type='$type'"; } if ($Abstract ) { $where[] = "Abstract LIKE '%$Abstract%'"; } if ($where) { $where[] automatically starts an array, so you don't need $where = array(); Quote Link to comment Share on other sites More sharing options...
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