I have a simple IFTTT recipe that triggers whenever an instagram photo is posted with the hashtag #dog (used as an example since it triggers nearly every second and makes it easy to troubleshoot). This works fine. The associated action is with Maker to submit a GET call to a website as pictured below.
This then runs a very simple PHP script as below.
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$dump = $conn->real_escape_string(var_export($_GET, true));
$sql = sprintf("INSERT INTO table (dump) VALUES ('%s');", $dump);
echo $sql;
if ($conn->query($sql) !== TRUE) {
echo "#######Error: ". $conn->error ."#######<br/>";
}
$conn->close();
This should simply 'dump' the $GET array into the MySQL table. When I use the browser to run the PHP script such as
http://www.example.com?data=test&data2=testing
I get the correct response in the DB. However, when triggered through IFTTT I get an empty array.
I've tried all the options for Content Type and both POST and GET Methods but nothing is submitted either way. I have found a workaround by putting the GET call together in the URL field, BUT this isn't tidy and I figure it should work so want to understand the problem. Also, I'd like to POST rather the GET if possible.
I assume the problem to be with the Body and it not being formatted correctly, but I've tried every combination I can think of and what you see here is how the example on IFTTT is displayed.
Does anyone have any ideas or a known solution.
Thanks,
You need to embed the data in the URL itself like you mentioned. That is how GET requests work. If you want to keep it tidy and nor embed the data in the URL, you can use $_POST in your PHP script and then use the body section in IFTTT to embed the data, as seen in the screenshot.
Related
I'm new to this and I know I'm probably doing this entire thing the wrong way, but I've been at it all day trying to figure it out. I'm realizing there's a big difference between programming a real project of my own rather than just practicing small syntax-code online. So, I lack the experience on how to merge/pass different variables/scopes together. Understanding how to fit everything within the bigger picture is a completely different story for me. Thanks in advance.
What I'm trying to do, is to make the function "selectyacht" output data in a different location from where it's being called (in viewship.php). The output data (in viewship.php) needs to be only certain fields (not everything) returned and those results will be scattered all over the html page (not in a table). In addition to that, I have this variable: "$sqlstatement" (in sqlconn.php) that I'm trying to bring outside the function because I don't want to repeat the connection function every time. I tried a global variable, as much as I shouldn't, and it thankfully it gave me an error, which means I have to find a better way.
Basically my struggle is in understanding how I should structure this entire thing based on two factors:
To allow the second conditional statement in sqlconn.php to be typed
as least often as possible for different "selectyacht" functions
that will come in the future.
To allow the connection instance in sqlconn.php to reside outside the function since it will be used many times for different functions.
Returning data in a different place from where it's being called in viewship.php because the call will be a button press, not the results to be shown.
This is probably very simple, but yet it eludes me.
P.S. Some of this code is a copy/paste from other resources on the internet that I'm trying to merge with my own needs.
sqlconn.php
<?php
$servername = "XXXXXXXX";
$username = "XXXXXXXX";
$password = "XXXXXXXX";
$dbname = "XXXXXXXX";
// Instantiate the connection object
$dbconn = new mysqli($servername, $username, $password, $dbname);
// Check if the connection works or show an error
if ($dbconn->connect_error) {
die("Connection failed: " . $dbconn->connect_error);
}
// Create a query based on the ship's name
function selectyacht($shipname) {
global $sqlstatement;
$sqlstatement = "SELECT * FROM ships WHERE Name=" . "'" . $shipname . "'";
}
// Put the sql statement inside the connection.
// Additional sql statements will be added in the future somehow from other functions
$query = $dbconn->query($sqlstatement);
// Return the data from the ship to be repeated as less as possible for each future function
if ($query->field_count > 0) {
while($data = $query->fetch_assoc()) {
return $data;
}
}
else {
echo "No data found";
}
// Close the connection
$dbconn->close();
?>
viewship.php
<html>
<body>
<?php include 'sqlconn.php';?>
<!-- ship being selected from different buttons -->
<?php selectyacht("Pelorus");?>
<br>
<!-- This is the output result -->
<?php echo $data["Designer"];?>
<?php echo $data["Length"];?>
<?php echo $data["Beam"];?>
<?php echo $data["Height"];?>
</body>
</html>
Mate, I am not sure if I can cover whole PHP coding standards in one answer but I will try to at least direct you.
First of all you need to learn about classes and object oriented programming. The subject itself could be a book but what you should research is autoloading which basically allows you to put your functions code in different files and let server to include these files when you call function used in one of these files. This way you will be able to split code responsible for database connection and for performing data operations (fetching/updating/deleting).
Second, drop mysqli and move to PDO (or even better to DBAL when you discover what Composer is). I know that Internet is full of examples based on mysqli but this method is just on it's way out and it is not coming back.
Next, use prepared statements - it's a security thing (read about SQL injection). Never, ever put external variables into query like this:
"SELECT * FROM ships WHERE Name=" . "'" . $shipname . "'";
Anyone with mean intentions is able to put there string which will modify your query to do whatever he wants eg. erase your database completely. Using prepared statements in PDO your query would look like this:
$stmt = $this->pdo->prepare("SELECT * FROM ships WHERE Name = :ship_name");
$stmt->bindValue(':ship_name', $shipname);
Now to your structure - you should have DB class responsible only for database connection and Ships class where you would have your functions responsible eg. for fetching data. Than you would pass (inject) database connection as an argument to class containing you selectYacht function.
Look here for details how implementation looks like: Singleton alternative for PHP PDO
For
'Returning data in a different place from where it's being called'
If I understand you correctly you would like to have some field to input ship name and button to show its details after clicking into it. You have 2 options here:
standard form - you just create standard html form and submit it with button click redirecting it to itself (or other page). In file where you would like to show results you just use function selectYacht getting ship name from POST and passing it to function selectYacht and then just printing it's results (field by field in places you need them)
AJAX form - if you prefer doing it without reloading original page - sending field value representing ship name via AJAX to other page where you use selectYacht function and update page with Java Script
I am trying some code to get value from URL through post method and search database table for that value and get info from the database and encode it into JSON response.
Here is my code :
<?php
//open connection to mysql db
$connection = mysqli_connect("localhost","root","","json") or die("Error " . mysqli_error($connection));
if (isset($_POST['empid'])) {
$k = $_POST['empid'];
//fetch table rows from mysql db
$sql = "select `salary` from tbl_employee where `employee_id` = $k ";
} else {
//fetch table rows from mysql db
$sql = "select `salary` from tbl_employee";
}
//fetch table rows from mysql db
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
//close the db connection
mysqli_close($connection);
?>
I used Postman extension on Chrome and pass the values but it is not returning. Instead it is returning the else part.
Postman Screenshot
Looking at your screen shot, you have not passed body key values, instead you passed params.
Click on Body Tab and then pass key & value pair.
As per your screenshot you are sending your empid through query parameter so you need to access that as follows
<?php
if (isset($_GET['empid'])) {
echo $_GET['empid'];
}else{
// else part
}
also for that you need to Request Url in Postman using GET method.
But as you have stated that you want to send empid through POST in postman, you have to send it through form-data in Postman and access it as $_POST["empid"];. following is the screenshot for your reference
else there is another option where you can send the POST data through body as row json and access it as
$rawPostData = file_get_contents('php://input');
$jsonData = json_decode($rawPostData);
and $post will contain the raw data. And you can send it through postman as in following screenshot.
You have to set the Body to "x-www-form-urlencoded" and adding the variables to be posted
Or try this SO question, its already been answered
I replicated the code and db on my system to figure out the problem. I also added some lines of code before if (isset($_POST['empid'])) { for diagnostics sake:
$method = $_SERVER['REQUEST_METHOD'];
echo $method."<br/>";
The application file is index.php deployed in json directory inside webroot.
When I send any request to http://localhost/json directory (either POST/GET), Apache redirects the request as a GET request to index.php (as configured in my Apache config file). I assume this is what you're experiencing.
But when I send the request to http://localhost/json/index.php, the request is accurately received and processed.
Therefore, I would say the solution is that you need to specify the php file and also set the empid parameter as part of the body in Postman (not as part of the url).
I think you should also check the post if emptyif (isset($_POST['empid']) AND ($_POST['empid']) != ""). to allow php to execute the line before else.Sometimes programming becomes unpredictable.
use if(isset($_REQUEST['empid'])) to test in POSTMAN...
Then use if(isset($_POST['empid'])) to test directly from app...
have a look Issue in POSTMAN https://github.com/postmanlabs/postman-app-support/issues/391
To get the value of a variable from the URL(query string), you need to use either $_GET or $_REQUEST.$_POST represents data that is sent to the script via the HTTP POST method.
So, in your code you just need to do this :
$_REQUEST['empid'] instead of $_POST['empid']
In POST method the data is sent to the server as a package in a separate communication with the processing script. Data sent through POST method will not visible in the URL.
Confirm that in postman Content-Type should be application/x-www-form-urlencoded in request header.
Postman reference doc : https://www.getpostman.com/docs/requests
Hey it sounds like you are just needing to do a GET request to your DB.
You are more than welcome to send in variables via a GET request as well.
GET http://localhost/json?empid=3
You can then get data from your GET request like so $_GET['empid']
I suggest a GET request because I see your not actually posting any data to your server, your just handing in a variable in which you want to use to query with.
I do understand that GET requests are less secure, but in your scenario your POST just doesn't seem to want to work. So a different tack might do you justice.
If you want a value from the URL, you need to use $_GET["empid"] instead $_POST["empid"]
Submitting a form through POST method
By post method of form submission, we can send number or length of data. Sensitive information like password does not get exposed in URL by POST method, so our login forms we should use POST method to submit data. This is how we collect data submitted by POST method in PHP
$id=$_POST['id'];
$password=$_POST['password'];
Collecting data submitted by either GET or POST method
If a page is receiving a data which can come in any one of the method GET or POST then how to collect it ? Here we are not sure how to collect the data. So we will use like this.
$id=$_REQUEST['id'];
$password=$_REQUEST['password'];
Looking at the URL you are requesting, you are sending a GET value within your POST request.
http://localhost/json?empid=3
As you can see here, the url holds the empid variable and so the is send to the server as beeing a GET variable ($_GET)
Use $_GET['empid'] to access this variable, while using $_POST to access the other variables.
You could also use $_REQUEST to access both GET and POST data by the same global.
My jqgrid loads perfectly data, but when I want to commit changes on it, nothing happens on my database.
I don't know exactly when happens the "enter key" event for saving the row, so I don't know where put this code:
jQuery("#lista").saveRow(id, function(){alert("changes saved")}, 'guardar_lista.php');
I already saw this example: JQgrid checkbox onclick update database
but I'm very hard headed about how to use ajax for send the info (sorry, I'm a newbie).
Can you give me a code example of how to send the info with ajax?
Here is my Editurl code:
<?
$dbhost="localhost";
$dbuser="root";
$dbpassword="";
$database="db_proyecto";
$db = mysql_connect($dbhost, $dbuser, $dbpassword) or die("Connection Error: " . mysql_error());
mysql_select_db($database, $db);
if($_POST['oper']=='edit'){
$invid=$_POST['id'];
$tax=$_POST['tax'];
$note=$_POST['note'];
$total=$_POST['total'];
$SQL="update invheader set note='"+$note+"' where invid="+$invid;
mysql_query($SQL,$db);
}
Thank you in advance #ruffin! (Sorry for the delay, I was very busy)
Well, you're going to have to post a little more code for us to know for sure. Your grid should have a value for your editurl, which is the page that's going to process your updated data (iirc). Just to be clear, that url is where you'd insert your mysql_query jive.
Note that you're either going to have jive coming in your $_GET or $_POST collection with the new values from the grid, with "_empty" for the id column if it's a new row (since it's not filled; this is a new row). (Yes, I realize you're updating, but just in case you add soon enough.)
Eg... ("FORM" each time means it's coming from the $_POST array)
FORM: PREFIX :: REV
FORM: FNAME :: Joe
FORM: MNAME :: Frazier
FORM: LNAME :: Test
FORM: SUFFIX :: suffix
FORM: oper :: add
FORM: id :: _empty
Decent code to review here:
http://www.trirand.com/jqgridwiki/doku.php?id=wiki:inline_editing
More on _empty and form editing here:
http://www.trirand.com/jqgridwiki/doku.php?id=wiki:form_editing
So we need your editurl, we need to know that the "php code for update data" is IN that page, and we probably should see what you've got on the page with the grid in full to give a SUPER WONDERFUL answer. ;^)
You should probably also write up a dummy page at your editurl that iterates through all the values from your $_GET and $_POST collections and writes them to a file so that you can check what the grid is sending you -- and that you're getting anything at all. Also, please please PUH-lease use mysql_real_escape_string() around $name and at least an int check for $id!
I know this is probably a really amateur question, but I can't figure this out and I don't know much about PHP or MySQL.
So I have a really simple script that basically allows a user to submit a couple lines of text and their zipcode, then it write it to a database and returns the results on the site. It's a shoutbox essentially.
My client wants the users to be able to filter the results by zipcode. So I have it all setup, and I have a search input where people type in their zipcode, search, and then the PHP returns the submissions from that zipcode.
While I can get the results to show by themselves echoed from the PHP script, how do I get them to display within a specified div within the site? I want it essentially to store a variable, the zipcode, that a user search by, and then use that once the page refreshes to display an updated list that filters out the results that aren't from that zipcode.
Thanks so much for the help.
Chris
Without getting into rewriting etc.:
When you redirect your page, add a query at the end of it that is the zipcode making sure that you don't send any whitespace and that you have an input in a given format:
search.php?zipcode=90210
Then on your search results page, fetch the variable passed and work as usual:
$zipcode = $_GET['zipcode'];
make sure that the zipcode is properly escaped and filtered for nasties.
Make the HTML page a PHP page (usually by making the extension .php). Replace the target div's content with <?php stuff to get and print results ?>.
If you have PHP installed and configured on the web server, it will take certain files (usually those ending in .php) and run them through PHP's interpreter. The PHP interpreter is invoked anytime there is a <?php in the document and it stops parsing content anytime it meets a ?> within the <?php block.
For instance, to get a web page to print "Hello World!" into a div through PHP, I would do:
<div><?php
print 'Hello World!';
?></div>
OK, I figured it out. I just stored my answer in a SESSION variable so that I could call it in my other PHP files.
Now I have one file processing the request, creating the session and the variable, and then redirecting the page. Then I have a PHP included in the page that grabs the SESSION variable and plugs it into my mysql_query to return the proper result!
Not sure if this is the best way of doing it, but it's working.... If anyone knows of a more elegant solution I would love to know it.
Thanks,
Chris
I suggest you do some tutorials on using PHP and database retrieval with mysqli.
<?php
/* Connect to a MySQL server */
$link = mysqli_connect(
'localhost', /* The host to connect to */
'user', /* The user to connect as */
'password', /* The password to use */
'world'); /* The default database to query */
if (!$link) {
printf("Can't connect to MySQL Server. Errorcode: %s\n", mysqli_connect_error());
exit;
}
/* Send a query to the server */
if ($result = mysqli_query($link, 'SELECT Name, Population FROM City ORDER BY Population DESC LIMIT 5')) {
print("Very large cities are:\n");
/* Fetch the results of the query */
while( $row = mysqli_fetch_assoc($result) ){
printf("%s (%s)\n", $row['Name'], $row['Population']);
}
/* Destroy the result set and free the memory used for it */
mysqli_free_result($result);
}
/* Close the connection */
mysqli_close($link);
?>
So I have a chatroom type of database where the text that a user inserts gets stored into a databse as their username in one field and their message in the other. I want to have my page output the database info, so that people can see each others messages.
How do I do this?
Also, is it possible to make a for loop that checks to see if the database has been updated with a new message, therefore it reloads the page? (Then the page outputs the database info again to update everyones messages)
Please help.. i'm so confused.
Take a look at MySQL functions in PHP manual. You need to connect to the server/database and run a select query to get the data from tables.
As for the loop: you could use JavaScript setInterval function and combine that with AJAX call to periodically poll for new records.
Like the others have said, you will want to connect to your database and then query the table that you have the data in.
while($row = mysql_fetch_assoc($results))
{
echo $row['username'] . " said: " . $row['message'] . "<br />";
}
I use mysql_fetch_assoc() instead of mysql_fetch_array() since the arrays are associative arrays (not indexed by integers, but rather by names (associations))
As for displaying the update on the page dynamically, that involves AJAX. Basically what that means is that your page will call out to a background script to get the new records from the database. This would require a new field in your 'messages' table, something like 'msg_delivered' that you could set to '1' when it has been fetched.
You should check out this if you are interested in making an AJAX chat client: http://htmltimes.com/javascript-chat-client-in-jquery.php
To read anything from a mysql database you would use the mysql_connect() and the mysql_query() functions
eg:
$link = mysql_connect('localhost', 'root', '');
$results = mysql_query('select * from messages');
while($row = mysql_fetch_array($results))
{
echo $row['username'] . ': ' . $row['message'].'<br />';
}
To display new messages the best way would be to use AJAX and poll the database from there, either loading a separate page into a DIV or getting XML back and placing into HTML tags. I would recommend using JQuery for these kinds of tasks. Check http://www.sitepoint.com/article/ajax-jquery/ for an example.