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Uncaught TypeError Message using Javascript variable as argument for PHP function that initiates mySQL query

I spent most of my day yesterday trying to solve this puzzle so today I've decided to reach out for some help. Before I begin, let me state that I am very aware that JavaScript is client-side and PHP is server-side and I use Ajax successfully for various other things. In this case, I can't find any S/O references that are similar to what I'm trying to accomplish. I would've thought this was very basic but I'm missing something here and could use some direction. Thank you.
In essence, I have a Javascript function that produces a JavaScript variable and then calls a PHP function that initiates a mySQL query (all on the same page). If I "hard-code" the argument in my PHP function call (e.g., sc_bdls = <?php echo Candidates::getBDLs(1000033); ?>;), the function runs perfectly and I receive the expected outcome (an array).
However, if I try and use a JavaScript variable in place of the argument, I receive a "Uncaught TypeError: Cannot read property 'id' of undefined" error. I have tried several iterations without success and I suspect I am missing something basic. Here is my code:
function tab_pos3(row){
var sc_id = row.toString();
var sc_bdls = [];
$.ajax({
type: "POST",
url: '/phpscripts/pass.php',
data: {row : row},
success:(function(data){
console.log("success");
alert(data);
})
});
sc_bdls = <?php echo Candidates::getBDLs($uid); ?>;
}
Here is the code from the pass.php file:
if(isset($_POST['row']))
{
$uid = $_POST['row'];
echo $uid;
}
Please notice that the console.log("success") and the alert(data) both show that the Ajax POST is working. Any thoughts on what might be going wrong? Thank you.
1st EDIT (comment from Swati)
I tried moving the PHP function call as suggested and then used $UID, $data, data, among others, and I still get the exact same error. Here is the edited code:
$.ajax({
type: "POST",
url: '/phpscripts/pass.php', //
data: {row : row},
success:(function(data){
console.log("success");
alert(data);
<?php $row = $_POST['row'];?>; // I tried this based on something I read.
sc_bdls = <?php echo Candidates::getBDLs(data); ?>;
})
The fact that the error doesn't change is gnawing at me. Could the argument be passed but in a format that is not being read right? Is the "TypeError" referring to data type?

Try moving the
sc_bdls = <?php echo Candidates::getBDLs($uid); ?>; to pass.php and read the returned data in ajax success function.
Please note php is executed before the browser sees it. And JS code is called client side.
This is how your function will look like
$.ajax({
type: "POST",
url: '/phpscripts/pass.php', //
data: {row : row},
success:(function(response){
console.log("success");
sc_bdls=response; // Just to show that response has the value you need.
alert(sc_bdls);
})
The pass.php will look like this
// I prefer using !empty here (unless you are expecting 0 as a valid input.
// This ensure $uid isset and is not empty.
// Also if $uid is supposed to be numeric you may want to add a validation here.
if(isset($_POST['row']))
{
$uid = $_POST['row'];
$response_array = Candidates::getBDLs($uid); //Your question says you are expecting an array
echo json_encode($response_array);
}

How do I properly use the json

I want to send data from php to a browser using JSON. I think I understand the process - see my example code below. But someone told me this is not the right way to do it. I have been researching for three days but because my English is poor I am not confident that I have found an answer.
What I am hoping for is a sample of code that will receive the JSON and pour it into html elements such as a div, and give it style via CSS, etc.
I really just want an example of how to do this so that I can learn from it and expand it myself for my own needs, but I am unconfident that this approach is correct and do not want to write more bad code.
Thanks
Javascript
$(document).ready(function() {
$.ajax({
type : 'POST',
url : 'server.php',
dataType:"json",
success : function (data) {
$("#orders").html(JSON.stringify(data));
}
});
});
PHP
<?php
$db = new PDO('mysql:host=localhost;dbname=Contact', 'root', '');
$statement=$db->prepare("SELECT * FROM myfeilds");
$statement->execute();
$results=$statement->fetchAll(PDO::FETCH_ASSOC);
$json=json_encode($results);
echo $json;
?>

$(document).ready(function() {
$.ajax({
type : 'POST',
url : 'server.php',
dataType:"application/json",
success : function (data) {
var jsns=JSON.parse(data);
$("#orders").html("");//clear the div
for(var i in data)
{
var id = data[i].id;
var name = data[i].name;
$("#orders").append("<span>"+id+"-"+name+</span>);//append each element
}
}
});
});
Since your server side php is already returning json.. you can just loop through the array of objects after converting it into JSON and then do DOM manipulation.
The best way to do it would be to use some templating engine like underscore.js

Display content from database on HTML page via Ajax and PHP

I'd like to post some content from a database on an html page. I'm not sure what the best way to do this is but my guess would be something that resembles the code below. Please let me know if there is a better way.
HTML:
$(document).ready(function(){
$.ajax({
type : "POST",
url : 'tablepageload.php',
data : 'test',
success: function(data) {
$('#echobox').html(data);
}
});
});
PHP:
if (isset($_POST['test'])) {
$sendtable = "SELECT `timein` FROM `timestamp` WHERE id='" . $latestrow . "' LIMIT 1";
$result = mysqli_query($link, $sendtable);
$row = mysqli_fetch_array($result);
echo $row['timein'];
};

Your workflow seems fine to me.
In essence, as already mentioned the work flow for what you're trying to achieve is:
Make request (your initial ajax call)
Process the request send response (your php script)
Handle the reponse (your 'success' callback)
Looking at your code i have some pointers.
Considering using the jquery .load() function.
If your ajax call is to do nothing more than populate a div you may aswell use this.
In terms of your sql query, i would recommend looking at:
Prepared statements (http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)
PDO (http://php.net/manual/en/book.pdo.php)
And if you want your latest row, order by 'id' DESC LIMIT 1.
Hope this proves helpful
EDIT: Also noticed a problem with your ajax call:
$.ajax({
type : "POST",
url : 'tablepageload.php',
//data : 'test',
//$_POST['test'] = "some_value",
//$_POST['another'] = "test"
data : {test:'some_value', another:'test'},
success: function(data) {
$('#echobox').html(data);
}
});

My ajax call not pulling in dynamic data from PHP/Mysql

Hello: I am working on a project where I am going to have divisions from a league listed as buttons on a page. And when you click on a button a different team list shows for each division. All divisions and teams are stored in a mysql database and are linked together by the "div_id". The plan was have the buttons use javascript or Jquery to send the 'div_id" to a function; which would then use ajax to access an external php file and then look up all the teams for that division using the div_id and print them on the page. I have been piecing this all together and getting the various pieces to work. But when I put it all together; it seems like the ajax part - does not pull in fresh data from the database if the data is changed. In fact, if I change the PHP file to echo some more data or something, it keeps using the original unaltered file. So, if the data is changed that is not updated, and if the file is changed that is not updated. I did find if I actually copied the file with a new name and then had my ajax call use that file instead; it would run it with new code and the new data at that time. But then everything is now locked in at that point and cannot get any changes.
So - I do not know much about ajax and trying to do this. I am not sure if this is totally normal for what I am using and for a dynamic changing team list, it cannot be done this way with ajax calling a PHP file.
OR - maybe there is something wrong with the ajax code and file I have which is making it behave this way? I will paste in the code of my ajax code and also the php file…
here is the ajax call:
var answer = DivId;
$.ajax({
type: 'GET',
url: 'path_to_file/gscript2.php',
data: 'answer=' + answer,
success: function(response) {
$('#ajax_content').html(response);
}
});
and here is the script.php file that it calls (removed db credentials):
<?php
if (!empty($_SERVER['HTTP_X_REQUESTED_WITH'])
&& strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'
) {
// AJAX request
$answer = $_GET['answer'];
$div_id=$answer;
echo "div id is: " . $div_id . "<br/>";
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database! Please try again later.');
mysql_select_db($dbname);
$result_g1 = mysql_query("SELECT * FROM teams WHERE div_id=$div_id");
while($row = mysql_fetch_array($result_g1, MYSQL_BOTH))
{
$team_id=$row[team_id];
$team_name=$row[team_name];
echo $team_id . " " . $team_name . "<br/>";
}
}
?>
So - to sum up - is there something wrong with this making it do this? Or is what it is doing totally normal and I have to find a different way?
Thanks so much...

Most likely your browser is caching.
Try adding cache: false as such:
$.ajax({
cache: false,
type: 'GET',
...
The jQuery documentation explains that by doing so, it simply adds a GET parameter to make every request unique in URL.
It works by appending "_={timestamp}" to the GET parameters.

I believe this is caused by your browser's cache mechanism.
Try adding a random number to the request so the browser won't cache the results:
var answer = DivId;
$.ajax({
type: 'GET',
url: 'path_to_file/gscript2.php?r=' + Math.random(),
data: 'answer=' + answer,
success: function(response) {
$('#ajax_content').html(response);
}
});
Or turning jQuery's caching option off by:
var answer = DivId;
$.ajax({
type: 'GET',
url: 'path_to_file/gscript2.php',
data: 'answer=' + answer,
success: function(response) {
$('#ajax_content').html(response);
},
cache: false
});
Or (globally):
$.ajaxSetup({ cache: false });

Sending data from javascript to php using via Ajax using jQuery. How to view the data back?

I know this question has been asked a lot as I have been googling since morning but I just can't think what's going on in my code.
I have a index.php (which is my website) that uses forms to receive information from user, then invokes a javascript file separately which in turn invokes another backend php file (this backend php file queries a database using mysql and does some stuff).
The problem is I am not sure what information is getting passed to the backend php from my js. I have just learnt jQuery and Ajax so I really don't know what is that small mistake I am making.
From my understanding, the backend php does its stuff and passes the value to javascript to be displayed on the web page. But my data is not getting sent / displayed. I am getting a error 500 internal server error.
Here are the pieces of code that are currently is question:
Javascript:
var data1 = {week:week, group:grp_name};
$.ajax({
dataType: "json",
type: "POST",
url : "php/remindUsers.php",
success : function(response){
alert ("success !");
},
error : function(response){
console.log(response);
alert("fail!");
}
})
});
PHP backend (remindUsers.php):
<?php
if (isset($_POST['week'])) {
$week = $_POST['week'];
}
if (isset($_POST['group'])) {
$group_name = $_POST['group'];
}
echo $week;
?>
I am ommiting out the sql code pieces because they work fine.
Edit: Now my status code is 200, response text is also ok . My response text shows a weird "enter' sign next to the actual response text expected. Is this normal ? And it is still going into the error block , not the success block of code.

I can not fully answer your question because I need more debug information about whats going on but theres 2-3 things about your code bugging me a little that might fix your bug.
First, use isset in your backend like this:
if (isset($_GET['your_input_name'])) {
$someData = $_GET['your_input_name'];
}
The isset part is very important here. Set it up and try it again. If you stop having a 500 error. Its probably because your data was never send to your backend or because your not checking the good input name.
Second, About input name. I can see in your code that you send:
var data1 = {week:week, group:grp_name};
So in your backend you should use the name of the value like this to retrieve your data:
$week = $_POST("week");
Third, I am not a json pro but maybe your json is not valid. Even if he is ok I suggest you build a "cleaner" one like this:
var data = [
{ 'name' : 'week', 'value' : week}
];
And finally, if you are using forms to send data to php then you can use something like that :
var myForm = $("#myForm").serializeArray();
$.ajax({
url: 'yourUrl',
type: "GET",
data: myForm,
dataType: 'json',
success: function(res){
//your success code
},
error: function(){
//your error code
}
});
I hope this helps.

You can't have these tags <body>,... in your PHP response over json.
It must be only:
<?php
$week = $_POST("data");
$json = json_decode($week);
echo json_encode($json);
?>
Remove the comment on
//data : {week :week}
And set a variable week with a valid value:
data : {week :week}
and so:
$.ajax({
dataType: "json",
type: "POST",
url : "php/remindUsers.php",
data : {week :week} ,
success : function(response){
console.log(response);
},
In order to see what is the shape of response.

You are doing a couple things wrong. First, you don't want to stringify your data before sending it to the server. You want to send JSON, so your commented line is the correct one. There is also a problem with the PHP. The data going to the server will look like:
{week: "Something"}
So in your PHP, you want to access the data like:
$_POST["week"];

USE THIS
PHP
$week = $_POST['data'];
$json = json_encode($week);
echo $json;
JS
$.ajax({
dataType: "json",
type: "POST",
url : "php/remindUsers.php"
//data : {week :week} ,
data: {data:{week:'week', group:'grp_name'}} ,
success : function(response){
alert ("success !");
},
error : function(response){
alert("fail!");
}
})

I would say wrap the php in a function and echo the json. Also its good to check if there was POSTed data, and if not return an error message. This is not tested, but will hopefully point you in the right direction.
<?php
function getJSON() {
if (isset($_POST["data"] && !empty($_POST['data']) ) {
$week = $_POST["data"];
$json = json_decode($week);
echo $json;
} else {
echo "There was a problem returning your data";
}
}
getJSON();
?>
Actually as I write this, I realized you could try these headers in your AJAX POST:
accepts: 'application/json',
contentType: 'application/json',
dataType: 'json'
Hope that helps.

It worked. I figured out the answer thanks to another SO post.
TIL : Even if server response is ok, the error condition will be triggered because the data returned to javascript from php is not json,since i had explicitly mentioned dataType: "json" in the ajax request.
Link here:
Ajax request returns 200 OK, but an error event is fired instead of success
Thanks folks for helping me and steering me in the right direction. Cheers!