I am having an issue with looping through an array that was passed from PHP through an Ajax request.
For some reason my javascript thinks that either every character is a part of the array or my response variable is just being passed as a string.
Here is my javascript:
<script>
$(function() {
$.ajax({
url: "/dev/editButton/get_names.php",
success: function(response) {
console.log(response);
}
});
});
</script>
And here is my PHP:
<?php
include '../portfolio/libraries/settings.php';
$connect = mysqli_connect($HOST, $DB_USER, $DB_PASS, $DATABASE);
$query = "SELECT * FROM AUTH_User";
$result = mysqli_query($connect, $query);
$names = array();
while ($row = mysqli_fetch_array($result)) {
array_push($names, $row['FirstName']." ".$row['LastName']);
}
echo json_encode($names);
?>
The response that I get looks like this:
["Test Person","Test2 Person"]
However, if I loop through this using javascript or just print out response[0] I get each character as part of the array. The first element would be [, next would be ", etc.
I would like Test Person to be one element and Test2 Person to be another.
Does anybody know what I am doing wrong? Thanks!
You need to use JSON.parse on the response. Wihtout calling that function you are just getting the index of characters in the JavaScript string.
var resultArray = JSON.parse(response);
resultArray[0]; //Should Be "test Person"
The result of the .ajax method is interpreted according to the Content-Type header of the response. If it is incorrect or not specified, the response variable will contain the raw json code as a string.
So one solution is change the PHP code by adding this line:
header("Content-Type: text/json");
Docs:
The type of pre-processing depends by default upon the Content-Type of
the response, but can be set explicitly using the dataType option. If
the dataType option is provided, the Content-Type header of the
response will be disregarded.
You can parse that text to an object, or you can let JQuery do that for you by specifying a datatype in the call. The response parameter will then hold the object instead of the raw json string.
Docs:
If json is specified, the response is parsed using jQuery.parseJSON
before being passed, as an object, to the success handler. The parsed
JSON object is made available through the responseJSON property of the
jqXHR object.
$(function() {
$.ajax({
url: "/dev/editButton/get_names.php",
datatype: "json",
success: function(response) {
console.log(response);
}
});
});
In this particular situation, you can use
success: function(response) {
response = eval(response);
console.log(response);
}
But this is bad practice.
Really the best solution here is to modify your ajax call as follow:
$(function() {
$.ajax({
url: "/dev/editButton/get_names.php",
datatype: 'json',
success: function(response) {
console.log(response);
}
});
});
The specified datatype, will request the returned data to be json, and the jquery will automatically parse it to a javascript object.
You must parse JSON to array. You can do this using the following code:
var arr = $.parseJSON(response);
Now arr[0] should be "Test Person".
You can do it the hard way, or this way:
First, you need to specify the return type for AJAX.
$(function() {
$.ajax({
url: "/dev/editButton/get_names.php",
dataType: "json",
success: function(response) {
console.log(response);
}
});
});
Alternatively, you could do it this way:
$(function() {
$.getJSON("/dev/editButton/get_names.php", function(response) {
console.log(response);
});
});
For this to work, you will need to specify the HTML headers accordingly in PHP:
<?php
include '../portfolio/libraries/settings.php';
$connect = mysqli_connect($HOST, $DB_USER, $DB_PASS, $DATABASE);
$query = "SELECT * FROM AUTH_User";
$result = mysqli_query($connect, $query);
$names = array();
while ($row = mysqli_fetch_array($result)) {
array_push($names, $row['FirstName']." ".$row['LastName']);
}
header("Content-Type: application/json");
echo json_encode($names);
exit();
?>
The exit(); is just for safety so you wouldn't ruin the valid JSON format.
JSON stands for JavaScript Object Notation, so you should not have to do anything complicated there. Here is a simple loop you could use for example:
for(var i in response) {
console.log(response[i]);
}
Alternatively, if the response is not an array but an object with properties, you could loop through the object properties by getting the right keys first:
var objKeys = Object.keys(response);
for(var i in objKeys) {
var key = objKeys[i];
console.log(response[key]);
}
I hope this helps!
Related
This question already has answers here:
How to parse JSON data with jQuery / JavaScript?
(11 answers)
Closed 5 years ago.
I created json with php. Data coming with ajax. But JSON.parse is giving an “undefined” object. Why ?
Php CODE
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
Ajax CODE
$.ajax({
type: "GET",
url: "http://localhost:8080/xxx.php/",
success: function (msg, result, status, xhr) {
var obj= JSON.parse(msg);
alert(obj.name);// giving undefined
},
error: function (error) {
}
});
json
[{"name":"eng","a":"sdf"}]
Your JSON is an array, meaning you'll have to point to the index of the object before accessing the property.
This code should work:
console.log(obj[0].name); //Returns "eng"
If your JSON array was something like this:
[{"name":"eng","a":"sdf"}, {"name":"esp", "a":"abc"}]
Then obj[1].name would return "esp".
You should obj[0].name
Because you are accessing the name property of the of the first element of the array.
A better way to get data from server
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode(array("data"=>$emparray));
Put all the json response on a key which is data here then at front end define that the server response is in JSON by dataType and then there is no need to parse data by JSON.parse()
msg.data.length will provide you the validaton on if the data received from server is empty or not which will prevent the undefined error
$.ajax({
type: "GET",
dataType: "JSON",
url: "http://localhost:8080/xxx.php/",
success: function (msg, result, status, xhr) {
var obj= msg.data;
if(msg.data.length){
alert(msg.data[0].name);// wll give name at 0 index
}
},
error: function (error) {
}
});
I try to get array from sql server using php , and parsing these array to javascript using ajax.
However , I have tried many solution by google , I can't get the array.
This is my php code:
<?php
include 'Connect.php';
$Store_int = $_GET['num'];
$sql = "SELECT * FROM `store` WHERE `Place_int` = " . $Store_int;
mysqli_select_db($link, "web");
$link->set_charset("utf8");
$result = mysqli_query($link, $sql);
$arr = array();
while ($row = mysqli_fetch_object($result)) {
$p = (string)$row->Name;
$arr[] = $p;
}
//print_r($arr);
$jsonArr = json_encode($arr, JSON_UNESCAPED_UNICODE);
echo $jsonArr;
mysqli_free_result($result);
mysqli_close($link);
?>
Array in php will encode and display:
["pen","pencil","apple","cat","dog"]
and the .js file
function getArr(store_int) {
var jsArray = new Array();
$.ajax({
url: "fromSQL_store.php",
data: {
num: $("#store_int").val()
},
type: "GET",
dataType: "json",
success: function (data) {
alert(num);
jsArray = JSON.parse(data.jsonArr);
}, error: function (data) {
alert("123");
}
});
//alert(jsArray.length);
return jsArray;
}
In .js , I will always get empty response(undefined) from php.
Because the ajax will answer "error" function..., and the error status is 200.
Your array will always return undfined as the AJAX call is async, your function returns jsArray before it is set. and You don't need JSON.parse() as you have defined dataType as json in your ajax call. Pass a function to your getArr() function and use your data in that function.
function getArr(store_int, outputFn){ // what do you use store_int for?
$.ajax({
url: "fromSQL_store.php",
data: {
num: $("#store_int").val()
},
type: "GET",
dataType: "json",
success: function(data) {
outputFn(data)
},error: function(data){
alert("123");
}
});
}
Then use it like
getArr('5', function (data) {
console.log(data)
})
Console output
Your problem lies here. You are attempting to access data.jsonArr which is always undefined. The JSON data sent is actually embodied by data variable.
success: function(data) {
alert(num);
jsArray = JSON.parse(data.jsonArr); // Replace by jsArray = data;
}
NOTE, You might also need to specify that the MIME media type that is being outputted is JSON. Putting this at the top of your PHP script should solve your problem.
<?php
header('Content-Type: application/json');
I hava data structure like this which is then returned to another file with AJAX:
$data = array();
$data['message'] = "You are searching: $domain!";
$data['domain:name'] = "domain.tld";
$data['domain:registrar'] = "Registrar Ltd.";
$data['domain:creation'] = "2015-26-05";
$data['domain:expiry'] = "2016-26-05";
$data['ns'] = "ns1.somedns.tld";
$data['owner']['name'] = "Owner Name";
$data['owner']['type'] = "Org";
echo json_encode($data);
That data is then append to html with AJAX like this:
$.ajax({
type: 'POST',
url: 'carnetEpp.php',
data: $(this).serialize(),
success: function (data) {
dataType: 'json',
//console.log(data);
$('#response').html(data);
$("#myModal").modal();
}
});
Now I want to pass that returned JSON object to PHP variable, so I can easy manipulate date with PHP. How do I do that? Or is best practice to do it with JS? Basically I want to print every key:pair value, so maybe for in is good choice.
And, I am not sure, should, or must I echo data in my script so AJAX can pick it up, or can I just pass data to variable and then fetch it in AJAX?
You need to add this code in success.
var obj = jQuery.parseJSON(data);
alert(obj.message);
OR
var obj = $.parseJSON(data);
alert(obj.message);
You will get the message sent from PHP.
before sending data in php, setup header for response:
$data = [
'key' => 'value',
'key2' => 'vlue2'
];
header('Content-Type: application/json');
echo json_encode($data);
then if u use jquery, $.getJson() it really cool solution for handle input json data.
I echo a simple associative array read from a mySQL table back to my jquery b.m.o the json_encode($array) method. However, my .ajax call fails back in the jquery and it seems due to the object not being a valid json object. I did some debugging in chrome and under the network tab verified the response preview- does indeed look to be in perfect json format:
{"UserName":"DehanL","UserPassword":"admin","UserEmail":"dehan#rocketmail.com"}
Here is my php:
<html>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName ="dbPodiumPro";
$inUsername = $_POST["name"];
$inPassword = $_POST["password"];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbName);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Use the dbPodiumPro
mysqli_select_db($conn,"dbPodiumPro");
$sql = "SELECT * FROM users WHERE UserName='$inUsername' AND UserPassword ='$inPassword' ";
$result = mysqli_query($conn,$sql);
// Fetch one row
$row=mysqli_fetch_assoc($result);
//Check to see if the username and password combo exists
if(!empty($row['UserName']) AND !empty($row['UserPassword']))
{
//$_SESSION['UserName'] = $row['UserPassword'];
//echo $_SESSION;
echo json_encode($row);
}
else { echo "Be gone imposter!"; }
$conn->close();
?>
</body>
</html>
Then the jquery:
$.ajax({
type: 'POST', // define the type of HTTP verb we want to use (POST for our form)
url: 'php/loginrequest.php', // the url where we want to POST
data: formData, // our data object
dataType: 'json' // what type of data do we expect back from the server
})
// using the done promise callback
.done(function (data) {
console.log("ajax done callback");
})
.fail(function (data) {
console.log(data, "ajax failed callback");
var IS_JSON = true;
try {
var json = $.parseJSON(data);
} catch (err) {
IS_JSON = false;
}
console.log(IS_JSON);
});
Your PHP starts like this:
<html>
<body>
Whatever else it outputs after that, it isn't going to be valid JSON.
I did some debugging in chrome and under the network tab verified the response preview
Look at the raw preview. You seem to be looking at the rendered HTML preview.
You don't need to parseJSON when you have added dataType as JSON option in the AJAX.
The data you are getting after ajax is completed is already in JSON format.
"json": Evaluates the response as JSON and returns a JavaScript object. The JSON data is parsed in a strict manner; any malformed JSON is rejected and a parse error is thrown. As of jQuery 1.9, an empty response is also rejected; the server should return a response of null or {} instead. (See json.org for more information on proper JSON formatting.)
Docs: http://api.jquery.com/jQuery.ajax/
The data inside fail method is not what you've passed from server. It is the information related to the request.
If you are encoding your password using MD5 and the likes, Json will not encode you can choose to unset the password from you your array or define the column you want to select from your database excluding password column.
Change your ajax as
$.ajax({
type:"POST",
url: 'php/loginrequest.php',
data:$("#form_id").serialize(),
dataType:"json",
success: function(data)
{
alert(data.UserName);
}
});
you can use both data:formData or data:$("#form_id").serialize(),
if you are getting alert message then everything is working fine
I am trying to pass an array() on PHP from JavaScript but PHP receives nothing. It always sets $str to "". Why?
JavaScript
var ArrayPassedID = [];
function pass(){
$.ajax({
url: 'http://mysite/index2.php?task=getPassed',
type:'get',
dataType:'json',
data: {id: JSON.stringify(ArrayPassedID)},
async: false,
success: function(response){
ArrayPassedID.push(response.id);
}
....
PHP
$str = "";
if(!empty($_POST["id"])){
$id = $_POST["id"];
$id = json_decode($id,true);
$str = implode(",",$id);
}
$data = query(SELECT id, response FROM `conversation` WHERE id not in ('".$str ."'));
$values = array();
$values['id'] = $data['id'];
$values['response'] = $data['response'];
return json_encode($values);
From Javascript, you send data with GET method of HTTP protocol.
From PHP, you retrieve data from global $_POST, which correspond to POST method.
Use same method in the two ways.
in ajax (js) :
type:"POST"