I'm trying to assign the return value of a php script to a js variable. I have this:
var email = jQuery("input[name=email]").val();
var emailRegex = /^([\w-\.]+#([\w-]+\.)+[\w-]{2,})?$/;
var exists='<?php
$query = "SELECT * FROM rss_members WHERE email_id=\"something#email\"";
$results = mysql_query($query);
$results = mysql_num_rows($results);
echo $results;
?>';
console.log(exists)
The query works and I get the correct results back, but I want to replace "something#email.com" with the email variable, but if I write something like email_id=\"'+email+'\"..., the query result comes back incorrect. What am I doing wrong?
You aren't taking in consideration execution time.
When you make a request to a server (enter www.google.com for example) the server gets a request, and replies with a HTML page. The server in your case is in PHP and it sends a HTML page with some javascript included in it. After the browser receives the HTML page, it then interprets it, and run javascript code.
So basically, your php code ran without access to the email variable. If you wanted to have access to server-side information after the page loaded, you would need to make an ajax request
Related
I have a database in phpMyAdmin that I have set up with XAMPP. I am working on a website that shows statistics from the user inputted scores in the database. Say that I would like to show the score percentile to the user after they submit their score: where do I write the query for that? In the HTML/PHP code? In phpMyAdmin? Somewhere else like a workbench or PopSQL?
I have successfully gotten the website to display the average score in any given table by writing this code to the HTML file:
<?php
$sql = "SELECT AVG(score) AS score FROM $input_subject";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_object($result) ;
echo nl2br("Average score for $input_subject: \n \n"
. round($row->score));
?>
It works, but when I search for tutorials for example for the percentile query or something like the CHECK function (to not accept anything less than 0 or more than 120) it seems that the queries are always written somewhere else than the HTML file.
Also, when I try to write the SQL code in phpMyAdmin, it always shows a bunch of error messages, even though I copy/pasted it in (changing the table names etc., of course).
So, do I need to look into some other programmes were to write the queries in or can I just write them into the HTML-file or in the phpMyAdmin? I'm a total newbie with this so anything helps!
Yes, you can write SQl queries(PHP) before your html code and also in between your html code, but make sure to change file extension to .php from .html otherwise PHP code will be printed on browser as it is.
"it seems that the queries are always written somewhere else than the HTML file."
we do this while using AJAX. we send data to server (a PHP file where we process data received) and server will give response. All this happens behind the scenes without page reloading.
in your case you can create a new PHP file lets say process.php, add your PHP code into it.
process.php
<?php
$sql = "SELECT AVG(score) AS score FROM $input_subject";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_object($result) ;
echo nl2br("Average score for $input_subject: \n \n"
. round($row->score));
?>
send user inputted scores from HTML file through ajax to PHP file (process.php) and response received from that file can be displayed in html file without page reloading.
you can go through the ajax api by following link jQuery-AJAX
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.
I've just designed my first form in HTML and a PHP page to display the results. In the form the user inputs some codes in response to some questions, a bit like a multiple choice, so for example, these are "ABC". The PHP page displays the code to the user as a link, which when clicked will go to a bookmark (a link within the same page) with the ID #ABC. This was achieved with simple manipulation of the PHP variable as follows:
<?php
$code = "ABC"
$part1 = '<a href="mywebpage.php#';
$part2 = '">Go to this code</a>';
$string = $part1.$code.$part2;
echo $string;
?>
(i.e. Link in the page says "go to this code" and when clicked will go to section with bookmark ABC)
This all works fine, but I simply need to know if there is a way of error trapping so that if a bookmark does not exist for the code entered, a message can be displayed to the user instead? Can this be done using the PHP variable, or do I need to use JavaScript? One work around may be to search the web page for the ID "#ABC'. Is it possible to do this? Another option would be to store an array of valid codes on the server then query this before setting the bookmark, but I want to keep it as simple as possible. Any help appreciated, thanks.
What you call a "bookmark" we call a hash. And when you say "go to a bookmark" you mean a hash change. Hash changes do not make an additional request to the server, it is all handled on the client-side, therefore this must be done with JavaScript and not PHP.
So let's just do some simple JavaScript on hash change window.onhashchange that will search for an element with that ID and if it's not found alert something.
window.onhashchange = function(){
if(!document.getElementById(location.hash){
alert("not found");
}
}
I have 2 php pages: query.php and result.php.
In query.php, I am executing a query (select) statement. It's returning a resultset
$rs = mysql_query($query);
Now I want to return this resultset from query.php to another page result.php and work with it. Like this:
In query.php:
return $rs
and in result.php:
$result = executeQuery($query) // we get the resultset in this variable
while ($row == mysql_fetch_array($result){
//do something
}
If the above is not recommended, please provide me with alternatives. But I want the query function and resultset in different pages.
You could just include results.php in your query.php page if you're just looking to keep the code separate in the source files but aren't actually required to redirect from one page to another:
In query.php:
$rs = mysql_query($query);
include "results.php";
In results.php:
while ($row == mysql_fetch_array($rs){
//do something
}
As far as trying to "return $rs" from one page to another that's not how PHP works. The return statement is only valid within a function. If you want to actually pass data from one PHP page to another and will be redirecting to that other page then you'll need to use either a session, a cookie, pass it in the URL (i.e. use GET) or use curl and add it as a POST var.
If this is really the way it must be, store the result set in a database somewhere or in a file and give each result a unique name. Then pass that name to the next page so it can be retrieved.
query.php will redirect to result.php?result_set=ab24sdfsdfklls for instance.
This has the added advantage that you can use the result_set as often as you want. Visitors can have multiple result sets during one visit. They can share the URL of the result set page with other people, etc.
Just be sure to eventually prune the data store as it will just keep on growing, but that's another matter entirely.
i want to pass a url value from a php script to another.I have a database in which i have stored some feeds.I have given some weight values to these feeds and a php script grabs a feed url randomly based on their weights.I want to take the feed url which has been grabbed by the script and then pass this url in another php script where it will be parsed with simplepie in order to show its content.
I am posting the codes of the two files right here:
this is the first script which grabs randomly the feed
http://pastebin.com/2ciQ87Es
this is the second script in which i want to pass the value and makes the parsing of the feed
http://pastebin.com/eN5qG29e
Have you something to recommend??
thanks in advance
Would a $_SESSION not suffice?
In the first script:
session_start();
$_SESSION['session_name'] = 'value';
In the second script:
session_start();
print $_SESSION['session_name'];
On second thoughts, could you not pass the value in a query string, to the second page.
second-page.php?key=value
You could wrap grabbing the feed url from the database in a function, and just include that file like any other php file, and then call that function.
//// feedgrabber.php
<?php
function grabber(){
$query = "SELECT * FROM `feeds`";
//takes all the feed that are declared
$result = mysql_query($query);
$data = array();
while($output = mysql_fetch_assoc($result)) {
$data[] = $output; // assigns feeds to an array called $data, one after the other, in they go!
}
return randomchoice($data); // finds a random feed by calling the function
}
?>
and then on the page where you need it:
require('feedgrabber.php');
$feed = grabber();