I need to extract the URL's from the php datas, how can i achieve this?
PHP
$query = 'SELECT * FROM picture LIMIT 3';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url.=$rec['pic_location'].";";
}
echo json_encode($url);
Ajax
<script type="text/javascript">
$(document).ready(function() {
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first: " + arr[0] + ", second: " + arr[1]);
alert(arr[0]);
$(".theImage").attr("src", +arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
the alert message isn't working its just printing H T ( i think these are http://... )
You're returning a string which is not parsed as JSON.
Just add dataType: "json" to the ajax settings.
And since you're reading it as an array in your javascript you should return it like so:
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
You are sending a string in your PHP and expecting an array as response in javascript. Change you PHP to
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
And javascript to
$.ajax({
url: "viewnew.php",
dataType: "JSON",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret[0]);
var arr = ret;
alert(arr);
alert("first: " + arr[0] + ", second: " + arr[1]); // THIS IS NOT WORKING!!!!
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
Related
I'm having some issues displaying an array which I created in a PHP file. The response data in question is data["vessel"]
I have some jQuery:
j$('select[name=vessel]').change(function(e) {
var tour_ID = j$('select[name=tour]').val();
var trip_Date = j$('input[name=trip_Date]').val();
var data = {
"action": "Count_Vessels",
"trip_Date": trip_Date,
"tour_ID":tour_ID
};
data = j$(this).serialize() + "&" + j$.param(data);
j$.ajax({
type: "POST",
dataType: "json",
url: "../include/booking_Modify.php",
data: data,
success: function(data) {
//console.log("vessel stack: " + data["vessel"][0]);
var arr=JSON.parse(data["vessel"]);
console.log("vessel stack: " + arr[0]);
console.log("Form submitted successfully.\nReturned json: " + data["json"]);
},
error: function (request) {
console.log(request.responseText);
}
});
});
The PHP:
function count_Vessels(mysqli $conn, $trip_Date, $tour_ID){
$return = $_POST;
$vessel_Stack = array();
$vessel_Query = "SELECT * FROM Vessel";
if(!$vessel_Results = $conn->query($vessel_Query)){
die('There was an error running the query [' . $conn->error . ']');
}
while( $vessel_Row = $vessel_Results->fetch_assoc() ){
$vessel_Stack[$vessel_Row['ve_ID']] = $vessel_Row['vessel_Name'];
}
$return['vessel'] = $vessel_Stack;
$return["json"] = json_encode($return);
echo json_encode($return);
}
when I display data["json"] in console, I get Returned json: {"vessel":{"1":"Thriller","2":"Jammin","3":"Thunderstruck","4":"Wildthing","6":"Joyride"}
Which is awesome, but I don't know how to do that using the data["vessel"] Any help would be greatly appreciated.
I solved my own riddle. Because data["vessel"] is an array, I had to loop through it. Doing so like this worked:
j$.each(data["vessel"], function(key, val) {
console.log('index ' + key + ' value ' + val);
});
Below is the ajax request.
$.post('delete.php', {'deletearray':deletearray, 'dir':dir}, function(deleted, undeleted){
if(undeleted == 0) {
alert('All ' + deleted + ' files delted from the server');
} else {
alert(deleted + ' files deleted and ' + undeleted + ' files could not be deleted');
}
}, 'json');
and here goes the delete.php
<?php
if(isset($_POST['deletearray'])) {
$files = $_POST['deletearray'];
$dir = $_POST['dir'];
$deleted = 0;
$undeleted = 0;
foreach($files as $file) {
if(unlink($dir.$file) && unlink($dir.'thumb/'.$file)) {
$deleted ++;
} else {
$undeleted ++;
}
}
echo json_encode($deleted, $undeleted);
}
return;
?>
Up on running the code it deletes the files successfully but no message displays.
I also tried changing the ajax request as:
$.post('delete.php', {deletearray:deletearray, dir:dir}, function(deleted, undeleted){
alert("php finished");
}, 'json');
still it does not display the message. So i guess something is wrong in the delete.php file. Please help.
First thing-
Use $_POST['deletearray'] instead of $_POST[deletearray]
Second thing-
You cannot return different variables from the PHP scrtipt, every thing you print there is returned in the ajax callback, so just write this-
PHP
json_encode(array('totalDeleted' => $deleted, 'totalUndeleted' => $undeleted));
AJAX
...
function(response){
response=JSON.parse(response);
console.log(response);
}
The best way to do jquery + ajax + php is as next:
jquery:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript">
function do_ajax() {
//set data
var myData = new Array();
myData.push({name:'deletearray',value:'deletearray'});
myData.push({name:'dir',value:'dir'});
//ajax post
$.ajax({
dataType: 'json',
url: 'delete.php',
type: 'post',
data: myData,
success: function(returnData) {
if(returnData.undeleted == 0) {
alert('All ' + returnData.deleted + ' files delted from the server');
} else {
alert(returnData.deleted + ' files deleted and ' + returnData.undeleted + ' files could not be deleted');
}
}
});
}
</script>
PHP:
<?php
$myData = $_POST;
if(isset($myData['deletearray']) AND isset($myData['dir'])) {
$files = $myData['deletearray'];
$dir = $myData['dir'];
$deleted = 0;
$undeleted = 0;
foreach($files as $file) {
if(unlink($dir.$file) && unlink($dir.'thumb/'.$file)) {
$deleted ++;
} else {
$undeleted ++;
}
}
print(json_encode(array('deleted' => $deleted, 'undeleted' => $undeleted)));
exit();
}
?>
You should use json_encode like following:
json_encode(array('deleted' => $deleted, 'undeleted' => $undeleted));
And you have to get vars with data.undeleted and data.deleted
$.post('delete.php', {'deletearray':deletearray, 'dir':dir}, function(data) {
if(data.undeleted == 0) {
alert('All ' + data.deleted + ' files delted from the server');
} else {
alert(data.deleted + ' files deleted and ' + data.undeleted + ' files could not be deleted');
}
}, 'json');
I'm new Jquery and AJAX and I've really been struggling with the syntax I've been trying to use other tutorials as reference but nothing seems to work. I feel I have the right idea but syntax is wrong somewhere please help.
Here is the Ajax side
var var_numdatacheck = <?php echo $datacheck; ?>;
var var_numcheck = parseInt(var_numdatacheck);
function activitycheck(){
$.ajax({
type: 'POST',
url: 'feedupdate.php',
data: {function: '3test', datacheck: var_numcheck},
dataType: "json",
success: function(data) {
var json = eval('(' + data + ')');
$('#datacheck').html(json['0']);
var var_numcheck = parseInt(msg);
//setTimeout('activitycheck()',1000)},
error:function(msg) {
console.log(msg);
}
});
}
$(document).ready(function() {
activitycheck();
});
Here is the php the AJAX calls
<?php
require "dbc.php";
$function = $_POST['function'];
$datacheck = $_POST['datacheck'];
$search="SELECT * FROM Feedtest ORDER BY id DESC";
$request = mysql_query($search);
$update= mysql_fetch_array($request);
$updateid = $update['id'];
$updatecheck = mysql_num_rows($request);
$data = array();
if ($function == $datacheck){
echo $updatecheck;
echo $datacheck;
}
if ($function == "3test" && $updatecheck > $datacheck ) {
$updatesearch="SELECT * FROM Feedtest WHERE id = '$updateid' ORDER BY id DESC";
$updatequery = mysql_query($updatesearch);
$data['id'] = $updateid;
while ($row = mysql_fetch_array($updatequery))
{
?>
<?php $data[]= $row['First Name']; ?>
<?php
}
echo json_encode($data);
}
?>
</div>
</ul>
first of all ,always use JSON.parse(data) instead of eval.It is considereda a good practice.
second thing is always try to debug your code by checking it in console or alerting.In your context,this is what is happening-:
$.ajax({
type: 'POST',
url: 'feedupdate.php',
data: {function: '3test', datacheck: var_numcheck},
dataType: "json",
success: function(data) {
var data = eval('(' + data + ')');
console.log("myData"+data)//debugging.check the pattern so that you can acces it the way you want!!!
for(var i=0;i< data.length;i++)
{
alldata += "<li>"+data[i][0]+"<li><hr>";
}
$('#datacheck').html(alldata);
});
}
For JSON.parse:
success: function(data) {
var data = JSON.parse(data);
console.log("myData"+data)//debugging.check the pattern so that you can acces it the way you want!!!
for(var i in data)
{
alldata += "<li>"+data[i].First Name+"<li><hr>";
}
$('#datacheck').html(alldata);
});
this is my code:
class_search.php
case 'users':
if(!empty($_REQUEST['user'])){
if(strlen($_REQUEST['user']) >= 3){
$_REQUEST['user'] = $this->sanitize($_REQUEST['user'], 'string');
$stmt = $this->sql->prepare('SELECT
id,
nome,
url
FROM
animes
WHERE
nome LIKE ?
LIMIT 10');
$stmt->execute(array('%'.$_REQUEST['user'].'%'));
$this->queries++;
$c = 0;
if($admin){
$result['users'] = array();
}
if($stmt->rowCount() > 0){
while($row = $stmt->fetch(PDO::FETCH_ASSOC)){
if($admin){
$result['users'][$c] = array('name'=>($prefix ? '[usr] ' : '').$row['nome'], 'id'=>$row['id']);
$c++;
}else{
$result[] = ($prefix ? '[usr] ' : '').$row['nome'];
}
}
}
}
}
break;
general.js
$('#top_search').typeahead({
source: function(typeahead, query) {
$.ajax({
url: baseurl + "/ajax_calls.php",
dataType: "json",
type: "POST",
data: {
call: 'top_search',
user: query
},
success: function(data) {
typeahead.process(data);
}
});
},
onselect: function(obj) {
location.href= baseurl + '/animes/'+obj;
}
})
ajax_calls.php
case 'top_search':
$status = $site->process_autosearch('users');
break
;
I am having problem with onselect, I need to select the row url in my MySQL and encode to json, because when I click in one result I am redirect to mysite.com/animes/name of anime/ (yes, with space) and I need to fix this.
Table animes in phpMyAdmin:
http://s18.postimage.org/3ulrcmss9/Animes_Table.jpg
Quickly video: http://www.screenr.com/plZ7
You would need to use urlencode().
In your class_search.php file, update this part:
if($admin){
$result['users'][$c] = array('name'=>($prefix ? '[usr] ' : '').urlencode($row['nome']), 'id'=>$row['id']);
$c++;
}else{
$result[] = ($prefix ? '[usr] ' : '').urlencode($row['nome']);
}
And to make the names looks good in the searche auto-complete as well, you need to modify your jQuery as well, wrapping the decodeURIComponent() function around the obj like so:
onselect: function(obj) {
location.href= baseurl + '/animes/' + decodeURIComponent(obj);
}
How can I pass data from a php of then rows back to ajax ?
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[]=$rec['pic_location'];
$name[]=$rec['name'];
$age[]=$rec['age'];
$gender[]=$rec['gender'];
}
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Ajax
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
dataType: "json",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first image url: " + arr[0][0] + ", second image url: " + arr[0][1]); // This code isnt working
alert("first image Name: " + arr[1][0] + ", second image name: " + arr[1][1]);
$(".theImage").attr("src", arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
My question is how can I display the values here ? The Alert message is giving me "Undefined" ?
You can do something along these lines.
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$res = mysql_query($query);
$pictures = array();
while ($row = mysql_fetch_array($res)) {
$picture = array(
"pic_location" => $row['pic_location'],
"name" => $row['name'],
"age" => $row['age'],
"gender" => $row['gender']
);
$pictures[] = $picture;
}
echo json_encode($pictures);
JS
...
$.ajax({
...
dataType: "json",
...
success: function(pictures){
$.each(pictures, function(idx, picture){
// picture.pic_location
// picture.name
// picture.age
// picture.gender
});
}
});
...
You can't put multiple echo statements for the AJAX response:
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Join your arrays and send a single response:
$arr = $url + $name + $age + $gender;
echo json_encode($arr);
You can easily do this using a single Array:
$pics = array();
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$pics[$rec['id']]['url'] = $rec['pic_location'];
$pics[$rec['id']]['name']=$rec['name'];
$pics[$rec['id']]['age']=$rec['age'];
$pics[$rec['id']]['gender']=$rec['gender'];
}
echo json_encode($pics);