Hello,
Is there a way to change where a user is taken when logging in? Also when the splash page is enabled, the login page from the splash page takes members to settings instead of the homepage. how do i fix this?
As you can see from the screenshot, there is no option to create an organization. Is this a bug or am I missing something? I'm a decent developer. If someone can tell me the mysql query to run to get this in the "+" dropdown menu that would be great! Thank so much! :)
i'm working on my next thang which involves adding fields to the database and then viewing those fields on each persons profile. Is there a built in function to get the id or profile id of the member whose profile i'm currently viewing. there are a couple functions to get currently logged in ID, but i can't find one to get the "persons" id that im currently looking at. I tried grabbing it from the url but that doesn't work correctly with php for some reason. There may be another way to do this, any help would be appreciated. I'm currently using $.post to get fields from database using a php script, but i can only get currently logged in ID. I need the ID of whatever profile is being viewed. php script below.
<?php
$servername = "localhost";
$username = "my user";
$password = "my pass";
$dbname = "my database name";
$conn = new mysqli($servername, $username, $password, $dbname);
function damnslashes($string)
{
$string=implode("/",explode("\/",$string));
return stripslashes(trim($string));
}
function damncomma($string)
{
$string=implode("},",explode("}",$string));
return stripslashes(trim($string));
}
require_once('inc/header.inc.php');
$url = (isset($_SERVER['HTTPS']) ? "https" : "http") . "://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]"; // trying to just get it from url :/
$urlArray = explode('=',$url);
$last = $urlArray[sizeof($urlArray)-1]; // <-- doesn't work
$id = getLoggedId(); < --- Function that gets logged in id, but that's no good :(
$return_arr = array();
$sql = "SELECT * FROM sys_accounts WHERE id = $id";
$result = $conn->query($sql);
header('Access-Control-Allow-Origin: *');
$row_array = array();
while($row = $result->fetch_assoc()) {
$row_array['FirstName'] = $row['FirstName'];
$row_array['MiddleName'] = $row['MiddleName'];
$row_array['LastName'] = $row['LastName'];
$row_array['Initiation_Date'] = damnslashes($row['Initiation_Date']);
print_r(json_encode($row));
}
then using jquery to disp correct fields...
?>
So here's a question, why can't i seem to access or use json data types? an ajax call w/ html data types works fine, but json is a no go? also, $.getjson does absolutely nothing? not even an error message. please someone an explanation.
Hello,
I'm a developer and I need to add the members to the site via php and SQL. Any tips will be helpful. I'm currently searching for the proper table to insert the queries... I'll also need to verify these members once they try to access the site so any comments there are appreciated. thanks! :)
I get the following error:
There are errors in the following MySQL queries:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''bx_groups_searchable_fields', 'group_name,group_desc', @iCategId, '_bx_groups_o' at line 1
('bx_groups_searchable_fields', 'group_name,group_desc', @iCategId, '_bx_groups_option_searchable_fields', 'list', 'a:2:{s:6:"module";s:9:"bx_groups";s:6:"method";s:21:"get_searchable_fields";}', '', '', 30);