i working on ssrs reports , need visible "export data feed" option in ssrs reports. updated tag in render section adding attribute visible="true" in reportconfig file, don't find solution. can 1 me how can solve this. thanks, sid
i need eliminate errors im getting. have shopping cart displayed on products page way of cart update page. below pair of errors (notices) every product listed: notice: undefined variable: obj in c:\xampp\htdocs\project2\browseproducts.php on line 89 notice: trying property of non-object in c:\xampp\htdocs\project2\browseproducts.php on line 89 when click on of add cart buttons, nothing added cart , get: notice: undefined index: stock in c:\xampp\htdocs\project2\cart_update.php on line 25 my code is: <?php $user = "root"; $password = ""; $database = "webstore"; mysql_connect("localhost", $user, $password); @mysql_select_db($database) or die("unable select database"); $current_url = base64_encode("http://".$_server['http_host'].$_server['request_uri']); $query = "select * products order category"; ...
after validation of select combo box have selected not able insert in database. tomcat gives following error java.sql.sqlexception: parameter index out of range (1 > number of parameters, 0). how caused , how can solve it? you error when call of setxxx() methods on preparedstatement , while sql query string not have placeholders ? this. for example wrong : string sql = "insert tablename (col1, col2, col3) values (val1, val2, val3)"; // ... preparedstatement = connection.preparestatement(sql); preparedstatement.setstring(1, val1); // fail. preparedstatement.setstring(2, val2); preparedstatement.setstring(3, val3); you need fix sql query string accordingly specify placeholders. string sql = "insert tablename (col1, col2, col3) values (?, ?, ?)"; // ... preparedstatement = connection.preparestatement(sql); preparedstatement.setstring(1, val1); preparedstatement.setstring(2, val2); preparedstatement.setstring(3, val3); note parameter i...
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