I'm trying to call a file from my server and return an HTML form. I asked a question here to get started but now I have another problem.
The textbox and submit button display, but since the data is JSON encoded and returned via AJAX to a DIV I'm not quite sure how to approach it.
Right now here's the result. Where I have "textbox and submit button" those elements are actually there. The other text appears around it.
testing
{"formHTML":"
"textbox here" " submit button here"<\/form>"}
Here's the code that would be on another server which calls to mine. This is the page that does the displaying
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
<?php
echo "testing";
?>
<script>
$.ajax({
type: 'GET',
url: 'form_deliverer.php',
data: "true",
success: function(response) { // on success..
$('#yourdiv').html(response); // update the DIV
}
})
</script>
<div id = "yourdiv">
//form is displayed here
</div>
Here is the page that gets called, form_deliverer.php
<?
$o = new stdClass();
$o->formHTML = "<form method='post'><input type='textbox' name='text'/><input type='submit' name='submit_text'/></form>";
echo json_encode($o);
?>
Because AJAX automatically updates the div, how can I decode the data? Should I even do that?
For reference, this displays the form properly without the extra text. However, since I'll be calling from another server and have to deal with same domain issues, I'll have to use JSONP
<?
if(isset($_GET['true'])){
echo "
<form method='post'>
<input type='textbox' name='text'/>
<input type='submit' name='submit_text'/>
</form>
";
}
?>
You won't need to decode the data, though you will have to treat the response as an object. Because you json_encode()ed an stdClass, your ajax call will basically get something like this in return:
{"formHTML": "<form method='post'><input type='textbox' name='text'/><input type='submit' name='submit_text'/></form>"}
To access the string, you could simply write
$("#yourdiv").html(response.formHTML);
However, if you are only passing in the string, you could simply json_encode() the string instead of creating an object and then encoding that. That way, you can use response directly in your javascript.
form_deliverer.php
<?
echo json_encode("<form method='post'><input type='textbox' name='text'/><input type='submit' name='submit_text'/></form>");
?>
javascript
$.ajax({
type: 'GET',
url: 'form_deliverer.php',
data: "true",
success: function(response) { // on success..
$('#yourdiv').html(response); // update the DIV
}
})
Related
I am trying to pass the link's text as a value to the next page so I can use it to search the database for the item and retrieve the information related to the value .I have tried using the POST method but regardless the information is not passed. This is the code I tried .
<form action="DetailedMenu.php" method = "POST" action = "<?php $_PHP_SELF ?>">
<?php
for($i=0;$i<sizeof($array);$i++) {
if($array[$i]["Food_Category"]=="starters") {
echo str_repeat(' ', 4); ?>
<a href="DetailedMenu.php" ><?php echo $array[$i]["Food_Name"];?></a>
<?php echo " " .str_repeat('. ', 25). "€".$array[$i]["Food_Price"]."<br>"; ?>
<input type="hidden" name="name" value="<?php echo $array[$i]["Food_Name"];?>">
<?php
}
}
?>
</form>
You don't need the form.
The easiest way to do what you're trying to do....
In addition to including the text in the content of the link, include it as a query string parameter.
for($i=0;$i<sizeof($array);$i++) {
if($array[$i]["Food_Category"]=="starters") {
...
<?php echo $array[$i]["Food_Name"];?>
...
}
}
I would actually recommend something more like this. I obviously don't know the names of your fields, so I've just taken a guess...
for($i=0;$i<sizeof($array);$i++) {
if($array[$i]["Food_Category"]=="starters") {
...
<?php echo $array[$i]["Food_Name"];?>
...
}
}
You'll be able to access "FoodID" as a parameter within your PHP, just as you would if it had been submitted from a form.
You may be looking for AJAX. AJAX lets you send the form data to a back end PHP file (that can then insert data into a DB, and/or get data from the DB) without refreshing the page.
In fact, when you are using AJAX you don't even need to use a <form> structure -- simple DIVs work just fine. Then you don't need to use event.preventDefault() to suppress the built-in form refresh.
Just build a structure inside a DIV (input fields, labels, etc) and when the user is ready to submit, they can click an ordinary button:
<button id="btnSubmit">Submit</button>
jQuery:
$('#btnSubmit').click(function(){
var fn = $('#firstname').val();
var ln = $('#lastname').val();
$.ajax({
type: 'post',
url: 'ajax_receiver.php',
data: 'fn=' +fn+ '&ln=' +ln,
success: function(d){
if (d.length) alert(d);
}
});
});
ajax_receiver.php:
<?php
$fn = $_POST['fn'];
$ln = $_POST['ln'];
//Do your stuff
Check out this post and especially its examples. Copy them onto your own system and see how they work. It's pretty simple.
I already have implemented an ajax that send a tr and td to a html table every time that a element from the table of the DB is updated, i'll like to send the ajax html data to a append and send it from there to the body and not directly from ajax to the body, any one knows hoy?
this is my php:
<?php if($total>0 && $search!='')
{
do {
echo "<tr><form method='post' action='modifystatusdown.php'>
<td style='width:20%; '><input type='text'><input type='hidden'>";
echo $fila['nombre'];
echo "</td><td style='width:20%;'>";
echo $fila['telefono'];
echo "</td><td style='width:20%;'>";
echo "<input type='Submit' name='delete' value='Atendido'></td></form></tr>";
}
while ($fila=mysqli_fetch_assoc($resultado));
} ?>
this is send it to this ajax:
$(document).ready(function()
{
function actualizar()
{
value = $('#value').html();
$.ajax(
{
type: "POST",
url: "../solicitudes/buscador.php",
success: function(data)
{
$('#value').html(data);
}
})
}
setInterval(actualizar, 3000);
})
finaly from ajax its send it to a table inside a div:
<div class="container">
<table id="value">
</table>
</div>
how could i send it to the append and then to the table? instead directly from ajax?
the idea is something like this, in a table of the database I have a status column which is initialized to 0 and every time someone requests a service from an application android status is changed to 1, that I mentioned above is working, I want to achieve is that every time status is equal to 1, then appears in a html page , just as already appears in the html page, the problem is that assigned an input field where you can type in the text field the service to be assigned to the user requesting and as the table where they are inserted into the html refreshes every 5 seconds and you can not write in the text because it cleared everytime refreshes the table automatically ,
You can combine both $.post and $.get
ie $.post('url', {datatoappend: 'append'}, function(data){
$.get('urlwithsentdata', function(data){
});
});
this works nicely
I have been going crazy for the last 2 weeks trying to get this to work. I am calling a MySQL Db, and displaying the data in a table. Along the way I am creating href links that DELETE and EDIT the records. The delete pulls an alert and stays on the same page. The EDIT link will POST data then redirect to editDocument.php
Here is my PHP:
<?php
foreach ($query as $row){
$id = $row['document_id'];
echo ('<tr>');
echo ('<td>' . $row [clientName] . '</td>');
echo ('<td>' . $row [documentNum] . '</td>');
echo "<td><a href='**** I NEED CODE HERE ****'>Edit</a>";
echo " / ";
echo "<a href='#' onclick='deleteDocument( {$id} );'>Delete</a></td>";
// this calls Javascript function deleteDocument(id) stays on same page
echo ('</tr>');
} //end foreach
?>
I tried (without success) the AJAX method:
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: 'edit_id='edit_id,
success: function(response){
$('#result').html(response);
}
});
}
</script>
I have been using <? print_r($_POST); ?> on editDocument.php to see if the id has POSTed.
I realize that jQuery/AJAX is what I need to use. I am not sure if I need to use onclick, .bind, .submit, etc.
Here are the parameters for the code I need:
POSTs the $id value: $_POST[id] = $id
Redirects to editDocument.php (where I will use $_POST[id]).
Does not affect other <a> OR any other tags on the page.
I want AJAX to "virtually" create any <form> if needed. I do not
want to put them in my PHP code.
I do not want to use a button.
I do not want to use $_GET.
I don't know what I am missing. I have been searching stackoverflow.com and other sites. I have been trying sample code. I think that I "can't see the forest through the trees." Maybe a different set of eyes. Please help.
Thank you in advance.
UPDATE:
According to Dany Caissy, I don't need to use AJAX. I just need to $_POST[id] = $id; and redirect to editDocument.php. I will then use a query on editDocument.php to create a sticky form.
AJAX is used when you need to communicate with the database without reloading the page because of a certain user action on your site.
In your case, you want to redirect your page, after you modify the database using AJAX, it makes little sense.
What you should do is put your data in a form, your form's action should lead to your EditDocument, and this page will handle your POST/GET parameters and do whatever database interaction that you need to get done.
In short : If ever you think you need to redirect the user after an AJAX call, you don't need AJAX.
You have a SyntaxError: Unexpected identifier in your $.ajax(); request here
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: 'edit_id='edit_id,
success: function(response){
$('#result').html(response);
}
});
}
</script>
it should be like this
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: {edit_id: edit_id},
success: function(response){
$('#result').html(response);
}
});
}
</script>
note the 'edit_id='edit_id, i changed, well for a start if you wanted it to be a string it would be like this 'edit_id = ' + edit_id but its common to use a object like this {edit_id: edit_id} or {'edit_id': edit_id}
and you could also use a form for the edit button like this
<form action="editDocument.php" method="POST">
<input type="hidden" name="edit_id" value="272727-example" />
<!-- for each data you need use a <input type="hidden" /> -->
<input type="submit" value="Edit" />
</form>
or in Javascript you could do this
document.location = 'editDocument.php?edit_id=' + edit_id;
That will automatically redirect the user
Given your comment, I think you might be looking for something like this:
Edit
$(document).ready(function() {
$('.editLink').click(function(e) {
e.preventDefault();
var $link = $(this);
$('<form/>', { action: 'editdocument.php', method: 'POST' })
.append('<input/>', {type:hidden, value: $link.data('id') })
.appendTo('body')
.submit();
});
});
Now, I don't necessarily agree with this approach. If your user has permission to edit the item with the given id, it shouldn't matter whether they access it directly (like via a bookmark) or by clicking the link on the list. Your desired approach also prevents the user from opening links in new tabs, which I personally find extremely annoying.
Edit - Another idea:
Maybe when the user clicks an edit link, it pops up an edit form with the details of the item to be edited (details retrieved as JSON via ajax if necessary). Not a new page, just something like a jQuery modal over the top of the list page. When the user hits submit, post all of the edited data via ajax, and update the sql database. I think that would be a little more user-friendly method that meets your requirements.
I was facing the same issue with you. I also wanted to redirect to a new page after ajax post.
So what is did was just changed the success: callback to this
success: function(resp) {
document.location.href = newURL; //redirect to the url you want
}
I'm aware that it defies the whole purpose of ajax. But i had to get the value from a couple of select boxes, and instead of a traditional submit button i had a custom anchore link with custom styling in it. So in a hurry i found this to be a viable solution.
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($row = mysql_fetch_array($result))
{
$entry = $row['Entry'];
$CID = $row['CID'];
$checked =$row['Checked'];
// echo $CID;
echo "<input type=\"text\" value=\"$entry\" name=\"textfield$CID;\" id=\"textfield$CID;\" onchange=\"showUser(this.value)\" />";
echo "<input type=\"checkbox\" value=\"\" name=\"checkbox$CID;\" id=\"checkbox$CID;\" value=\"$checked\"".(($checked == '1')? ' checked="checked"' : '')." />";
echo "<br>";
}
?>
<div id="dynamicInput"></div>
<input type="submit" id="checklistSubmit" name="checklistSubmit" class="checklist-submit"> <input type="button" id="CompleteAll" name="CompleteAll" value="Check All" onclick="javascript:checkAll('checklist', true);"><input type="button" id="UncheckAll" name="UncheckAll" value="Uncheck All" onclick="javascript:checkAll('checklist', false);">
<input type="button" value="Add another text input" onClick="addInput('dynamicInput');"></form>
It is populated from the database based on the users session_id, however if the user wants to create a new list item (or is a new visitor period) he can click the button "Add another text input" and a new form element will generate.
All updates to the database need to be done through AJAX\JQUERY and not through a post which will refresh the page.
I really need help on this one. Getting my head around this kind of... Updating method kind of hurts!
Thanks.
You will need to catch the click of the button. And make sure you stop propagation.
$('checklistSubmit').click(function(e) {
$(e).stopPropagation();
$.post({
url: 'checklist.php'
data: $('#checklist').serialize(),
dataType: 'html'
success: function(data, status, jqXHR) {
$('div.successmessage').html(data);
//your success callback function
}
error: function() {
//your error callback function
}
});
});
That's just something I worked up off the top of my head. Should give you the basic idea. I'd be happy to elaborate more if need be.
Check out jQuery's documentation of $.post for all the nitty gritty details.
http://api.jquery.com/jQuery.post/
Edit:
I changed it to use jquery's serialize method. Forgot about it originally.
More Elaboration:
Basically when the submit button is clicked it will call the function specified. You want to do a stop propagation so that the form will not submit by bubbling up the DOM and doing a normal submit.
The $.post is a shorthand version of $.ajax({ type: 'post'});
So all you do is specify the url you want to post to, pass the form data and in php it will come in just like any other request. So then you process the POST data, save your changes in the database or whatever else and send back JSON data as I have it specified. You could also send back HTML or XML. jQuery's documentation shows the possible datatypes.
In your success function will get back data as the first parameter. So whatever you specified as the data type coming back you simply use it how you need to. So let's say you wanted to return some html as a success message. All you would need to do is take the data in the success function and place it where you wanted to in the DOM with .append() or something like that.
Clear as mud?
You need two scripts here: one that runs the AJAX (better to use a framework, jQuery is one of the easiest for me) and a PHP script that gets the Post data and does the database update.
I'm not going to give you a full source (because this is not the place for that), but a guide. In jQuery you can do something like this:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() { // DOM is ready
$("form#checklist").submit(function(evt) {
evt.preventDefault(); // Avoid the "submit" to work, we'll do this manually
var data = new Array();
var dynamicInputs = $("input,select", $(this)); // All inputs and selects in the scope of "$(this)" (the form)
dynamicInputs.each(function() {
// Here "$(this)" is every input and select
var object_name = $(this).attr('name');
var object_value = $(this).attr('value');
data[object_name] = object_value; // Add to an associative array
});
// Now data is fully populated, now we can send it to the PHP
// Documentation: http://api.jquery.com/jQuery.post/
$.post("http://localhost/script.php", data, function(response) {
alert('The PHP returned: ' + response);
});
});
});
</script>
Then take the values from $_POST in PHP (or any other webserver scripting engine) and do your thing to update the DB. Change the URL and the data array to your needs.
Remember that data can be like this: { input1 : value1, input2 : value2 } and the PHP will get something like $_POST['input1'] = value1 and $_POST['input2'] = value2.
This is how i post form data using jquery
$.ajax({
url: 'http://example.com',
type: 'GET',
data: $('#checklist').serialize(),
cache: false,
}).done(function (response) {
/* It worked */
}).fail(function () {
/* It didnt worked */
});
Hope this helps, let me know how you get on!
Well basically what im trying to do is create a method that can take any datatype and essentially convert it into php recognized data or JSON for that matter. So say i wanted to pass an array, a 2d array, or just some basic string. I would call this function pass in the parameters and sent it off to php. Now i would reverse engineer this function to be able to translate what was sent and get it as a php array, or just a string. I'm good at php arrays but iam bad at javascript ones, here lies my dilemma. If anyone could give me some insight into the javascript side, i can take if from there. The most important however is how to send an array with javascript with php that is DYNAMIC, that is, its dimension can change and its length. Say for example this function handles different forms of different number of input fields, it needs to post this form's input fields into an array, convert to JSON and send.
Appreciate any help i can get or a guide in the right direction, thank you.
Here is an example using jQuery, AJAX, JSON and PHP JSON encode. I hope it will help you to understand the flow.
test.php:
<script type="text/javascript" src="jquery-1.4.2.js"></script>
<script type="text/javascript" src="jsFile.js"></script>
<form action='_test.php' method='post' class='ajaxform'>
<input type='text' name='txt1' value='Test Text 1'>
<input type='text' name='txt2' value='Test Text 2'>
<input type='submit' value='submit'>
</form>
<br /><br />
<div id='testDiv1'>txt1's text comes here....</div>
<div id='testDiv2'>txt2's text comes here....</div>
_test.php:
<?php
$arr = array( 'testDiv1' => $_POST['txt1'], 'testDiv2' => $_POST['txt2'] );
echo json_encode( $arr );
?>
jsFile.js
jQuery(document).ready(function(){
jQuery('.ajaxform').submit( function() {
$.ajax({
url : $(this).attr('action'),
type : $(this).attr('method'),
dataType: 'json',
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
return false;
});
});