query from a textbox

[daniel0816]

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:

 

 

<?php
error_reporting(E_ALL);
$connect = mysql_connect("dbinfo", "dbinfo", "dbinfo");
//select database
mysql_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

 

[cyberRobot]

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

[Psycho]

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>";

?>