How call another php page after jQuery / Ajax success method? - php

Question

I have following Ajax Success method :
success: function(data, el){
var parent = el.find(".jFiler-jProgressBar").parent();
el.find(".jFiler-jProgressBar").fadeOut("slow", function(){
$("<div class=\"jFiler-item-others text-success\"><i class=\"icon-jfi-check-circle\"></i> Success <b class=\"text-danger\"> Delete</b></div>").hide().appendTo(parent).fadeIn("slow");
});
console.log(data);
},
In this success method you see that I have a link which is :
Delete
and
console.log(data)
This console.log(data) is returning following String :
mpic_list_573ecaafae8220.41741946 |76|40
Now I want call a php page e.g : delete_upload_image.php after click on this link Delete with a string mpic_list_573ecaafae8220.41741946 and 76 and 40
I mean :
In delete_upload_image.php page I can get 3 string which is passed when I click on the delete link.
`

Answer 1

You must either bind the click to a function where you send the user to the desired page or directly make a link in the button.
First, if you want to send the differents params separately, you must split them.
var params = data.split('|');
And then use it as you want, for example bind the click method to the new link to create a new ajax call.
$('#delete_upload_image').on('click', function() {
var params = data.split('|');
var url = "delete_upload_image.php?param1=" + params[0] +"&param2=" + params[1] + "&param3=" + params[2];
// Make AJAX call as you want to url
$.get(url);
});

Answer 2

Catch your parameters in your data using split method :
var stringParam = data.split('|')[0];
var int1Param= data.split('|')[1];
var int2Param= data.split('|')[2];
then put your parameters inside your link :
$("<div class=\"jFiler-item-others text-success\"><i class=\"icon-jfi-check-circle\"></i> Success <b class=\"text-danger\"> Delete</b></div>").hide().appendTo(parent).fadeIn("slow");

Answer 3

you must put your data in your link tag:
<a href=\"#delete_upload_image\" data-params='" + data + "'>Delete</a>
then you can send delete requests to "delete_upload_image.php" by below codes:
$(function(){
$('body').on('click', '[href=#delete_upload_image]', function() {
var params = $(this).data('params').split('|');
var url = "delete_upload_image.php?param1=" + params[0] +"&param2=" + params[1] + "&param3=" + params[2];
// Make AJAX call as you want to url
$.get(url);
});
});

Answer 4

now I understand what is your problem;
You want to get the parameters from console.log.
but you can not get the console.log value directly.
What you can do is:
hook the console.log function so that you store when it logs by this code:
var store = '';
var oldf = console.log;
console.log = function(){
store = JSON.stringify(arguments);
oldf.apply(console, arguments);
}
use below function to get parameters from console_store variable and delete image:
function delete_image()
{
var params = store.split('|'),
queryString = $.param({param0:params[0], param1:params[1], param2:params[2]}),
url = "delete_upload_image.php?" + queryString;
$.get(url);
}
call the function on link click:
$('body').on('click', '[href=#delete_upload_image]', delete_image);