At the lessons i learn how to pass result from a sql request to js via JSON/AJAX. I need the value of the row from this request in my js but it doesnt work. Via console i have an error: Uncaught SyntaxError: Unexpected token <
part of PHP:
<?php
//get all the course from db and reply using json structure
//connection to db
$mysqli = new mysqli("localhost", "root", "", "my_hyp");
$id = $_POST['id'];
if (mysqli_connect_errno()) { //verify connection
exit(); //do nothing else
}
else {
# extract results mysqli_result::fetch_array
$query = " SELECT * FROM course WHERE course_category='$id'";
//query execution
$result = $mysqli->query($query);
//if there are data available
if($result->num_rows >0)
{
$myArray = array();//create an array
while($row = $result->fetch_array(MYSQL_ASSOC)) {
$myArray[] = array_map('utf8_encode', $row);
}
$response = array();
$response['rows'] = $row;
$response['query'] = $myArray;
echo json_encode($response);
}
//free result
$a=num_rows;
$result,$a->close();
//close connection
$mysqli->close();
}
?>
first part of Script:
$.ajax({
method: "POST",
//dataType: "json", //type of data
crossDomain: true, //localhost purposes
url: "./query/cate_has_courses.php", //Relative or absolute path to file.php file
data: {id: i},
success: function(response) {
console.log(JSON.parse(response));
var course=JSON.parse(response.query);
var row=JSON.parse(response.rows);
Seem you use JSON.parse in wrong way.
The JSON.parse must be done one time only. The result of JSON.parse is store in course the the access to the data is due by response.query or respons.row .. and so on and not by JSON.parse(respose.query)
Related
I want to call php file from javascript, and this php file will update id=1
like this way:
javascript:
if(lastTemp >= document.getElementById("TempSet").value){
var jsonData2 =$.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
onclick onto a button the js starts an ajax request to the php file. The php file then gets all entries of one table of a database with php-loop. adding them to an array then the php file parse it to a JSON and echos it back to the ajax request. on success the ajax request should output an alert but neither do i get an error nor a alert.
After adding some changes according to the comments. it is now showing the error message 2 error messages randomly:
Fehler: {"readyState":0,"responseText":"","status":0,"statusText":"error"}
Fehler: {"readyState":4,"responseText":"<br />\n<b>Fatal error</b>: Uncaught Error: Cannot use object of type mysqli_result as array in C:\\xampp\\htdocs\\integration\\get.php:32\nStack trace:\n#0 C:\\xampp\\htdocs\\integration\\get.php(13): getLevel1()\n#1 {main}\n thrown in <b>C:\\xampp\\htdocs\\integration\\get.php</b> on line <b>32</b><br />\n","status":200,"statusText":"OK"}
php request (mysqli attributes are left out on purpose):
$func = $_POST['func'];
if ($func == "getLevel1"){
getLevel1();
}
$result = array();
function getLevel1(){
// Create connection
$conn = new mysqli(servername, username, password, dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name FROM capability_level1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$result[] = '<button onclick="capability('. $row["id"] .')">' . $row["name"]. '</button></br>';
}
echo json_encode($result);
} else {
echo json_encode("0 results");
}
$conn->close();
}
js ajax call:
async function getLevel1() {
return $.ajax({
type: "POST",
dataType: "json",
url: "get.php",
data: {
func: "getLevel1"
},
success: function(data) {
alert(JSON.stringify(data));
console.log(data);
},
error: function(data) {
alert("Fehler: "+ JSON.stringify(data));
}
});
}
You need to put the json encoding when you have a complete array to be encoded: after the while:
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$result[] = '<button onclick="capability('. $row["id"] .')">' . $row["name"]. '</button></br>';
}
echo json_encode($result);
} else {
Also note that you probably have to change your data type to Json (and send a json to php) to be able to return it. Actually your Ajax is waiting for text to be returned (based on the data type)
For your further error: it is related to the point that you are fetching the rows from mysql using the wrong function. See this question for more details on how to fix it.
I'm looking to make an ajax call to a PHP script to get data from MySQL, create a json array and pass it back to the success function of the ajax call, where i will then use it as parameters for a JavaScript function.
This is my ajax call,
$('button[name="message"]').click(function() {
var $row = $(this).closest("tr"); // Find the row
var $tenant_id = $row.find(".col-md-1 id").text(); // Find the tenants ID
var $landlord_id = "<?php echo $id; ?>"
$.ajax({
url : "./message.php",
type : "POST",
async : false,
data: {
landlord_id: $landlord_id,
tenant_id : $tenant_id
},
success: function(data){
console.log(data);
var messages = data;
insertChat(messages.sender_id, messages.body, messages.timestamp);
}
})
});
And this is my PHP file,
<?php
session_start();
require_once('../dbconnect.php');
// update tenants table to show deposit returned
if(isset($_POST['tenant_id'])){
$tenant_id = $_POST['tenant_id'];
$landlord_id = $_POST['landlord_id'];
$sql = "SELECT * from messages WHERE messages.sender_id OR messages.receiver_id = '$tenant_id' AND messages.sender_id OR messages.receiver_id = '$landlord_id'";
$result = mysqli_query($conn, $sql) or die("Error in Selecting " . mysqli_error($conn));
//create an array
$messages = array();
while($row =mysqli_fetch_assoc($result))
{
$messages[] = $row;
}
echo json_encode($messages);
}
?>
If anybody has a link to a tutorial or the individual parts that would be fantastic. I don't even know if the process i have outlined above is correct.
If anybody could tell me the correct way to go about this that would be of great help!
Thanks
Just a few things to adjust your javascript side (I won't explain the php sql injection issue you have... but please research prepare, bind_param and execute):
Since you are returning an ARRAY of $messages from php (json_encoded), you need to loop on those in your success handler.
Add dataType: 'JSON' to your options, so it explicitly expects json returned from php.
And you were missing a couple semicolons ;)
Adjustments added to your code:
$('button[name="message"]').click(function() {
var $row = $(this).closest("tr");
var tenant_id = $row.find(".col-md-1 id").text();
var landlord_id = "<?php echo $id; ?>";
$.ajax({
url : "./message.php",
type : "POST",
data: {
landlord_id: landlord_id,
tenant_id : tenant_id
},
dataType: 'JSON',
success: function(data){
console.log(data);
if (typeof data !== undefined) {
for(var i = 0; i < data.length; i++) {
insertChat(data[i].sender_id, data[i].body, data[i].timestamp);
}
}
}
});
});
I want to take data from database and save it in an array.
Like this
var locations = [ ['Current', 18.53515053, 73.87944794, 2],
['VimanNagar', 18.5670762, 73.9084194, 1]
];
First of all I have created a php page
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "citytrans";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM driver_location";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo json_encode($row);
}
} else {
echo "0 results";
}
$conn->close();
?>
which gives me below result
{"driver_id":"1","driver_lat":"18.53515053","driver_lng":"73.87944794","driver_code":"122"}{"driver_id":"2","driver_lat":"18.53640175","driver_lng":"73.88206482","driver_code":"133"}
Now I want to convert this into an array using Jquery (I want to decode it ), I just want drivers_lat and drivers_lng value from my jSON data fetched form the database show above.
I am using below code to parse the data form json
jQuery.ajax({
url: baseurl + "getdriverlocation.php",
type: "JSON",
async: false,
success: function(data){
var myArray = JSON.parse(data);
console.log(myArray.driver_lat)
}
});
but it is giving me error (shown below)
SyntaxError: JSON.parse: unexpected non-whitespace character after
JSON data at line 1 column 92 of the JSON data
I just want the two values from json data and save it in an array variable
Please help
Use this one..
jQuery.ajax({
url: baseurl + "getdriverlocation.php",
type: "JSON",
async: false,
success: function(data){
var myArray = jQuery.parseJSON(data);// instead of JSON.parse(data)
jQuery(myArray).each(function( index, element ) {
console.log(element.driver_lat)
});
}
});
In your php you should do :
if ($result->num_rows > 0) {
// output data of each row <- no, build your data, then make only 1 output
$output = array();
while($row = $result->fetch_assoc()) {
$output[] = $row;
}
echo json_encode($output);
}
Then in your jQuery, parse the whole json-decoded array...
Your json data is invalid.
You must put comma bettween two JSON Objects
Your respons must be
{"driver_id":"1","driver_lat":"18.53515053","driver_lng":"73.87944794","driver_code":"122"},
{"driver_id":"2","driver_lat":"18.53640175","driver_lng":"73.88206482","driver_code":"133"}
As i identified your Response JSON format is invalid, response JSON format should like this in order to parse into JSON via JSON.parse()
[{"driver_id":"1","driver_lat":"18.53515053","driver_lng":"73.87944794","driver_code":"122"},
{"driver_id":"2","driver_lat":"18.53640175","driver_lng":"73.88206482","driver_code":"133"}]
Try this
$arrTmp = array();
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$arrTmp[] = $row;
}
}
echo json_encode($arrTmp);
And maybe the jQuery tools bellow for old browsers
$.parseJSON(data);
I am trying to retrieve a row from mysql db using ajax, code bellow:
jQuery.ajax({
type: 'POST',
url: 'Connection.php',
dataType: 'text',
data: {'query_id' : query_id},
success: function(response){
data = response;
alert(data['username']); //print undefined!!!
},
error: function(xhr, ajaxOptions, thrownError){
alert("thrownError");
}
});
Here is my mysql php code:
<?php
$con = mysql_connect('****','****','****');
mysql_select_db("eBay",$con);
$username = $_SESSION['username'];
$query_id = $_POST['query_id'];
$myquery = "SELECT * FROM `Output` WHERE `username` =" '$username';
$query = mysql_query($myquery);
if ( ! $query ) {
echo mysql_error();
}
$data = mysql_fetch_array($query);
echo ($data);
mysql_close($server);
?>
In the response I get, I have undefined array cell. Any idea?
to return just the username from php:
$data = mysql_fetch_assoc($query)['username'];
in your javascript you should be able to do alert(data) instead of searching for the username tuple in an array.
Ideally you should return your data back in json instead of text that way it can remain structured and easier to access in javascript. You will need to change your ajax option to dataType: 'json'
and the PHP code to :
$data = json_encode(mysql_fetch_assoc($query));
You are getting array in $data variable so you should use json_encode function to get all values from resulting row like this-
$myquery = "SELECT * FROM `Output` WHERE `username` ='".$username."'";
foreach ($myquery as $test)
{
$field1_value =$test->name_of_field1_in_db;
$field2_value =$test->name_of_field2_in_db;
$field3_value =$test->name_of_field3_in_db;
}
$all_values_in_array=array('field1'=>$field1_value,'field2'=>$field2_value,'field3'=>$field3_value,);
echo json_encode(echo json_encode($arr_provider);
and get all value on ajax suucess function like this--
success: function(response){
var pro = jQuery.parseJSON(response);
var field1_val=pro.field1; //var field1 contain value of field1 in db
var field2_val=pro.field2;
var field3_val=pro.field3;
},
hope it will help you.
Best of Luck