I am having issues with reading Json_encode response in a java script.
The PHP file reads values from database and sends the results as Json_encode array to html.
<?php
include("connect.php");
try {
$conn = new PDO("mysql:host=$servername;dbname=mydb", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "Call print_awb (#output1,#output2,:input_awb_ref_id)";
if (isset($_POST)) {
$p_in_awb_ref_id =reset($_POST["var_p_in_awb_ref_id"]);
}
$stmt = $conn->prepare($sql);
$stmt->bindParam(':input_awb_ref_id',$p_in_awb_ref_id, PDO::PARAM_INT);
$stmt->execute();
$out_awb_ref_id = $conn->query("SELECT #output1")->fetch(PDO::FETCH_ASSOC);
$out_agent_id = $conn->query("SELECT #output2")->fetch(PDO::FETCH_ASSOC);
$output = array(
"out_awb_ref_id" => $out_awb_ref_id,
"out_agent_id" => $out_agent_id,
);
echo json_encode($output);
$stmt->closeCursor();
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
?>
The HTML CODE
<script type="text/javascript">
function get_parameters(){
var quote = ABCD1234;
quote.toString();
window.alert(quote);
$.ajax({
type: "POST",
url: "printawb.php",
data: {var_p_in_awb_ref_id:quote},
dataType: "text",
success: function (result) {
alert(result);
var a = result.out_awb_ref_id;
alert (a);
alert(result['out_awb_ref_id']);
alert(result['out_agent_id'])
},
error:function (jqXHR, status, err){
//Fail
layer_1.html(html);
}
});
return vars;
};
</script>
The response I have is
HVP000062
{"out_awb_ref_id":{"#output1":"MIR"},"out_agent_id":{"#output2":"rtPreston"}}
undefined
undefined
All Need is the values in the 2nd line of the result ie. "MIR" and "rtPreston"
I have tried couple of things:
1. changed the calling function type as 'JSON' but the response was 'Object' without values
2. Tried converting to jsonString
3. tried reading values using JSON.parse(jsonString);
4. result['out_ref_awb_id']
None of them work. Can someone help me how I can get values of these? I can then use them to populate on a html page.
Many thanks
Add this as a first line in the HEAD section of your HTML template
<meta content="text/html;charset=utf-8" http-equiv="Content-Type">
<meta content="utf-8" http-equiv="encoding">
Then try this code:
<script type="text/javascript">
function get_parameters(){
var quote = ABCD1234;
quote.toString();
window.alert(quote);
$.ajax({
type: "POST",
url: "printawb.php",
data: {var_p_in_awb_ref_id:quote},
dataType: "text",
success: function (result) {
var data = jQuery.parseJSON(result);
var param1 = data.out_agent_id;
var param2 = data.out_awb_ref_id;
alert(param1["#output2"]);
alert(param2["#output1"])
},
error:function (jqXHR, status, err){
//Fail
layer_1.html(html);
}
});
return vars;
};
</script>
Related
I want to call php file from javascript, and this php file will update id=1
like this way:
javascript:
if(lastTemp >= document.getElementById("TempSet").value){
var jsonData2 =$.ajax({
url: "setpp.php",
dataType: "json",
async: false
}).responseText;
var obj2 = JSON.parse(jsonData2);
console.log(obj2);
}
else {
}
php file:
<?php
$DATABASE_HOST = 'localhost';
$DATABASE_USER = 'use';
$DATABASE_PASS = 'pass';
$DATABASE_NAME = 'database';
// Try and connect using the info above.
$db = mysqli_connect($DATABASE_HOST, $DATABASE_USER, $DATABASE_PASS,
$DATABASE_NAME);
if (!$db){
die("Connection Failed: ". mysqli_connect_error());
}
$db_update = "UPDATE setpoint_control SET status='ON' WHERE id=1";
$result = mysqli_query($db, $db_update);
?>
<?php
$data = array();
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_array($result)){
array_push($data, $row['status']);
}
}
echo json_encode($data);
?>
the code is executed and the status in database table is changed but I got error in console : SyntaxError: JSON.parse: unexpected character at line 4 column 2 of the JSON data
How can I solve this issue which I think I need to rewrite json_encode but I don't know how?
$.ajax({
type: 'post',
dataType: 'json',
cache: false,
url: 'setpp.php',
success: function (response) {
$.each(response, function(i, item) {
alert(item);
});
},
error: function () {
alert("error");
},
});
example php answer setpp.php
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
array_push($data, $row['status']);
}
die(json_encode($data));
} else {
$answer = array(
'No Records'
);
die(json_encode($answer));
}
I think the problem is the value returned by setpp.php.
remember to die(), otherwise the php answer will not be correct
I'm having problems with this ajax/jquery programming.
I've tried many different things but nothing has worked.
Ajax posts selItem to ajaxsql.php, this works!
The sql query in ajaxsql.php works, cause it outputs this if i call the script directly in the browser: [{"forumname":"SDE forum","user":"michael","txt":"Jeg hedder Michael!"}]
The problem is that the ajax function shows an alert box with Error[object Object]
forum.php script:
<script type="text/javascript">
function ForumChat(selItem) {
$.ajax({
type: "POST",
url: 'ajaxsql.php',
data: { selectedItem : selItem.value },
dataType: "json",
success: function(data) {
alert(data);
$('#txtarea').html(data);
},
error: function(data) {
alert('Error' + data);
}
});
}
</script>
ajaxsql.php script:
<?php
if(!isset($_SESSION))
{
session_start();
}
include('class.php');
//$sel = $_POST['selectedItem'];
$sel = "SDE forum";
$sql = " SELECT * FROM forum WHERE user = '".$_SESSION['currentuser']."' AND forumname = '".$sel."' ";
$result = mysqli_query($_SESSION['con'], $sql);
while($row = mysqli_fetch_array($result))
{
$forumname = $row['forumname'];
$user = $row['user'];
$txt = $row['text'];
$return[] = array("forumname" =>$forumname, "user" =>$user, "txt" =>$txt);
}
echo json_encode($return);
?>
because ajaxsql.php returns object ..
what you can do in your ajax is
success: function(response) {
$('#txtarea').html('');
$.each(response.data, function(){
console.log(this);
$('#txtarea').append(data);
});
},
I have read many answers on stack overflow but I can't find an apt answer. I want to send multiple variables from php file to a javascript file. I want to use those variables later separately. So please explain with a simple example of how to get the variables from php file and how to use them separately later.
This is my js.
<script>
function here(card_numb) {
alert("pk!");
$.ajax({
url: 'details.php',
type: "GET",
dataType: 'json',
data: ({
card_number: card_numb
}),
success: function(data) {
console.log('card_number:'+data.card_number+'book_issued:'+data.book_isued);
}
});
}
I'm getting the alert 'pk!'. But $.ajax ain't working.
This is details.php
<?php
if(isset($_GET['card_number'])){
$card_number = $_GET['card_number'];
$query = "Select * from users where card_number = '".$card_number."'";
$query_run = mysqli_query($link,$query);
$row_numb =#mysqli_num_rows($query_run);
if($row_numb == 0){
echo "<div class='bdiv1'>No such number found!</div>";
} else{
$row=mysqli_fetch_assoc($query_run);
$book1 = $row['user_name'];
$arr = array('isued_book' => $book1,'card_number' => $card_number);
echo json_encode($arr);
exit();
}
}
?>
Thank you!
somthing.js - ur jspage
<script>
function here(card_numb) {
$.ajax({
url: 'details.php',
type: 'GET',
dataType: 'json',
data: {
card_number: card_numb
},
success: function(data) {
console.log('card_number:'+data.card_number+'book_issued:'+data.isued_book);
}
});
}
success: function(result){
console.log('variable1:'+result.var1+'variable2:'+result.var2+'variable3:'+result.var3);
} });
details.php
<?php
if(isset($_GET['card_number'])){
$card_number = $_GET['card_number'];
$query = "Select * from users where card_number = ".$card_number;
$query_run = mysqli_query($link,$query);
$row_numb =#mysqli_num_rows($query_run);
if(!$query_run){
echo "<div class='bdiv1'>No such number found!</div>";
} else {
$row=mysqli_fetch_assoc($query_run);
$book1 = $row['user_name'];
$arr = array('isued_book' => $book1,'card_number' => $card_number);
echo json_encode($arr);
exit();
}
if the currect value get in $row you can get the result in console
I'm trying to make an ajax call to get result from my database, but i'm facing an error.
My javascript:
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script language="Javascript">
setTimeout(makeAjaxCall, 1000);
function makeAjaxCall(){
$.ajax({
type: "post",
url: "call/update",
cache: false,
data: {action: 'getUpdate', term: '<?php echo $id;?>'},
success: function(json){
try{
var obj = jQuery.parseJSON(json);
alert( obj['STATUS'] + obj['results']);
}catch(e) {
alert('Exception while request..');
}
},
error: function(){
alert('Error while request..');
}
});
}
</script>
And my controller's method:
public function update()
{
if (isset($_POST['action'])){
if ($_POST['action'] == 'getUpdate'){
pollNewData();
}
}
function pollNewData(){
$term = $_POST['term'];
$query = $this->db->query("SELECT * FROM users where guid <> '' and user_id = '$term'");
$res = $query->result();
echo json_encode(array('STATUS'=>200, 'results'=>$res));
}
}
i have this error on chrome debugs tool:
500 (Internal Server Error)
You have several issues. Below is the working code:
public function update()
{
if(!function_exists('pollNewData')){ // don't redeclare if already exists
function pollNewData($db){ // pass $db
$term = $_POST['term'];
$query = $db->query("SELECT * FROM users where guid <> '' and user_id = '$term'");
$res = $query->result();
echo json_encode(array('STATUS'=>200, 'results'=>$res));
}
}
if (isset($_POST['action'])){
if ($_POST['action'] == 'getUpdate'){
pollNewData($this->db); // pass $this->db
}
}
}
Changes:
Moved the function definition to before it is called - it must exist before calling.
The $this context is not set in the function, so pass the $db object as an argument.
When defining functions inside a class method, you must have a function_exists() check because on the second call, it will try to redeclare the function and produce a fatal error.
For future debugging you should turn errors on:
error_reporting(E_ALL);
ini_set('display_errors', '1');
Some suggestions:
url: "<?php echo base_url();?>call/update",
//pollNewData();
echo $this->pollNewData(); //call like this and echo it out to the ajax
//echo json_encode(array('STATUS'=>200, 'results'=>$res));
return json_encode(array('STATUS'=>200, 'results'=>$res)); //return it instead to the calling function
I am doing an ajax call like this:
function myCall() {
var request = $.ajax({
url: "ajax.php",
type: "GET",
dataType: "html"
});
request.done(function(data) {
$("image").attr('src',data);
});
request.fail(function(jqXHR, textStatus) {
alert( "Request failed: " + textStatus );
});
}
This is my ajax.php:
<?php
$connection = mysql_connect ("",
"", "");
mysql_select_db("");
// QUERY NEW ONE
$myquery = "SELECT * FROM mytable ORDER BY rand() LIMIT 1";
$result = mysql_query($myquery);
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
echo $currenturl,$currentnam, $currenturl,$currentimage;
}
mysql_close($connection);
?>
My data variable from the ajax call now contains all variables at once:
($currenturl,$currentnam, $currenturl,$currentimage)
How can I separate them so I can do something like:
request.done(function(data) {
$("id").attr('src',data1);
$("name").attr('src',data2);
$("url").attr('src',data3);
$("image").attr('src',data4);
});
jQuery :
$.ajax({
type:"POST",
url:"ajax.php",
dataType:"json",
success:function(response){
var url = response['url'];
var name = response['name'];
var image = response['image'];
// Now do with the three variables
// $("id").attr('src',data1);
// $("name").attr('src',data2);
// $("url").attr('src',data3);
// $("image").attr('src',data4);
},
error:function(response){
alert("error occurred");
}
});
From your code:
echo $currenturl,$currentnam, $currenturl,$currentimage;
Replace the above line with the code below:
$array = array('url'=>$currenturl, 'name'=>$currentname, 'image'=>$currentimage);
echo json_encode($array);
instead of string return an array i.e. use json type for returning value
i.e instead of
echo $currenturl,$currentnam, $currenturl,$currentimage;
use
echo json_encode array('current' => $currenturl,'currentnam' => $currentnam, 'currenturl' => $currenturl,'currentimage' => $currentimage);
and also write 'dataType' as 'json' in ajax