How to pass ajax sucess result to another php page - php

In my code below I want display $("#searchresults").html(data) this result to other page.
$.ajax({
type: "POST",
url: base_url + 'front/searchresult',
data: data,
success: function(data) {
alert("test");
var val = $("#searchresults").html(data);
window.location.assign("<?php echo base_url()?>front/search/" + val);
}
});

what exactly is in the data variable you receive from your post? is it a json object? is it plain text?
if it is html, I think you should consider placing the result in a div on the current page, and hide items you don't want to see after searching
relocating after ajax requests is not the way to go. Is it an option in your case, to use a form and change the action attribute of the form to your new location?
<form action="front/search/">
<input type="text" name="data">
</form>

Related

Multiple submit buttons on one page but distinguishable (jquery)

I am trying to write a code that 'stores items for later' - a button that has url of the item as hidden input, on submit it calls a php script that does the storage in a db. I am more into php, very little knowledge of anything object-oriented, but I need to use jquery to call the php script without moving over there
The problem is how to assign the x and y variables when I have multiple forms on one page
I was only able to write the following
$("form").bind('submit',function(e){
e.preventDefault();
var x = $("input[type=hidden][name=hidden_url]").val();
var y = $("input[type=hidden][name=hidden_title]").val();
$.ajax({
url: 'save_storage.php?url='+x+'&tit='+y,
success: function() {
alert( "Stored!");
location.reload();
}
});
});
It works fine if you have something like...
<form method="post" action="#">
<input type="hidden" id="hidden_url" name="hidden_url" value="<?php echo $sch_link; ?>"/>
<input type="hidden" id="hidden_title" name="hidden_title" value="<?php echo $sch_tit; ?>"/>
<input type="submit" id="send-btn" class="store" value="Store" />
</form>
..once on the page, I've got about 50 of them.
These are generated via for-loop I suppose I could use $i as an identifier then but how do I tell jquery to assign the vars only of the form/submit that was actually clicked?
You'll have to scope finding the hidden fields to look within the current form only. In an event handler, this will refer to the form that was being submitted. This will only find inputs matching the given selector within that form.
$("form").bind('submit',function(e){
e.preventDefault();
var x = $(this).find("input[type=hidden][name=hidden_url]").val();
var y = $(this).find("input[type=hidden][name=hidden_title]").val();
$.ajax({
url: 'save_storage.php',
data: {
url: x,
tit: y
},
success: function() {
alert( "Stored!");
location.reload();
}
});
});
As #Musa said, it's also better to supply a data key to the $.ajax call to pass your field values.
Inside your form submit handler, you have access to the form element through the this variable. You can use this to give your selector some context when searching for the appropriate inputs to pass through to your AJAX data.
This is how:
$("form").bind('submit',function(e) {
e.preventDefault();
// good practice to store your $(this) object
var $this = $(this);
// you don't need to make your selector any more specific than it needs to be
var x = $this.find('input[name=hidden_url]').val();
var y = $this.find('input[name=hidden_title]').val();
$.ajax({
url: 'save_storage.php',
data: {url:x, tit: y},
success: function() {
alert( "Stored!");
location.reload();
}
});
});
Also, IDs need to be unique per page so remove your id attribute from your inputs.

get the value of <select> without submitting on the same page using php

Hope someone can help me..
i made my program more simpler so that everybody will understand..
i want my program to get the value of the without submitting, i know that this can only be done by javascript or jquery so I use the onChange, but what I want is when i select an option the value should be passed immediately on the same page but using php..
<select id="id_select" name="name" onChange="name_click()">
<option value="1">one</option>
<option value="2">two</option>
</select>
<script>
function name_click(){
value_select = document.getElementById("id_select").value;
}
</script>
and then i should pass the value_select into php in post method.. i dont know how i will do it.. please help me..
You cannot do this using PHP without submitting the page. PHP code executes on the server before the page is rendered in the browser. When a user then performs any action on the page (e.g. selects an item in a dropdown list), there is no PHP any more. The only way you can get this code into PHP is by submitting the page.
What you can do however is use javascript to get the value - and then fire off an AJAX request to a php script passing the selected value and then deal with the results, e.g.
$(document).ready(function() {
$('#my_select').on('change', do_something);
});
function do_something() {
var selected = $('#my_select').val();
$.ajax({
url: '/you/php/script.php',
type: 'POST',
dataType: 'json',
data: { value: selected },
success: function(data) {
$('#some_div').html(data);
}
});
}
With this code, whenever the selected option changes in the dropdown, a POST request will be fired off to your php script, passing the selected value to it. Then the returned HTML will be set into the div with ID some_div.
not sure ..but i guess ajax is what you need..
<script>
function name_click(){
value_select = $("#id_select").val();
$.post('path/to/your/page',{"value":value_select},function(data){
alert('done')
})
}
</script>
PHP
$value=$_POST['value']; //gives you the value_select data
Post with ajax as Alex G was telling you (+1) and then handle the post with PHP. You can define a callback in Javascript which will run when the page responds.
My suggestion go with jquery. Try with this
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js">
<script>
$(document).ready(function(){
$("#id_select").change(function(){
var url = 'http:\\localhost\getdata.php'; //URL where you want to send data
var postData = {'value' : $(this).value};
$.ajax({
type: 'POST',
url: url,
data : postData,
dataType: 'json',
success: function(data) {
//
},
error: function(e) {
console.log(e.message);
}
});
})
})
</script>
In getdata.php
<?php
var $value = $_POST['value'];
// you can do your logic
?>

How to pass multiple parameters by jQuery.ajax() to PHP?

I have a simple load more style script that works fine on the index page, where only one parameter is sent via ajax
$(function() {//When the Dom is ready
$('.load_more').live("click",function() {//If user clicks on hyperlink with class name = load_more
var last_msg_id = $(this).attr("id");//Get the id of this hyperlink this id indicate the row id in the database
if(last_msg_id!='end'){//if the hyperlink id is not equal to "end"
$.ajax({//Make the Ajax Request
type: "POST",
url: "index_more.php",
data: "lastmsg="+ last_msg_id,
beforeSend: function() {
$('a.load_more').html('<img src="loading.gif" />');//Loading image during the Ajax Request
},
success: function(html){//html = the server response html code
$("#more").remove();//Remove the div with id=more
$("ul#updates").append(html);//Append the html returned by the server .
}
});
}
return false;
});
});
With this HTML/PHP
<div id="more">
<a id="<?php echo $msg_id; ?>" class="load_more" href="#">more</a>
</div>
However, I want to add another php variable so that it can also work with particular categories, I have no problems writing the HTML and PHP but I am new to Jquery and struggling to edit the script to include the additional parameter if it is set. This is the HTML that I am thinking of using, just struggling with editing the JQuery
<div id="more"class="<?php echo $cat_id;?>">
<a id="<?php echo $msg_id;?>" class="load_more2" href="#">more</a>
</div>
As always any help is much appreciated!
You can set
data = {onevar:'oneval', twovar:'twoval'}
And both key/value pairs will be sent.
See Jquery ajax docs
If you look under the data section, you can see that you can pass a query string like you are, an array, or an object. If you were to use the same method you already are using then your data value would be like "lastmsg="+ last_msg_id + "&otherthing=" + otherthing,
You can pass multiple URL params in the data portion of your ajax call.
data: "lastmsg="+ last_msg_id +"&otherparam="+ other_param
On the PHP side, you'd just process these as you already are.
You can use this code:
$.ajax({//Make the Ajax Request
type: "POST",
url: "index_more.php",
data: {var1: "value1", var2: "value2"},
beforeSend: function() {
$('a.load_more').html('<img src="loading.gif" />');//Loading image during the Ajax Request
},
success: function(html){//html = the server response html code
$("#more").remove();//Remove the div with id=more
$("ul#updates").append(html);//Append the html returned by the server .
}
});
Try It:
data: JSON.stringify({ lastmsg: last_msg_id, secondparam: second_param_value});
You can add more parameters separating them by comma (,).

Fetch input value from certain class and send with jquery

<input id="u1" class="username">
<input id="u2" class="username">
<input id="u3" class="username">
...
How to fetch input value with "username" class and send with ajax jquery to php page.
i want to recive data like simple array or simple json. (i need INPUT values and not ids)
var inputValues = [];
$('input.username').each(function() { inputValues.push($(this).val()); });
// Do whatever you want with the inputValues array
I find it best to use jQuery's built in serialize method. It sends the form data just like a normal for submit would. You simply give jQuery the id of your form and it takes care of the rest. You can even grab the forms action if you would like.
$.ajax({
url: "test.php",
type: "POST",
data: $("#your-form").serialize(),
success: function(data){
//alert response from server
alert(data);
}
});
var values = new Array();
$('.username').each(function(){
values.push( $(this).val());
});

Sending a value from a dropdown box to PHP via jQuery

I'm trying to take values from a dropdown two boxes and send them to a PHP file which will draw an appropriate field from a mySQL database depending on the combination chosen and display it in a div without refreshing the page using AJAX. I have the second part sorted, but I'm stuck on the first part.
Here is the HTML: http://jsfiddle.net/SYrpC/
Here is my Javascript code in the head of the main document:
var mode = $('#mode');
function get() {$.post ('data.php', {name: form.him.value, the_key: #mode.val()},
function(output) {$('#dare').html(output).show();
});
}
My PHP (for testing purposes) is:
$the_key = $_POST['the_key'];
echo $the_key;
After I have it in PHP as a variable I can manipulate it, but I'm having trouble getting it there. Where am I going wrong? Thanks for your replies!
You need a callback function as well to have the server response to the POST.
$.post('ajax/test.html', function(data) {
$('.result').html(data);
});
This snippet will post to ajax/test.html and the anonymous function will be called upon its reply with the parameter data having the response. It then in this anonymous function sets the class with result to have the value of the server response.
Help ? Let me know and we can work through this if you need more information.
Additionally, $.post in jQuery is a short form of
$.ajax({
type: 'POST',
url: url,
data: data,
success: success
dataType: dataType
});
your jquery selectors are wrong:
html:
<select id="mode">
jquery selector:
$("#mode").val();
html:
<select name="player">
jquery selector:
$("select[name=player]").val();
You want to add a callback to your ajax request, its not too hard to do, here ill even give you an example:
$.ajax({
url: "http://stackoverflow.com/users/flair/353790.json", //Location of file
dataType: "josn",//Type of data file holds, text,html,xml,json,jsonp
success : function(json_data) //What to do when the request is complete
{
//use json_data how you wish to.;
},
error : function(_XMLHttpRequest,textStatus, errorThrown)
{
//You fail
},
beforeSend : function(_XMLHttpRequest)
{
//Real custom options here.
}
});​
Most of the above callbacks are optional, and in your case i would do the following:
$.ajax({
url: "data.php",
dataType: "text",
data : {name: ('#myform .myinput').val(),the_key: $('#mode').val()},
success : function(value)
{
alert('data.php sent back: ' + value);
}
});​
the ones you should always set are url,success and data if needed, please read The Documentation for more information.

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