php - Ajax Get Syntax -
i 've read few tutorials ajax http request, somehow couldn't understand enough build syntax need.
var path = $(this).data('path'); $.get('http://example.ro/index.php?page=contract.php&path='+path, function(){ });
what try achieve change url (http://example.ro/index.php?page=contract.php) , without reloading page,so can use path value later use in query or other action,in same page.but url doesnt change (it must &path=value @ end),so can't catch value like:
if(isset($_get['id'])){ /do }
is syntax wrong? or cant in way? please!
update
$(document).on('click','tr.listcontractrow', function(e){
var path = $(this).data('path'); $.ajax({ type: 'get', url: 'contract.php', data: {id: path}, success: function(result) { console.log(result) }, error: function() { } }); });
the variable path value taken php content,and want take ajax , return can catch php.
right syntax ajax request made jquery:
$.ajax({ type: 'get', url: 'yourphpscript.php', data: { key1: value1, keyn: valuen }, success: function(result) { // executed on status 200 ok console.log(result) // check sent page }, error: function() { //to executed on status 500 error } });
this pretty basic syntax.
let's assume on page. , want call ajax request, it, store in page, , send later php. submit or ajax.
so step1:
$.ajax({ type: 'get', url: 'givemepath.php', data: { key1: value1, keyn: valuen }, success: function(result) { // pretty ugly, functional. window.path = result.path; // or if plan submit form prepare hidden field id myhiddenfield $.('#myhiddenfield').val(result.path); }, error: function() { //to executed on status 500 error } });
your givemepath.php
<?php $path = ...some php code.... header("content-type: application/json"); echo json_encode(array("path" => $path)); ?>
later when want use again value in page
$.ajax({ type: 'get', url: 'sendpathbacklater.php', data: { path: window.path, otherdata: otherdata }, success: function(result) { //sent it! }, error: function() { //to executed on status 500 error } });
or submit form.
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