I am not familiar with JSON handling in a web dev context, so would appreciate a little guidance.
I have a login utilising a web form - if a successful login has been made, a JSON array is returned as follows:
{
"result": "success",
"message": "Login Successful",
"user": {
"name": "Foo Bar",
"email": "foo#bar.com",
"unique_id": "59bea8b7d56a63.8888888"
}
}
My DB Operations returns the JSON to my Functions.PHP. I then do the following:
$response["result"] = "success";
$response["message"] = "Login Successful";
$response["user"] = $result;
$json = json_encode($response);
It's here that I am stuck because I want to send the encoded JSON to the web page that will be open on success, as I want to make further DB queries based on the user unique_id.
I have tried on page 1:
$response["result"] = "success";
$response["message"] = "Login Successful";
$response["user"] = $result;
$json = json_encode($response);
header('Location: http://example.co.uk/quiz/dashboard.php/'.$json);
Page 2
<?php
$data_get = $_REQUEST['user'];
?>
<script type="text/javascript">
var mydata =<?php echo $data_get; ?>;
</script>
But I have not been able to echo out the user data. I want to be able to retrieve the encoded array on page 2 and then decode it and store the name/email/unique_id in variables to use when needed on page 2.
You line:
header('Location: http://example.co.uk/quiz/dashboard.php/'.$json);
seams to be the problem.
If you really want to transfer the whole json string via GET request you need to define a name for it. try this:
header('Location: http://example.co.uk/quiz/dashboard.php?data='.$json);
And then on page 2 this is how you receive it:
$response = json_decode($_GET['data']);
After that you can access the $response array same as on page 1.
Btw: it might be easier to store the data in a session.
In your page 1
$response["result"] = "success";
$response["message"] = "Login Successful";
$response["user"] = $result;
$json = json_encode($response);
header('Location: http://example.co.uk/quiz/dashboard.php/?data='.$json);
In Page 2 you will get values like that
$data = json_decode($_GET['data']);
echo $data->message. "</br>";
echo $data->result."</br>";
echo $data->user."</br>";
Or if you want assign json to javascript variable then
<script type="text/javascript">
var mydata =<?php echo $_GET['data']; ?>;
</script>
With what you are doing try this in file 2:
$decoded=json_decode(array_keys($_REQUEST)[0]);
echo $decoded->user;
Related
I new in programming. Currently, I develop a system that registration part. The registration part is successfully saved to the database. What I want to know is how to popup an alert dialog with one button e.g "Ok" after registration was successful and redirect to another page, such as home page. Now I only echo "successfully saved"
Below is my current code
<?php
require "DbConnect.php";
$name = $_POST['name'];
$badgeid = $_POST['badgeid'];
$position = $_POST['position'];
$department = $_POST['department'];
$factory = $_POST['factory'];
$reviewer = $_POST['reviewer'];
$title = $_POST['title'];
$year = $_POST['year'];
$month = $_POST['month'];
$suggestionwill = $_POST['suggestionwill'];
$present = $_POST['present'];
$details = $_POST['details'];
$benefit = $_POST['benefit'];
$sql_query = "INSERT INTO topsuggest (name,badgeid,position,department,factory,
reviewer,title,year,month,suggestionwill,present,details,benefit) VALUES('$name','$badgeid','$position','$department','$factory','$reviewer','$title','$year','$month','$suggestionwill','$present','$details','$benefit')";
if(mysqli_query($conn,$sql_query))
{
echo "<p id='msg'></p>";
}
else
{
echo "Error!! Not Saved".mysqli_error($con);
}
?>
Just use php header and use javascript to alert a message .
if(mysqli_query($conn,$sql_query))
{
echo "<script>alert('Successfuly Saved');</script>";
header('Location: PATH TO BE REDIRECTED');
}
For a example
if(mysqli_query($conn,$sql_query))
{
echo "<script>alert('Successfuly Saved');</script>";
header('Location: ../Insert/Index.php');
}
Please note that space between Location: is compulsory
After inserting data you can simply redirect to your interested page with a success message like:
header("location:page_of_interest.php?msg=Record Inserted");
and on page_of_interest.php you can simply check for msg and show if it is set like:
if(isset($_GET['msg'])){
echo $_GET['msg'];
}
I am trying to call JavaScript function in php and pass it value of a php json array type variable as an argument. I found from search on SO forum one way to do this is to echo/print_r the variable value to a js var inside a js script within php code. I am trying do it this way but I am not able to recover from 'unexpected token: identifier error ' while doing so.
I am trying to figure out the reason of syntax error but couldn't. I tried different ways what I found; by putting quotes single/double around php part within the script, without quotes as some places I found solution with quotes some places without but no one seems working.
Here is my code. It will be very helpful if someone sees it and point what is causing this error.
<script>
dspChrt(WData);
.......
</script>
<HTML>
<?php
$WData;
require("Connection.php");
try {
$stmt = $conn->prepare("Select humidity, temperature FROM weatherdata");
$stmt->execute();
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC);
foreach($stmt->fetchAll() as $k=>$v) {
$WData = json_encode($v);
//print_r($WData);
}?>
<script>
var Wdata = <?php print_r($WData);?>
dspChrt(WData);
consol.log(WData);
</script>
<?php
}
catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
?>
</HTML>
First of all you need to parse the JSON using JSON.parse.
Also you need to change the sequence of php and javascript code.
If you want to assign php data to Javascript variable, please retrieve data using php first and write javascript code below it.
For example :
<?php
$v = array(1,2,3);
$data = json_encode($v);
?>
<script>
var WData = JSON.parse('<?php echo $data; ?>');
dspChrt(WData);
</script>
You should encode your PHP into JSON to pass it to JavaScript.
And you should prepare your data first.
<?php
$data = array('xxx'=>'yyy');
?>
<script>
var data = <?php echo json_encode($data); ?>;
//then in js, use the data
</script>
for your code, there are too many errors to be fixed:
<HTML>
<?php
require("Connection.php");
$stmt = $conn->prepare("Select humidity, temperature FROM weatherdata");
$stmt->execute();
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC);
$WData = array();
foreach($stmt->fetchAll() as $k=>$v) {
$WData[] = $v;
}
?>
<script>
var WData = <?php echo json_encode($WData);?>;
console.log(WData);
dspChrt(WData);
</script>
</HTML>
I am trying to pass a javascript variable into an SQL WHERE query and I keep getting null in return.
On-click of a button, the buttonClick function is ran:
<script>
var var1;
function buttonClick(elem){
var1 = elem.src //this gets the url from the element
var path = var1.slice(48); //this cuts the url to img/art/9/1.jpg
ajax = theAjax(path);
ajax.done(processData);
ajax.fail(function(){alert("Failure");});
}
function theAjax(path){
return $.ajax({
url: 'info.php',
type: 'POST',
data: {path: path},
});
}
function processData(response_in){
var response = JSON.parse(response_in);
console.log(response);
}
</script>
Here is the code stored in the info.php file:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$json = json_encode($result3);
echo $json
?>
As you can see, once I click the button, the buttonClick() function is ran and a variable stores the image path or src. That path variable is send to theAjax() function where it is passed to the info.php page. In the info.php page, the SQL WHERE query is ran and returned to the processData() function to be parsed and printed in the developer console. The value printed shows null.
Below is a picture of what I am trying to get from the database:
1.Check that path is correct or not? you can check inside jquery using console.log(path); or at PHP end by using print_r($_POST['path']);
2.Your Php code missed connection object as well as record fetching code.
<?php
if(isset($_POST['path'])){
$path = $_POST['path'];
$conn = mysqli_connect ('provide hostname here','provide username here','provide password here','provide dbname here') or die(mysqli_connect_error());
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'");
$result = []; //create an array
while($row = mysqli_fetch_assoc($result3)) {
$result[] = $row; //assign records to array
}
$json = json_encode($result); //encode response
echo $json; //send response to ajax
}
?>
Note:- this PHP query code is wide-open for SQL INJECTION. So try to use prepared statements of mysqli_* Or PDO.
mysqli_query() required 1st parameter as connection object.
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'"); // pass your connection object here
I think your issue is that you're trying to encode a database resource.
Try adjusting your PHP to look like the following:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$return_data = [];
while($row = mysqli_fetch_assoc($result3)) {
$return_data[] = $row;
}
$json = json_encode($return_data);
echo $json
?>
I have a form that currently is able to auto complete base on user input, it queries the MySQL database and successfully lists all possible matches in the table and give suggestions. Now I want to handle rows that do not exist. I am having trouble to get my PHP file to echo the error. Here is what I have so far:
I'm guessing in my auto search function in my javascript in main.php I need to return the error message to the page?
search.php
<?php
//database configuration
$host = 'user';
$username = 'user';
$password = 'pwd';
$name = 'name';
//connect with the database
$dbConnection = new mysqli($host,$username,$password,$name);
if(isset($_GET['term'])){
//get search term
$searchTerm = '%'.$_GET['term'].'%';
//get matched data from skills table
if($query = $dbConnection->prepare("SELECT * FROM nametbl WHERE name LIKE ? ORDER BY name ASC")) {
$query->bind_param("s", $searchTerm);
$query->execute();
$result = $query->get_result();
//$row_cnt = $result->num_rows;
//echo $row_cnt;
if($result -> num_rows){
while ($row = $result->fetch_assoc()) {
$data[] = $row['name'];
}
//return json data
echo json_encode($data);
mysqli_close($dbConnection);
}
else { echo '<pre>' . "there are no rows." . '</pre>'; }
}
else {
echo '<pre>' . "something went wrong when trying to connect to the database." . '</pre>';
}
}
?>
main.php
<div id="gatewayInput">
<form method="post">
<input type="text" id="name" name="name" placeholder="Name..."><br><br>
<?php
include("search.php");
?>
</div>
...
...
...
<script src ="../../../jqueryDir/jquery-3.2.1.min.js"></script>
<script src ="../../../jqueryDir/jquery-ui.min.js"></script>
<script type="text/javascript">
//auto search function
$(function() {
$( "#name" ).autocomplete({
source: 'search.php'
});
});
1.your method type is post in the form
in main.php
and in the search.php, you have used "if(isset($_GET['term'])){"
this needs to be fixed I guess. either both needs to be POST or GET.
Again you are using include method which the whole code in search.php will be made in-line and treated as a one file main.php. so you need not use GET or Post method.
How does get and Post methods work is
3.1) you have a html or PHP which submits the data from browser(main.php), and this request is being served by an action class(search.php)
example :- in main.php
3.2) now in search.php you can use something like if(isset($_POST['term'])){
You can use num_rows (e.g. if ($result -> num_rows)) to see if the query returned anything.
I don't know why but the script tag is not working, the SELECT query is working but i am not getting the prompt from the javascript.
it is not redirecting anywhere only a blank screen is seen
$qry1="SELECT area, aadhar FROM user where username='$user'";
$result1 = $connector->query($qry1);
if($result1){
$row1=mysql_fetch_array($result1);
$userarea= $row1['area'];
$useraadhar=$row1['aadhar'];
}?>
<body>
<script type="text/javascript">
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
if(inputarea==userarea){
<?php/
//date
$today = date("Y-m-d");
//Create INSERT query
$qry = "INSERT INTO complain (user,category,regno,course,lecturer,room,details,address,datein) VALUES ('$userid','$category','$reg','$coursename','$lectname','$roomno','$details','$address','$today')";
//$result = #mysql_query($qry);
$result = $connector->query($qry);
//Check whether difjslk the query was successful or not
if($result) {
$errmsg_arr[] = 'Complain succesfully added, please wait for your response';
$errflag = true;
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: _new_complains.php");
exit();
}
header("location: _new_complains.php");
exit();
}else {
die("Query failed, couldn't add the new record");
header("location: _new_complains.php");
exit();
}
?>
}
You are sending data (for example body tag) before header(), therefore PHP creates an error. You just don't see it. Header needs to come before anything is sent to the browser (even a space).
You have multiple JS syntax errors:
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
Never EVER dump out raw text from PHP into a Javascript context. You're generating code that looks like
var inputarea=foo;
var userarea=bar;
var useradhar=baz';
The data will be seen as undefined variables, and you've got a stray ' in there. All of these errors will KILL the entire <script> block.
Always use json_encode() to dump from PHP->JS:
var inputarea = <?php echo json_encode($coursename); ?>;
This will GUARANTEE that you're producing correct Javascript code. The above line would produce
var inputarea = 'foo';
and be perfectly valid and executable code.