将表单数据输入到MySql表中，然后在PHP中直接搜索该表[关闭]

Question

After submitting a form and inserting the data into a table, I want to search this table and set a session variable to use on another database insert. Below is the PHP which inserts and then searches the newly updated table.

require_once("db_connect.php");

if($db_server) {

     mysql_select_db($db_database) or die ("<p>Couldn't find database.</p>");
     $query = "INSERT INTO routes (user_id, title, description, startlat, startlng, endlat, endlng) VALUES ('$user_id', '$title', '$desc', '$startlat.', '$startlng', '$endlat', '$endlng')";
     if (mysql_query($query)) {
          $get_id = mysql_query("SELECT * FROM routes WHERE user_id, title = '$user_id', '$title'");
          $route_id = mysql_result($get_id, 0,'id');
          $_SESSION['route_id'] = $route_id;
          echo $route_id; 
     } else {
          $message = ("Insert failed. " . mysql_error() . "<br/>" . $query);
     }  
} else {
     $message = "Error: could not connect to the database.";
}
mysql_close($db_server);

I get this warning:

 Warning: mysql_result() expects parameter 1 to be resource, boolean given in... 

which refers to this line of the above code:

 $route_id = mysql_result($get_id, 0,'id');

Basically what I am attempting to do is get the 'id' of the newly inserted row (the id is the primary key and is auto incremented) and set it as the $route_id so that on my next page I can insert this value into another table as a route_id which is used to link the two tables together.

Answer 1

SELECT * FROM routes WHERE user_id, title = '$user_id', '$title'

The one above is an invalid query. That is why your function fails and doesn't return the expected resource. Its Proper syntax is

SELECT * FROM routes WHERE user_id= '$user_id' and title= '$title'