I've been looking everywhere for the answer.
I have the JSON string
"[{"id":"0"}]"
I've tried
obj['id'] and obj.id
but that doesn't work
$.ajax({
url: 'php/checkdoctorappointmentonday.php',
data: 'doctorName=' + doctorName + '&dayOfEvent=' + date1,
type: "POST",
success: function (json) {
obj = JSON.parse(json.data)[0];
b = obj.id;
}
});
return true;
}
Am I missing anything?
This is the php used to get the result
Edit:
<?php
$doctorName = $_POST['doctorName'];
$dayOfEvent = $_POST['dayOfEvent'];
// Query that retrieves events
$query = "SELECT COUNT(id) AS 'id'
FROM doctoravailability
WHERE start >='$dayOfEvent' AND start < DATE_ADD('$dayOfEvent', INTERVAL 1 DAY)
AND title = '$doctorName'
AND backgroundColor = 'red'
";
// connection to the database
try {
$bdd = new PDO("mysql:host=$servername;dbname=$dbname",$username,$password);
} catch(Exception $e) {
exit('Unable to connect to database.');
}
// Execute the query
$resultat = $bdd->query($query) or die(print_r($bdd->errorInfo()));
// sending the encoded result to success page
echo json_encode($resultat->fetchAll(PDO::FETCH_ASSOC));
?>
As per comments, the object is actually an array containing an object
var str = "[{\"id\":\"0\"}]";
var obj = JSON.parse(str)[0];
alert(obj.id);
http://jsfiddle.net/6ae0bgag/
obj["id"] would have also worked, its the same as obj.id
Does your JSON string actually have the start and end " in it, or have you just added them there to illustrate that it is a string?
Assuming that our data string actually has the ", then you just want to use the syntax
[{"id":"0"}]
Try this:
JS
$.ajax({
url: 'php/checkdoctorappointmentonday.php',
data: {
doctorName: doctorName,
dayOfEvent: date1
},
type: "POST",
dataType: 'json',
success: function (data) {
console.log(data);
b = data.0.id;
}
});
PHP
<?php
$doctorName = $_POST['doctorName'];
$dayOfEvent = $_POST['dayOfEvent'];
// Query that retrieves events
$query = "SELECT COUNT(id) AS 'id'
FROM doctoravailability
WHERE start >='$dayOfEvent' AND start < DATE_ADD('$dayOfEvent', INTERVAL 1 DAY)
AND title = '$doctorName'
AND backgroundColor = 'red'
";
// connection to the database
try {
$bdd = new PDO("mysql:host=$servername;dbname=$dbname",$username,$password);
} catch(Exception $e) {
exit('Unable to connect to database.');
}
// Execute the query
$resultat = $bdd->query($query) or die(print_r($bdd->errorInfo()));
// sending the encoded result to success page
return $resultat->fetchAll(PDO::FETCH_ASSOC);
?>
Related
I am new to php/ajax/jquery and am having some problems. I am trying to pass json to the php file, run some tasks using the json data and then issue back a response. I am using the facebook api to get log in a user,get there details, traslate details to json, send json toe the server and have the server check if the users id already exists in the database. Here is my javascript/jquery
function checkExisting() {
FB.api('/me', function(response) {
console.log('Successful login for: ' + response.id );
var json = JSON.stringify(response);
console.log(json);
$.ajax({
url: "php.php",
type: "POST",
data: {user: json},
success: function(msg){
if(msg === 1){
console.log('It exists ' + response.id );
} else{
console.log('not exists ' + response.id );
}
}
})
});
}
Here is my php file
if(isset($_POST['user']) && !empty($_POST['user'])) {
$c = connect();
$json = $_POST['user'];
$obj = json_decode($json, true);
$user_info = $jsonDecoded['id'];
$sql = mysql_query("SELECT * FROM user WHERE {$_GET["id"]}");
$count = mysql_num_rows($sql);
if($count>0){
echo 1;
} else{
echo 0;
}
close($c);
}
function connect(){
$con=mysqli_connect($host,$user,$pass);
if (mysqli_connect_errno()) {
echo "Failed to connect to Database: " . mysqli_connect_error();
}else{
return $con;
}
}
function close($c){
mysqli_close($con);
}
I want it to return either 1 or 0 based on if the users id is already in the table but it just returns a lot of html tags. . The json looks like so
{"id":"904186342276664","email":"ferrylefef#yahoo.co.uk","first_name":"Taak","gender":"male","last_name":"Sheeen","link":"https://www.facebook.com/app_scoped_user_id/904183432276664/","locale":"en_GB","name":"Tadadadn","timezone":1,"updated_time":"2014-06-15T12:52:45+0000","verified":true}
Fix the query part:
$sql = mysql_query("SELECT * FROM user WHERE {$_GET['id']}");
Or another way:
$sql = mysql_query("SELECT * FROM user WHERE ". $_GET['id']);
Then it's always better to use dataType in your ajax
$.ajax({
url: "php.php",
type: "POST",
data: {user: json},
dataType: "jsonp", // for cross domains or json for same domain
success: function(msg){
if(msg === 1){
console.log('It exists ' + response.id );
} else{
console.log('not exists ' + response.id );
}
}
})
});
Where is $jsonDecoded getting assigned in your PHP? Looks unassigned to me.
I think you meant to say:
$obj = json_decode($json, true);
$user_info = $obj['id'];
And your SELECT makes no sense. Your referencing $_GET during a POST. Maybe you meant to say:
$sql = mysql_query("SELECT * FROM user WHERE id = {$user_info}");
I want to get a parameter from an url. The url looks like this:
www.example.com/?v=12345
I want to get the parameter and query my mysql database to get the right data with ajax.
So i have my ajax call here:
$.ajax({
type:"POST",
url:"ajax2.php",
dataType:"json",
success:function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error:function(response){
alert("error occurred");
}
});
As you can see, the data which i want to get are in a json array and will be saved in javascript variables.
This is my php file:
<?php
// Connection stuff right here
$myquery = "SELECT * FROM mytable **WHERE id= **$myurlvariable**;
$result = mysql_query($myquery);
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array = array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
echo json_encode($array);
}
?>
The part where i want to query the right variable is bolded. I don't know how to query that. And Furthermore how to even get the url parameter in the proper form.
Can anybody help? Thank you!
You can get the query string using JavaScript and send it in the AJAX request.
Getting the query string(JavaScript) -
function query_string(variable)
{
var query = window.location.search.substring(1);
var vars = query.split("&");
for (var i=0;i<vars.length;i++) {
var pair = vars[i].split("=");
if(pair[0] == variable){return pair[1];}
}
return(false);
}
//Getting the parameter-
v = query_string('v'); // Will return '12345' if url is www.example.com/?v=12345
This needs to be passed as data in the AJAX call.
$.ajax(
{
type: "POST",
dataType: "json",
url: "ajax2.php",
data: "v="+v,
success: function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error: function(jqXHR,textStatus,errorThrown){
//alert(JSON.stringify(jqXHR));
//alert(textStatus);
//alert(errorThrown);
alert(JSON.stringify(jqXHR)+" "+textStatus+" "+errorThrown);
//alert("error occurred");
}
}
);
This can be accessed as $_POST['v'] in the php form.
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
...
And in php form, before you echo out the json response, change the content type. Something like this-
header("Content-Type: application/json");
echo json_encode($array);
If there is a database error, then it has to be handled.
So do this -
<?php
// Connection stuff right here
header("Content-Type: application/json");
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
$result = mysql_query($myquery) or die(json_encode(Array("error": mysql_error()));
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array[]= array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
}
echo json_encode($array);
}else{
echo json_encode(Array("error": "No POST values"));
}
?>
So this way, if the query has not executed properly, then you will know what exactly the error is.
Without any error checking, just the important part:
$myquery = "SELECT * FROM mytable WHERE id=" . $_POST['v'];
I have jquery pop form . It takes one input from the user ,mapping_key , Once the user enters the mapping key ,i make an ajax call to check if there is a user in the database with such a key.
This is my call .
Javascript:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
success: function(response) {
alert(response)
}
});
PHP:
$sql = "select first_name,last_name,user_email,company_name from registered_users where mapping_key = '$mapping_key'";
$res = mysql_query($sql);
$num_rows = mysql_num_rows($res);
if($num_rows == 0)
{
echo $num_rows;
}
else{
while($result = mysql_fetch_assoc($res))
{
print_r($result);
}
}
Now i want to loop through the returned array and add those returned values for displaying in another popup form.
Would appreciate any advice or help.
In your php, echo a json_encoded array:
$result = array();
while($row = mysql_fetch_assoc($res)) {
$result[] = $row;
}
echo json_encode($result);
In your javascript, set the $.ajax dataType property to 'json', then you will be able to loop the returned array:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
dataType : 'json',
success: function(response) {
var i;
for (i in response) {
alert(response[i].yourcolumn);
}
}
});
change
data : {"mapping_key":mapping_key} ,
to
data: "mapping_key=" + mapping_key,
You have to take the posted mapping_key:
$mapping_key = $_POST['mapping_key'];
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = '$mapping_key'";
or this:
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = $_POST['mapping_key']";
I am trying to pass some values to my PHP page and return JSON but for some reason I am getting the error "Unknown error parsererror". Below is my code. Note that if I alert the params I get the correct value.
function displaybookmarks()
{
var bookmarks = new String();
for(var i=0;i<window.localStorage.length;i++)
{
var keyName = window.localStorage.key(i);
var value = window.localStorage.getItem(keyName);
bookmarks = bookmarks+" "+value;
}
getbookmarks(bookmarks);
}
function getbookmarks(bookmarks){
//var surl = "http://www.webapp-testing.com/includes/getbookmarks.php";
var surl = "http://localhost/Outlish Online/includes/getbookmarks.php";
var id = 1;
$.ajax({
type: "GET",
url: surl,
data: "&Bookmarks="+bookmarks,
dataType: "jsonp",
cache : false,
jsonp : "onJSONPLoad",
jsonpCallback: "getbookmarkscallback",
crossDomain: "true",
success: function(response) {
alert("Success");
},
error: function (xhr, status) {
alert('Unknown error ' + status);
}
});
}
function getbookmarkscallback(rtndata)
{
$('#pagetitle').html("Favourites");
var data = "<ul class='table-view table-action'>";
for(j=0;j<window.localStorage.length;j++)
{
data = data + "<li>" + rtndata[j].title + "</li>";
}
data = data + "</ul>";
$('#listarticles').html(data);
}
Below is my PHP page:
<?php
$id = $_REQUEST['Bookmarks'];
$articles = explode(" ", $id);
$link = mysql_connect("localhost","root","") or die('Could not connect to mysql server' . mysql_error());
mysql_select_db('joomla15',$link) or die('Cannot select the DB');
/* grab the posts from the db */
$query = "SELECT * FROM jos_content where id='$articles[$i]'";
$result = mysql_query($query,$link) or die('Errant query: '.$query);
/* create one master array of the records */
$posts = array();
for($i = 0; $i < count($articles); $i++)
{
if(mysql_num_rows($result)) {
while($post = mysql_fetch_assoc($result)) {
$posts[] = $post;
}
}
}
header('Content-type: application/json');
echo $_GET['onJSONPLoad']. '('. json_encode($posts) . ')';
#mysql_close($link);
?>
Any idea why I am getting this error?
This is not json
"&Bookmarks="+bookmarks,
You're not sending JSON to the server in your $.ajax(). You need to change your code to this:
$.ajax({
...
data: {
Bookmarks: bookmarks
},
...
});
Only then will $_REQUEST['Bookmarks'] have your id.
As a sidenote, you should not use alert() in your jQuery for debugging. Instead, use console.log(), which can take multiple, comma-separated values. Modern browsers like Chrome have a console that makes debugging far simpler.
I'm trying to develop an application which gets the the response from the MySQL database using ajax post and update in list selector, but the list is displaying empty, can some one help me out from this please.....
code for .js:
SecondAssistant.prototype.setup = function() {
this.selectorChanged = this.selectorChanged.bindEventListener(this);
Mojo.Event.listen(this.controller.get('firstselector'), Mojo.Event.propertyChange, this.selectorChanged);
this.names = [];
try {
new Ajax.Request('http://localhost/projects/testingasdf.php', {
method: 'post',
parameters: {
'recs': getallrecords,
'q': q
},
evalJSON: 'true',
onSuccess: function(response){
var json = response.responseJSON;
var count = json.count - 1;
for(i=0; i<count; i++){
this.names.push({
label: json[i].name,
value: '0'
});
}
this.controller.modelChanged(this.model);
}.bind(this),
onFailure: function(){
Mojo.Controller.errorDialog('Failed to get ajax response');
}
});
}
catch (e){
Mojo.Controller.errorDialog(e);
}
this.controller.setupWidget("firstselector",
this.attributes = {
label: $L('Name'),
modelProperty: 'currentName'
},
this.model = {
choices: this.names
}
);
};
code for php:
<?php
header('Content-type: application/json'); // this is the magic that sets responseJSON
$conn = mysql_connect('localhost', 'root', '')// creating a connection
mysql_select_db("test", $conn) or die('could not select the database');//selecting database from connected database connection
switch($_POST['recs'])
{
case'getallRecords':{
$q = $_POST['q'];
//performing sql operations
$query = sprintf("SELECT * FROM user WHERE name= $q");
$result = mysql_query($query) or die('Query failed:' .mysql_error());
$all_recs = array();
while ($line = mysql_fetch_array($result,MYSQL_ASSOC)) {
$all_recs[] = $line;
}
break;
}
}
echo json_encode($all_recs);
// Free resultset
mysql_free_result($result);
// closing connection
mysql_close($conn);
?>
I would move the model updating code out of the SecondAssistant.prototype.setup method and have it fire somewhere in SecondAssistant.prototoype.activate.
Also call modelChanged
this.controller.modelChanged(this.model);
There is a typo on bindEventListener - should be bindAsEventListener and the return of the bind should be a different object:
this.selectorChangedBind = this.selectorChanged.bindAsEventListener(this);