daniel0816 Posted August 30, 2013 Share Posted August 30, 2013 I am trying to write a query that executes when the user types in a number into a textbox and then clicks the search button. The query is supposed to get the JS_Status from the JobStatus table with the JS_ID that matches the value that has been submitted through the textbox and display it in an alert box. The code I have runs but it gives me this weird message in the alert box saying Resource id#3 which I am hoping someone could tell me what that is. Here is my code: <?phperror_reporting(E_ALL);$connect = mysql_connect("dbinfo", "dbinfo", "dbinfo");//select databasemysql_select_db("dbinfo", $connect); $refNum = $_POST["refNum"];$status = mysql_query("SELECT JS_Status FROM JobStatus WHERE $refNum = JS_ID");if($status){ echo '<script language="javascript" type="text/javascript"> alert(\''.$status .'\'); </script>';}else { die ("could not run query"); } ?> Thanks again Quote Link to comment Share on other sites More sharing options...
cyberRobot Posted August 30, 2013 Share Posted August 30, 2013 Have you tried echoing $status on the PHP side to see what it contains? Side note: perhaps you were going to do this later, but it's recommended that you sanitize/validate any user-supplied information before using it in things like MySQL queries. Since $_POST["refNum"] is supposed to be a number, you could utilize ctype_digit(): http://php.net/manual/en/function.ctype-digit.php Quote Link to comment Share on other sites More sharing options...
Psycho Posted August 30, 2013 Share Posted August 30, 2013 When you execute a query, the return value is a resource identifier that points to the results of the query. You need to use one of the mysql_fetch functions to get the actual data. And, that data would be an array - so it would not be usable in JavaScript until you get the value you need from the array first. The less used method would be to mysql_result() to get a particular value from the resource identifier. <?php error_reporting(E_ALL); $connect = mysql_connect("dbinfo", "dbinfo", "dbinfo"); //select database mysql_select_db("dbinfo", $connect); $refNum = intval($_POST["refNum"]); $query = "SELECT JS_Status FROM JobStatus WHERE JS_ID = $refNum"; $result = mysql_query($query); if(!$result) { $message = "There was a problem getting the results"; } elseif(!mysql_num_rows($result)) { $message = "There was no matching record"; } else { $status = mysql_result($result, 0); } echo "<script language=\"javascript\" type=\"text/javascript\"> alert('{$status}'); </script>"; ?> Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.