I'm looking to pass an array from a separate database within wordpress and echo it out into the content, having a bit of trouble understanding how.
Here is my current code in functions.php:
function get_lead() {
$wpdb = new wpdb('root','password','database', 'localhost');
if ($wpdb->connect_errno) {
die("Failed to connect to MySQL: (" . $db->connect_errno . ") " . $db->connect_error);
}
// Get lead info
if(isset($_SESSION['lead_id'])) {
global $lead;
$table_name = $wpdb->prefix . 'customers';
$lead = $wpdb->get_row($wpdb->prepare("SELECT * FROM $table_name WHERE id = '".$_SESSION['lead_id']."'", ARRAY_A));
$error = $wpdb->print_error();
return $lead;
}
}
I originally tried this in a custom php file included in a custom wordpress header via require_once 'filename.php';, here is the code for that:
if(isset($_SESSION['lead_id'])) {
$lead_query = $db->query("SELECT * FROM customers WHERE id = '".$_SESSION['lead_id']."'");
$lead = $lead_query->fetch_assoc();
$lead_query->free();
}
I know this gets the data fine and passing the variable if I place it in header.php and echo out there, but I can't get WordPress to load it into the content within the editor, it's just an empty if I echo $lead['postcode']; there for example.
I've tried setting up a shortcode but can't seem to return an array with that, and the function action doesn't seem to work either, I just get NULL when I var_dump($lead);
It's a 3 step form with a thank you page at the end. I want to echo one piece of data into the first step of the form and two pieces of data onto the thank you page.
Hope this makes sense, any help would be great :)
Related
I am using similar syntax in my blog. However, On my forum, nothing happens! This has been such an infuriating thing to tackle, as everything seems to be working exactly as my blog did. Here's my code I pass through and call the delete_post page
CHUNK FROM VIEWPOST.PHP
while($row = mysqli_fetch_array($result)){
echo '<tr>';
echo '<td class="postleft">';
echo date('F j, Y, g:i a', strtotime($row['forumpost_Date'])) . "<br>" .$row['user_Name']. "<br>" .$row['forumpost_ID'];
echo '</td>';
echo '<td class="postright">';
echo $row['forumpost_Text'];
echo '</td>';
if(isset ($_SESSION['loggedin']) && ($_SESSION['user_AuthLvl']) == 1){
echo '<td class="postright">';
echo '<a class= "btn btm-default" href="#">Edit</a>';
echo '<a class= "btn btm-default" href="delete_post.php?forumpost_ID='.$row['forumpost_ID'].'">Delete</a>';
echo '</td>';}
else if(isset ($_SESSION['loggedin']) && ($_SESSION['user_ID']) == $row['forumpost_Author']){
echo '<td class="postright">';
echo '<a class= "btn btm-default" href="#">Edit</a>';
echo '<a class= "btn btm-default" href="delete_post.php?forumpost_ID='.$row['forumpost_ID'].'">Delete</a>';
echo '</td>';}
echo '</tr>';
}echo '</table>';
DELETE POST FUNCTION
<?php
include ('header.php');
include ('dbconnect.php');
//A simple if statement page which takes the person back to the homepage
//via the header statement after a post is deleted. Kill the connection after.
if(!isset($_GET['forumpost_ID'])){
header('Location: index.php');
die();
}else{
delete('hw7_forumpost', $_GET['forumpost_ID']);
header('Location: index.php');
die();
}
/********************************************
delete function
**********************************************/
function delete($table, $forumpost_ID){
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
?>
Now it is showing the ID's as intended, it just simply does not delete the post. It's such a simple Query, I don't know where my syntax is not matching up!
EDIT FOR DBCONNECT.PHP
<?php
/*---------------------------------------
DATABASE CONNECT PAGE
A simple connection to my database to utilize
for all of my pages!
----------------------------------------*/
$host = 'localhost';
$user = 'ad60';
$password = '4166346';
$dbname = 'ad60';
$connectDB = mysqli_connect($host, $user, $password, $dbname);
if (!$connectDB){
die('ERROR: CAN NOT CONNECT TO THE DATABASE!!!: '. mysqli_error($connectDB));
}
mysqli_select_db($connectDB,"ad60") or die("Unable to select database: ".mysqli_error($connectDB));
?>
Ok, I saw this and I would like to suggest the following:
In general
When you reuse code and copy paste it like you have done there is always the danger that you forget to edit parts that should be changed to make the code work within the new context. You should actually not use code like this.
Also you have hard coded configuration in your code. You should move up all the configuration to one central place. Never have hard coded values inside your functional code.
Learn more about this in general by reading up about code smell, programming patterns and mvc.
To find the problem
Now to fix your problem lets analyse your code starting with delete_post.php
First check if we actually end up inside delete_post.php. Just place an echo "hello world bladiebla" in top of the file and then exit. This looks stupid but since I can't see in your code if the paths match up check this please.
Now we have to make sure the required references are included properly. You start with the include functionality of php. This works of course, but when inside dbconnect.php something goes wrong while parsing your script it will continue to run. Using require would fix this. And to prevent files from loading twice you can use require_once. Check if you actually have included the dbconnect.php. You can do this by checking if the variables inside dbconnect.php exist.
Now we know we have access to the database confirm that delete_post.php received the forumpost_ID parameter. Just do print_r($_GET) and exit. Check if the field is set and if the value is set. Also check if the value is actually the correct value.
When above is all good we can go on. In your code you check if the forumpost_ID is set, but you do not check if the forumpost_ID has an actual value. In the above step we've validated this but still. Validate if your if
statement actually functions by echoing yes and no. Then test your url with different inputs.
Now we know if the code actually gets executed with all the resources that are required. You have a dedicated file that is meant to delete something. There is no need to use a function because this creates a new context and makes it necessary to make a call and check if the function context has access to all the variables you use in the upper context. In your case I would drop the function and just put the code directly within the else statement.
Then check the following:
Did you connect to the right database
Is the query correct (echo it)
Checkout the result of mysqli_query
Note! It was a while ago since I programmed with php so I assume noting from the codes behavior. This is always handy. You could check the php versions on your server for this could also be the problem. In the long run try to learn and use MVC. You can also use frameworks like codeigniter which already implemented the MVC design pattern.
You have to declare $connectDB as global in function.
function delete($table, $forumpost_ID){
global $connectDB;
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
See the reference about variable scope here:
http://php.net/manual/en/language.variables.scope.php
please try to use below solution.
<?php
include ('header.php');
include ('dbconnect.php');
//A simple if statement page which takes the person back to the homepage
//via the header statement after a post is deleted. Kill the connection after.
if(!isset($_GET['forumpost_ID'])){
header('Location: index.php');
die();
}else{
delete('hw7_forumpost', $_GET['forumpost_ID'], $connectDB);
header('Location: index.php');
die();
}
/********************************************
delete function
**********************************************/
function delete($table, $forumpost_ID, $connectDB){
$table = mysqli_real_escape_string($connectDB, $table);
$forumpost_ID = (int)$forumpost_ID;
$sql_query = "DELETE FROM ".$table." WHERE id = ".$forumpost_ID;
$result = mysqli_query($connectDB, $sql_query);
}
?>
I wish this solution work for you best of luck!
The following code works well when the correct URL is used like...
http://example.com/user.php?id=joe
Where I need help is where do I add an if statement and how should it be written to provide either a default landing page or better, an input field so the visitor can retype the user name and submit it to have the correct page load.
<?php
$pdo = new PDO('mysql:host=localhost;dbname=users', 'root', '');
// retrieve member's data
$ps = $pdo->prepare("SELECT * FROM members WHERE id = ?");
if(isset($_GET["id"])) {
$ps->execute(array($_GET["id"]));
}
$result = $ps->fetch(PDO::FETCH_ASSOC);
extract($result);
echo $First_Name . ' - ' . $Email;
Any help?
if(!empty($_GET['id'])) {
// YOUR PREVIOUS CODE HERE
} else {
// REDIRECT CODE HERE
}
I assume you'd want to wrap all the code with the if statement so it doesn't run unless an id is present.
<?php
if( isset( $_GET['id'] ) ) {
//retrieve member data here and render normal page
}
else {
//render landing page
}
?>
It might be wise to put the code for rendering the two different pages into different functions to help with organization as well.
I want to use one single page with pre-defined divs, layout etc. as basis so that when a product is clicked on from elsewhere it loads that product info onto the page?
They way im doing it ill be sitting here till about 2020 still typing out product info onto pages.
EDIT*************
function product ()
{
$get = mysql_query ("SELECT id, name, description, price, imgcover FROM products WHERE size ='11'");
if (mysql_num_rows($get) == FALSE)
{
echo " There are no products to be displayed!";
}
else
{
while ($get_row = mysql_fetch_assoc($get))
{
echo "<div id='productbox'><a href=product1.php>".$get_row['name'].'<br />
'.$get_row['price']. '<br />
' .$get_row['description']. '<br />
' .$get_row['imgcover']. "</a></div>" ;
}
}
}
In addition one problem I have with that code is that the <a href> tag only goes to product1.php. Any ideas how I can make that link to blank product layout page that would be filled with the product info that the user has just clicked on, basically linking to itself on a blank layout page.
Thanks any help would be great!
Thanks Maxyy
Since you dont have code this is a general way of doing this. What you want is a template for the product page
Query the database
load the data into a variable
make a script that will print out the data from the variable into a product page
somescript.php
<?php
$productid = $_REQUEST['productid']; //Of course do sanitation
//before using get,post variables
//though you should be using mysqli_* functions as mysql_* are depreciated
$result = mysql_query("select * from sometable where id='{$productid}");
$product = mysql_fetch_object($result);
include("productpage.php");
productpage.php
<div class="Product">
<div class="picture"><img src="<?php echo $product->imghref;?>" /></div>
<div class="price"><?php echo $product->price;?></div>
</div>
so on and so fourth. Included scripts use whatever variables are currently in the scope of the calling function
If you are meaning to load the products into the same page without doing another page load you will need to use ajax. Which is javascript code that use XHR requests to return data from a server. You can either do pure javascript or a library like jQuery to simplify the process of doing a xhr request by using $.ajax calls.
I know this question has been asked over 4 years ago, but since there's been no answer marked as right, I thought I might chip in.
First, let's upgrade from mysql and use mysqli - my personal favorite, you can also use PDO. Have you tried using $_GET to pull the id of whatever product you want to see and then displaying them all together or one at a time?
It could look something like this:
<?php // start by creating $mysqli connection
$host = "localhost";
$user = "username";
$pass = "password";
$db_name = "database";
$mysqli = new mysqli($host, $user, $pass, $db_name);
if($mysqli->connect_error)
{
die("Having some trouble pulling data");
exit();
}
Assuming the connection was made successfully we move on to checking for an ID being set. In this case I check it via an URL param assumed to be id. You can make it more complex, or take a different approach here.
if(isset($_GET['id']))
{
$id = htmlentities($_GET['id']);
$query = "SELECT * FROM table WHERE id = " . $id;
}
else
{
// if no ID is set, just bring all the results down
// then you can modify how, and which table the results
// are being used.
$query = "SELECT * FROM table ORDER BY id"; // the query can be changed to whatever you would be prefer
}
Once we have decided on a query we go on to start querying the database for information. I have three steps:
Check query >
Check table for records >
Loop through roles and create an object for each.
if($result = $mysqli->query($query))
{
if($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
// you can set up your element here
// you can set it up in whatever way you want
// to see your product being displayed, by simply
// using $row->column_name
// each column becomes an object here. So your id
// column would be pulled using $row->id
echo "<h1>" . $row->name . "</h1>";
echo "<p>" . $row->description . "</p>";
echo "<img src=" . $row->image_path . ">";
// etc ...
}
}
else
{
// if no records match the selected ID
echo "Nothing to see here...";
}
}
else
{
// if there's a problem with the query
echo "A slight problem with your query.";
}
$mysqli->close(); // close connection for safety
?>
I hope this answers your question and can help you if you are still stuck on this problem. This is the bare skeleton of what you can do with MySQLi and PHP, you could always use some Ajax to make the page more interactive, and user-friendly.
Adding content to a page on click needs to be done in either Javascript or in JQuery.
You can use ajax call to retrive the needed data from php page, Syntax is here.
Or you can also load a php page to a div content with .load() function in JQuery, Syntax is here.
Im using ThickBox 3.1 and regretting it since it is no longer supported and Im new to programming so I could use all the help I can get.
Im almost done setting up several Thickboxes on my site but my last one requires a variable to be passed from the parent page to ThickBox and I just can't seem to pick it up in my php script. I have a parent page that is getting the variable from a currently selected drop down menu. When I hover over the link that will open the ThickBox Modal I see:
http://localhost/forms/modal_product.html?strUser=100&TB_iframe=true&height=300&width=590
which is great because I need the strUser variable. So the ThickBox opens I enter some information into a short form just like my other ThickBoxes but nothing happens. When I look under firebug to see the AJAX response it informs me that:
<b>Notice</b>: Undefined index: strUser in <b>C:\xampp\htdocs\forms\item_add.php</b> on line <b>3</b
Ive tried seeing what $_SERVER['QUERY_STRING']; comes up with and it is NULL.
I think my question is generic in that I just don't know where my _GET variable is going? Can anyone with experience with ThickBox help me out? Ive read my butt off and I realize in older versions of ThickBox it was harder to pass variables and it required hacks but I should be able to do this the way I have it. What am I doing wrong? Is there an easier way to pass variables or am I just missing something obvious?
Also when I go to my form page directly and not as a link through the parent window, if I manually put a GET variable into the URL it still gives me the same errors which leads me to believe I'm just missing something basic. Here is my php code.
<?php
include '../dbc.php';
$getman = ($_GET['strUser']);
$manid1 = mysql_query("SELECT manufacturer_id FROM manufacturers WHERE man_name='$getman'");
$manid11 = mysql_fetch_array($manid1);
$manid21 = $manid11[0];
if ($_POST) {
// Collect POST data from form
$item_num = stripslashes($_POST['item_num']);
$descript = stripslashes($_POST['descript']);
$quanti = stripslashes($_POST['quanti']);
$fdaa = stripslashes($_POST['fdaa']);
}
$params = $_SERVER['QUERY_STRING'];
$domain = $_SERVER['SCRIPT_NAME'];
$queryString = $_SERVER['REQUEST_URI'];
$stmnt2 = mysql_query("INSERT INTO products (product_id, item_number, description, quantity_per_unit, fda_approved, manufacturer_id) VALUES ('NULL', '" . $item_num . "', '" . $descript . "' ,'" . $quanti . "', '" . $fdaa . "' , '" . $manid21 . "')");
$resp['status'] = 'success';
if ($stmnt2) {
$resp['errmessage'] = "Item submitted. Submit another item or click close.";
} else {
$resp['errmessage'] = $params;
}
echo json_encode($resp);
?>
I worked it out. I had to pull the variable in Jquery from the URL, put it into a hidden field and then run my php script.
I am working on something it has 2 pages. One is index.php and another one is admin.php.I am making CMS page where you can edit information on the page yourself. Then it will go to the database, where the information is stored. I also have to have it where the user can update the information on the page. I am getting a little bit confused here.For instance here I am calling the database and I am starting a function called get_content:
<?php
function dbConnect(){
$hostname="localhost";
$database="blank";
$mysql_login="blank";
$mysql_password="blank";
if(!($db=mysql_connect($hostname, $mysql_login, $mysql_password))){
echo"error on connect";
}
else{
if(!(mysql_select_db($database,$db))){
echo mysql_error();
echo "<br />error on database connection. Check your settings.";
}
else{
return $db;
}
}
function get_content(){
$sql = "Select PageID,PageHeading,SubHeading,PageTitle,MetaDescription,MetaKeywords From tblContent ";
$query = mysql_query($sql) or die(mysql_error());
while ($row =mysql_fetch_assoc($query,MYSQL_ASSOC)){
$title =$row['PageID'[;
$PageHeading =$row['PageHeading'];
$SubHeading = $row['SubHeading'];
$PageTitle = $row['PageTitle'];
$MetaDescription =$row['MetaDescription'];
$MetaKeywords = $row['MetaKeywords'];
?>
And then on the index page and I am going to echo it out in the spot that someone can change:
<h2><?php echo mysql_result($row,0,"SubHeading");?>A Valid XHTML and CSS Web Design by WG.</h2>
I do know that the function is not finished I am still working on that part. What I am wondering is am I echoing it out right or I am way off. This is my first time messing with CMS in php and I am still learning it. I am working with navicat and text pad on this, yes I know it is old school but that is what I am being shown with. But my index is a form not a blog. I have seen many of CMS pages for blogs not to many to be used with forms. Any input will be considered thanks for reading my question.
Your question is a bit confusing and your code very incomplete. I'ts hard to say if you do it the right way since I don't see the rest of the script. You need to connect to the database there as well and get your data. The $row variable only exists in the while statement inside you function get_content() though.
You could complete the get_content() and use it in the index.php as well. Remember that the variables you define inside a function only is available there though. If you need the data outside that function you need to return the values you need and save them to some other variable there. Put if you do the same as you've started doing in the get_content() function in index.php, then you just have to echo the variables you define. Like this:
<h2><?php echo $SubHeading; ?></h2>
or you could also do it like this somewhere inside the php tags:
echo '<h2>{$SubHeading}</h2>';
I hope that answers your question.
EDIT:
What you need in the index.php page is exactly what you seem to be doing in the admin file. You need to connect to db using mysql_connect() and select db with mysql_select_db(). You then need to select the data from the db using the appropriate query with $query = mysql_query($sql). If it's more then one row you want to display you need to put it in a while loop otherwise (which seems to be the case here) you just need to do one $row = mysql_fetch_assoc($query). After that you can get the data using $row['column_name']. If you have more than one row you can just use $row['column_name'] in side the while loop to get each consecutive row's data.
Here is an example index.php:
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password') or
die('Could not connect: ' . mysql_error());
mysql_select_db('database_name')) or die('Could not select database: ' .
mysql_error());
$sql = "SELECT SubHeading FROM tblContent WHERE PageID='1' LIMIT 1;";
$query = mysql_query($sql);
$row = mysql_fetch_assoc($query);
echo '<h2>{$row[\'SubHeading\']}</h2>';
mysql_close();
?>
This is just what you need to display the SubHeading from you database. You probably also need to handle your form and save the submitted data to the database in your admin.php file.