I have worked on API (zomato) that gives me restaurant details. I want to insert them into my local database, but I have a problem with passing the variable to PHP because it's too much big for $_GET to handle it. I tried to use $_POST but The output of the post was empty.
// JS code
function showCafes(str){
var xhttp;
xhttp = new XMLHttpRequest();
console.log(str);
xhttp.open("GET","https://developers.zomato.com/api/v2.1/search?entity_type=city&q=t&start="+str+"&count=20" , true);
xhttp.send();
var restaurants="";
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
var r=JSON.parse(this.responseText);
var rest ={
name : r.restaurant[1].restaurant.name
};
$.post("addFromApi.php", rest);
window.location.href="addFromApi.php";
// PHP code
<?php
print_r($_POST);
?>
I expected from the PHP code to print the name of the first element in it.
// Sample output From API
{"results_found":1,
"results_start":0,
"results_shown":1,
"restaurants":
[{"restaurant":{
"R":{"res_id":18692654},
"id":"18692654",
"name":"East Village"}
I have solved it with making a form from js like this
document.getElementById("cafes").innerHTML += '<form id="rests" action="addFromApi.php" method="post"><input type="hidden" name="q" value="'+restaurants+'"></form>';
document.getElementById("rests").submit();
//resturants is the variable that I wanted to pass to php.
Please tell me there is such a link. /add.php?id=2 How can I do it without reloading the page, but that the php file is executed.
use ajax
from:
https://www.w3schools.com/xml/ajax_intro.asp
function loadDoc() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = this.responseText;
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
}
You can do it with ajax.
http://api.jquery.com/jquery.ajax/
Javascript will execute that php file.
Ajax demo:
https://www.w3schools.com/xml/ajax_intro.asp
Don't overuse w3schools. I'm giving you this for demo purposes
My application gathers some client related information via JavaScript and submits it via AJAX to a php page.
See the code below:
index.php
<html>
<head>
<script>
function postClientData(){
var client_data = new Array(screen.availHeight, screen.availWidth);
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
if(this.responseText == client_data[0]){
document.getElementById("message").innerHTML = "Client data successfully submitted!";
}
}
};
var parameters = "ajax.php?screen_height=" + client_data[0] + "&screen_width=" + client_data[1];
xmlhttp.open("GET", parameters, true);
xmlhttp.send();
}
</script>
</head>
<body onload="postClientData()">
<span id="message"></span></p>
</body>
</html>
ajax.php
<?php
echo $_REQUEST["screen_height"];
//Does something else...
?>
I was wondering if I could merge the ajax.php content to my index.php and eliminate ajax.php. When I try adding the code, I probably get into a endless loop since I don't get my "success" message.
How can I solve this issue?
Correct, IMO I would definitely keep this specific logic separated in the ajax.php file.
If you do really want to merge, add it to the top of index.php (before printing data):
if (isset($_GET['screen_height'])) && isset($_GET['screen_width']) {
// Execute specific logic and exit to prevent sending the HTML.
exit;
}
I have 2 php files, the first is BAConsult.php which is the main file, and the other is BAConsultRecordsAJAX.php. This line of code is in BAConsultRecordsAJAX.php:
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse=explode(",",$row['skincarecurrentlyinuse']);
}
In BAConsult.php I have this:
echo $skincareinuse;
Is it possible to bring over the $skincareinuse value from the BAConsultRecordsAJAX.php page, to the $skincareinuse variable on the BAConsult.php page?
Such that, lets say whenever I do a click on the table row, the value of $skincareinuse will be updated and shown by the echo, without the page refreshing?
[EDITED TO SHOW MORE OF MY CODES]
This is BAConsult.php file, where my main code is.
<?php echo $jsonvariable; ?>//Lets say i want to store the JSON data into this variable
function showconsultationdata(str) { //face e.g and checkboxes for that date selected.
if (str == "") {
document.getElementById("txtHint2").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint2").innerHTML = xmlhttp.responseText;
var a = JSON.parse($(xmlhttp.responseText).filter('#arrayoutput').html());
$("textarea#skinconditionremarks").val(a.skinconditionremarks);
$("textarea#skincareremarks").val(a.skincareremarks);
var test = a.skincareinuse;
//I want to store this ^^ into the php variable of this page.
}
};
xmlhttp.open("GET","BAConsultRecordsAJAX.php?q="+str,true);
xmlhttp.send();
}
}
This is my BAConsultRecordsAJAX.php page.
$q = $_GET['q']; //get dateconsulted value
$consult="SELECT * FROM Counsel where nric='$_SESSION[nric]' and dateconsulted='$q'";
$consultresult = mysqli_query($dbconn,$consult);
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse=explode(",",$row['skincarecurrentlyinuse']);
$skincondition=explode(",",$row['skincondition']);
$queryResult[] = $row['skincareremarks'];
$queryResult[] = $row['skinconditionremarks'];
}
$skincareremarks = $queryResult[0];
$skinconditionremarks = $queryResult[1];
echo "<div id='arrayoutput'>";
echo json_encode(array('skincareremarks'=>$skincareremarks
'skinconditionremarks'=>$skinconditionremarks
'skincareinuse'=>$skincareinuse,
'skincondition'=>$skincondition));
echo "</div>";
You don't actually want to transfer the variable from BAConsultRecordsAJAX.php to BAConsult.php, but to BAConsult.js (I suppose that's the name of your JS file).
In reality, even that's what you wanted to do, it's impossible, because your main PHP is processed before the page even loads. What you can do, however, is overwrite the rendered value with a new one with the use of JavaScript.
To do that, send an AJAX request to BAConsultRecordsAJAX.php requesting the variable's value as shown below:
In JavaScript:
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
// Check if the AJAX request was successful
if (xhttp.readyState === 4 && xhttp.status === 200) {
var td = document.getElementById("your table td value's id");
var td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(""); // You don't need to send anything to the PHP file
In PHP:
<?php
echo $skincareinuse;
?>
[EDIT]:
If you are using inline JavaScript use the following code:
<script type = "application/javascript">
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
// Check if the AJAX request was successful
if (xhttp.readyState === 4 && xhttp.status === 200) {
var td = document.getElementById("your table td value's id");
td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(""); // You don't need to send anything to the PHP file
</script>
[EDIT 2]:
To pass a JSON object to PHP file via an AJAX request you have to make it a string. That requires the following code:
var JSONstring = JSON.stringify(yourJSONobject);
Then you can pass it in your PHP file with the AJAX request by putting it inside send() as shown:
xhttp.send("json_object=" + JSONstring);
Then in PHP, in order to use it, you have to decode it:
<?php
$json_object = json_decode($_POST["json_object"]);
// Now it's ready to use
?>
[About the code]:
Notes about the first file:
First of all, you have put your plain JavaScript code inside a PHP file and therefore it will not work. You have to wrap it in <script type = "application/javascript></script>
I don't have a clue what you are trying to do here:
Code:
var a = JSON.parse($(xmlhttp.responseText).filter('#arrayoutput').html());
It seems like you are trying to filter out and parse the innerHTML of #arrayoutput.
If the response is a JSON string, not HTML, so you absolutely can't do that. The logic of this above line is flawed.
How I would write your code:
function showconsultationdata(str) {
var xmlhttp;
if (!str) {
$("#txtHint2").empty();
} else {
xmlhttp = new XMLHttpRequest();
// Providing support for a 15-year-old browser in 2016 is unnecessary
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
$("#txtHint2").html(xmlhttp.responseText);
var a = JSON.parse(xmlhttp.responseText);
$("textarea#skinconditionremarks").val(a.skinconditionremarks);
$("textarea#skincareremarks").val(a.skincareremarks);
var test = a.skincareinuse; // The variable you want
// Its saved in "text" and can use something like:
// $("#element").html(test); to have it replace your current text
}
};
xmlhttp.open("GET","BAConsultRecordsAJAX.php?q="+str,true);
xmlhttp.send();
}
}
Notes about the second file:
I don't know if the query works for you, but it probably shouldn't, because:
$_SESSION is an associative array and you pass nric, whereas it should be "nric".
As $_SESSION is an array, the proper method to insert its value to your query is: {$_SESSION["nric"]}.
To ensure that you don't become the victim of an SQL Injection, use parameterised queries or at least sanitise somehow the received data, because GET is relatively easy to hack. Check the improved code later on how to do it.
How I would write your code:
$q = $_GET['q']; //get dateconsulted value
$dbconn = mysqli_connect("localhost", "root", "") or die("Error!");
mysqli_select_db($dbconn, "database") or die("Error!");
$consult = "SELECT * FROM Counsel where nric='{$_SESSION["nric"]}' and dateconsulted = ?";
if ($stmt = mysqli_prepare($dbconn, $consult)) { // By using a prepared statement
mysqli_stmt_bind_param($stmt, "s", $q); // you eliminate the chances to have
if (mysqli_stmt_execute($stmt)) { // an SQL Injection break your database
$consultresult = mysqli_stmt_get_result($stmt);
if (mysqli_num_rows($consultresult) > 0) {
while ($row = mysqli_fetch_array($consultresult)) {
$skincareinuse = explode(",",$row['skincarecurrentlyinuse']);
$skincondition = explode(",",$row['skincondition']);
$queryResult[] = $row['skincareremarks'];
$queryResult[] = $row['skinconditionremarks'];
}
$skincareremarks = $queryResult[0];
$skinconditionremarks = $queryResult[1];
$array = array('skincareremarks'=>$skincareremarks
'skinconditionremarks'=>$skinconditionremarks
'skincareinuse'=>$skincareinuse,
'skincondition'=>$skincondition);
echo "<div id='arrayoutput'>";
echo json_encode($array);
echo "</div>";
mysqli_stmt_close($stmt);
mysqli_close($dbconn);
exit;
}
}
}
Obviously you may change the values to suit your requirements, you must have jquery installed, the time value could be anything you want
$.ajaxSetup({ cache: false });
setInterval(function() {
var url = "pagethatrequiresrefreshing.php";
$('#divtorefresh').load(url);
}, 4000);
Using include will most likely work. From the docs:
When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward.
BAConsultRecordsAJAX.php
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse[] = explode(",",$row['skincarecurrentlyinuse']);
}
include "BAConsult.php";
I'm trying to send parametres from a .php file to my Javascript but I can't even manage to send a String.
Javascript fragment:
var params = "action=getAlbums";
var request = new XMLHttpRequest();
request.open("POST", PHP CODE URL, true);
request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
request.setRequestHeader("Content-length", params.length);
request.setRequestHeader("Connection", "close");
request.send(params);
request.onreadystatechange = function() {
var phpmessage = request.responseText;
alert(phpmessage);
};
PHP fragment:
$deviceFunction = $_POST["action"];
if ($deviceFunction == "") $deviceFunction = $_GET["action"];
// Go to a function depending the action required
switch ($deviceFunction)
{
case "getAlbums":
getAlbumsFromDB();
break;
}
function getAlbumsFromDB()
{
echo "test message!";
}
The alert containing phpmessage pops up but it's empty (it actually appears twice). If I do this the alert won't even work:
request.onreadystatechange = function() {
if(request.status == 200) {
var phpmessage = request.responseText;
alert(phpmessage);
}
};
The readystatenchange event will be called each time the state changes. There are 5 states, see here: https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest#readyState
Rewrite your JS:
request.onreadystatechange = function () {
if (request.readyState == 4) {
console.log('AJAX finished, got ' + request.status + ' status code');
console.log('Response text is: ' + request.responseText);
}
}
In your code, you only check for the returned status code. The code above will check for the ready state and then output the status code for debbuging.
I know that this answer is more a comment than an answer to the actual question, but I felt writing an answer in order to include nicely formatted code.
I faced a similar problem working with Django. What I did:
I used a template language to generate the javascript variables I needed.
I'm not a PHP programmer but I'm going to give you the idea, let me now if works. The following isn't php code, is just for ilustrate.
<?php
<script type="text/javascript" ... >
SOME_VARIABLE = "{0}".format(php_function()) // php_function resolve the value you need
</script>
?>
The I use SOME_VARIABLE in my scripts.
Please specify your onreadystatechange event handler before calling open and send methods.
You also should make your choice between GET and POST method for your request.
If you want to popup your message only when your request object status is OK (=200) and readyState is finished whith the response ready (=4), you can write :
request.onreadystatechange = function() {
if (request.readyState==4 && request.status==200) {
var phpMessage = request.responseText;
alert(phpMessage);
}
};