Ajax, PHP, SQL, and JavaScript -
i ran issue cannot find answer, ask questions here, rather stumped. assistance shall appreciated.
this php receives ajax call.
<?php session_start(); $_session["my_data"] = $_post['action']; $db_hostname = 'localhost'; $db_username = 'username'; $db_password = 'password'; $link2 = mysqli_connect($db_hostname,$db_username,$db_password) or die('unable establish db1 connection'); mysqli_select_db($link2, '$db_username'); $orderquery = mysqli_query($link2, "select * table id='".$_session['my_data']."'"); $orderquery = mysqli_fetch_assoc($orderquery); $orderinfo = " <table class='table table-striped'> <tbody> <tr> <td>#: </td> <td>". $_session['my_data'] ."</td> </tr> <tr> <td> full name: </td> <td>". $orderquery['firstname'] . " " . $orderquery['lastname'] ."</td> </tr> <tr> <td> address: </td> <td> ". $orderquery['shipping_address_1'] ."<br> ". $orderquery['shipping_city'] . " " . $orderquery['shipping_zone'] . " " . $orderquery['shipping_postcode'] ." </td> </tr> <tr> <td> card expiry date month: </td> <td> 08 </td> </tr> </tbody> </table><br> "; echo $orderinfo/* . $_post['action']*/; ?>
and script makes call.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script> <script> var mywindow; function myfunction() { mywindow = window.open('', '_blank'); mywindow.document.write("<link rel='stylesheet' type='text/css' href='stylesheet.css'>"); var ordernum; ordernum = document.getelementsbyclassname('summary_value')[0].innerhtml; ordernum = ordernum.replace("#", ""); $.ajax( { type : 'post', data : {'action':ordernum}, url : 'process.php', success: function ( data ) { mywindow.document.write( data ); }, error: function ( xhr ) { alert( "error" ); } }); mywindow.document.write("<br>"); mywindow.document.write(document.getelementbyid("payinfor").innerhtml); } </script> <button onclick='myfunction()' class="btn btn-default">print pay info</button>
it simple issue, can't see it.
i know ajax working because displays of information ($_session['my_data'])
, thinking it's sql statements, syntax looks correct.
mysqli_select_db()
expects parameter 1 connection , parameter 2 database name. in case passing '$db_username'
. code connect database named '$db_username'
because of single quotes. change database name instead. either $db_name
or 'database_name'
should work. while in development mode, try enabling error_reporting(e_all)
catch errors this.
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