Echoed The MySQL Query
I'm currently building a time booking system, and what I have done is created a few tables in my database and populated these with some test data.
The problem I'm having is that I know for a fact that when I click on the 1st Sept for example the first couple of time slots have been taken,so for a test I've used the word 'booked', this works, if the time slot is not taken then it should say 'available', this doesn't do it.
I have echoed the MySQL query and I've test this on the MySQL console and it works, but yet it says 'booked' when its really not.
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Basically I am using PHP 5.1.2 with Apache 2.0.5 on a FreeBSD 5.4 box with Mysql 4.1.1 running. I am attempting to write information to a Mysql table called Jabber. I can connect successfully, query data, and write data pretty much without problem. My problem is this. I am trying to do some duplicate checking on this table so I query the dbase with the following: /* Searching for username in Authreg database and assiging it to variable */ $query1 = "SELECT * FROM `$table` WHERE username='$username'"; $query1_results = mysql_query($query1) or die("Query Failed"); $existing = mysql_result($query1_results, 0,0); Problem is this. When a user already exsits in database I have a conditional if statement that looks like such: /* Conditional statement that searches for exsiting username */ if ($username == $existing) { die ("Please select a unique username"); } else { .... } And it errorr out telling me to, "Please select a unique username" as stated above. When I am adding a new user to the database above I get the following error in my logs or in the browser depending on what --display-errors= is set to in my php.ini.: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 4 in /usr/local/www/sfg/add_jabber/add_test.php on line 38 I know this error is happening as a result of the following line when the user is not present in the database: $existing = mysql_result($query1_results, 0,0);
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Phpcoin And Mysql Query
This maybe off-topic as it is mainly SQL, but I thought I would ask here as I have had many helpful replies from here in the past and also the query is being run from a php script (although examples here are from where I was testing it in phpMyAdmin) I'm using a piece of software called 'phpcoin' along with the popular 'IPN' mod. I am having some problem with it and have got it down to an mySQL function functioning incorrectly. This is the structure of the tables involved: phpcoin_clients: Field Type Null Default cl_id int(11) No 0 cl_join_ts varchar(10) Yes NULL cl_status varchar(20) No pending cl_company varchar(50) No cl_name_first varchar(20) No cl_name_last varchar(20) No cl_addr_01 varchar(50) No cl_addr_02 varchar(50) No cl_city varchar(50) No cl_state_prov varchar(50) No cl_country varchar(50) No cl_zip_code varchar(12) No cl_phone varchar(20) No cl_email varchar(50) No cl_user_name varchar(20) No cl_user_pword varchar(50) No cl_notes text No cl_groups int(11) No 0 phpcoin_clients_contacts: Field Type Null Default contacts_id int(11) No contacts_cl_id int(11) No 0 contacts_name_first varchar(20) No contacts_name_last varchar(20) No contacts_email varchar(50) No The sql query is as follows: SELECT cl_id from phpcoin_clients, phpcoin_clients_contacts WHERE (cl_email='email@domain.com') OR (contacts_email='email@domain.com') AND contacts_cl_id=cl_id
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MySQL Workaround Query ?
How is it possible to work around a security issue with an Apache server running phpSuExec and which doesn't allow the creation of arbrirary MySQL username logins ? I'd have thought that most 'payment & services' sites would use the following sytem and all referencing a MySQL database to check details; 1) Details input by user first time. 2) Upon return, input login name and password. 3) Click on button for number of days (rerouting to PayPal,WorldPay or whoever) for services. I am informed that it is possible to access the cPanel by using a cPanel script but I can't find script examples anywhere.
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