mikosiko

Finally!!... good to have it back

^^^

Read carefully what you wrote (in red) and compare it with what you are doing in your select... you should be able to figure it out

In that code you have 2 issues: 1.- You can not mix mysql_ and mysqli_ APIs 2.- mysqli_query() (and in general all the "mysqli_" procedural methods/functions) need to have the database link as a parameter.... read http://php.net/manual/en/mysqli.query.php here are some usage's examples http://php.net/manual/en/mysqli-result.fetch-array.php

food for the thought: (meaning for you to read and investigate the causes of the error and do not create a worst bandaid) http://dev.mysql.com/doc/refman/5.6/en/gone-away.html http://dev.mysql.com/doc/refman/5.6/en/server-system-variables.html#sysvar_wait_timeout http://dev.mysql.com/doc/refman/5.6/en/server-system-variables.html#sysvar_max_allowed_packet and a question: What type/size are your columns called SName (in both tables)?

In case you are willing to explore the alternative Mysql also provide the FEDERATE storage engine to solve your scenario https://dev.mysql.com/doc/refman/5.5/en/federated-storage-engine.html

In your code you are using the variable $conn as your database connection for both your SELECT and for the INSERT. as mac_gyver mentioned before you should be using 2 different variable names for your database connections and use them in the right place

If you go directly to that directory look for the laravel.bat or similar an try to execute it there... The error message is clear enough... Did you try restarting your computer and try?

And you did this? "Make sure to place the ~/.composer/vendor/bin directory in your PATH so the laravel executable is found when you run the laravel command in your terminal."

Are you using this code just as a learning exercise and for one time execution? (hope so) or are you planning using it to create a database and associated table(s) for Each project? If the lastest is the case then IMHO you are using an incorrect approach, you should have only one database and define your tables to manage multiples projects/rfis

$POST is not the same as $_POST

you are generating a <tbody>...</tbody> tags inside of your foreach loop... take those out of the loop and test

Are you sure that you are including the bootstrap CSS in your code, are you?

Obvious queations: - are you sure that your sql query is executing correctly - it is producing any results? - what is the contain of $data after the loop? - what is the contain after json_encode? And last but not least - where are you defining $outp?

After modifying the php.ini did you restarted the Web server? (Apache or whatever) You don't notice anything missing in this line? $my_db = "notes_db" And in this one.. Is that exactly what you wrote using string literals or your were trying to use the previously defined variables? (like $my_host or $my_user, etc.) $con = mysqli_connect("localhost","my_user","my_password","my_db");