I have a simple code-form like this:
$.ajax({
url: "http://myurl/test.php",
type: "POST",
dataType: "html",
data: {
action: "login"
},
success: function (data) {
console.log(data)
}
});
When on PHP server page, I try to print $_POST["action"] yet this field is blank. Why does it work fine if I use $.post(url,data,function)?
$.ajax({
url: "http://myurl/test.php",
type: "POST",
dataType: "html",
data: {
method: "login"
},
success: function (data) {
console.log(data);
}
});
login is a php function
You may use method instead of type.
$.ajax({
url: "http://myurl/test.php",
method: "POST", // <-- Use method, not type
dataType: "html",
data: {
action: "login"
},
success: function (data) {
console.log(data)
}
});
Problem is solved.....
There is a problem with the callback on success event...and yesterday I focused my attention only on the request php in the developers tool....when I used Curl (with the right request) all work correctly and I forgot to check client code in success callback....
Never use post method with developers console,it's very easy to lose more time :=)
Related
I am calling a ajax function on button click.
$.ajax({
type: 'GET',
url: "javascript.php?orderid=CF450AA4",
//data: "orderid=CF450AA4",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (msg, status) {
alert("successful");
console.log(msg);
},
error: function (msg, status) {
console.log("failure");
console.log(msg);
alert("failure");
}
});
}
I have the javascipt.php file where the code is written for server side and is working fine.
I want to recode it. I want that when someone clicks on button the url change to index.php?orderid=CF450AA4 and result is delivered in the variable. How to do this
i did it by doing..
location.hash = 'url';
Hope will work for you..
I am attempting to get our knockout form to submit to a php script and am getting undefinedIndex errors. I am pretty sure it is the way we are sending the data over in our ajax function.
Here is the ajax:
$.ajax({
url: '/orders/add',
type: 'post',
data: {payload:ko.toJSON(allModel)},
contentType: 'application/json',
success: function (result) {
alert(result);
}
});
Here is the PHP (we use laravel)
return json_decode($_POST["payload"]);
Pete is correct. You need to use just one data field. If you want a variable, define it before the $.ajax post
var dataPayload = ko.toJSON(allModel);
$.ajax({
url: '/orders/add',
type: 'post',
data: {payload: dataPayload},
contentType: 'application/json',
success: function (result) {
alert(result);
}
});
I have a piece of jQuery code:
var newData = checkCP(somedata);
if(newData=="hi"){
alert("Welcom");
}
function checkCP(jsData){
$.ajax({
type: "POST",
url: "Process.php",
data: jsData,
dataType: "json",
success: function(data){
if(data.match==1)
return "hi";
else
return "bye";
}
});
}
I don't know why the welcome alert never shows up. I checked everything; the PHP file returns 1, but apparently before it waits on the AJAX response it passes the
if(new=="hi"){
alert("Welcom");
}
Is there any way to wait for the AJAX response, then read the rest of codes in jQuery?
Yes, you can set the 'async' option in $.ajax() to false.
$.ajax({
type: "POST",
async: false,
url: "Process.php",
data: jsData,
dataType: "json",
success: function(data){
if(data.match==1)
return "hi";
else
return "bye";
}
Firstly, please don't use 'new' as a variable name. 'new' already means something.
Now onto the actual question...
when you call checkCP jquery does the ajax post successfully and automatically waits for a response before execuiting the success function. the thing is that the success function is a stand-alone function and returning a value from that does not return anything from checkCP. checkCP is returning null.
As proof try sticking an alert in your success function.
Something like this should do the trick:
function deal_with_checkCP_result(result)
{
if(result=="hi"){
alert("Welcom");
}
}
function checkCP(jsData,callback){
$.ajax({
type: "POST",
url: "Process.php",
data: jsData,
dataType: "json",
success: function(data){
if(data.match==1)
callback( "hi");
else
callback( "bye");
}
});
}
Pass deal_with_checkCP_result as callback
i have one AJAX function getting results from php file as bellow
$.ajax({
type: "POST",
url: "GeteT.php",
cache:false,
data:"id="+ encodeURIComponent(1),
dataType:'json',
success: function(json)
{
g_foo = json.foo;
}
});
now i want to use the value of g_foo but i can't i try using console.log but i am facing strange problem. like
if i use function like that
$.ajax({
type: "POST",
url: "GeteT.php",
cache:false,
data:"id="+ encodeURIComponent(1),
dataType:'json',
success: function(json)
{
g_foo = json.foo;
console.log(g_foo);
}
});
now i can see the value return from php file
if now i use function like that
$.ajax({
type: "POST",
url: "GeteT.php",
cache:false,
data:"id="+ encodeURIComponent(1),
dataType:'json',
success: function(json)
{
g_foo = json.foo;
}
});
console.log(g_foo);
now i got error undefined g_foo;
thanks
As ajax is asynchronous, there's no way of making sure that g_foo is available as soon as you call the $.ajax() request, hence it's not available outside that callback.
You could however, specify async:false to that .ajax() call, but it will most likely block the browser, attempting to wait for the request, something I don't think you want to happen.
To use it outside you have two ways:
1)make the call syncronous like this (this is not a best practice as the browser has to wait for the call to continue the flow):
$.ajax({
type: "POST",
url: "GeteT.php",
cache:false,
data:"id="+ encodeURIComponent(1),
dataType:'json',
async: false,
success: function(json)
{
g_foo = json.foo;
}
});
console.log(g_foo);
2) call a function on success and continue the flow from there using whatever data has been returned from the function (this is the best practice)
$.ajax({
type: "POST",
url: "GeteT.php",
cache:false,
data:"id="+ encodeURIComponent(1),
dataType:'json'
success: function(json)
{
g_foo = json.foo;
yourfunction(g_foo);
}
});
function yourfunction(g_foo){
//here g_foo is set
}
Hi i have problem with some things
Here is the code
function get_char_val(merk) {
$.ajax({
type: "POST",
url: "char_info2.php",
data: { name: merk },
dataType: "html",
success: function(data){return(data);}
});
}
alert(get_char_val("str"));
when alert comes out it's output the undefined please help fast i'm making this two days xC
get_char_val does not return anything. The success callback needs to have alert(data) in it to return the data from AJAX.
$.ajax is asynchronous - meaning it doesn't happen in order with other code, that is why the callback exists.
Your return; statement will return the value to the anonymous function you pass into the success handler. You cannot return a value like this, you need to invoke another callback instead.
function get_char_val(merk, cb) {
$.ajax({
type: "POST",
url: "char_info2.php",
data: { name: merk },
dataType: "html",
success: function(data){cb.apply(this, data);}
});
}
get_char_val("str", function(data) {
alert(data);
});
You can either set the async:false config parameter in the ajax call options or use the call back.
Using Async:false -
function get_char_val(merk)
{
var returnValue = null;
$.ajax
(
{
type: "POST",
async:false,
url: "char_info2.php",
data: { name: merk },
dataType: "html",
success: function(data){returnValue = data;}
}
);
return returnValue;
}
alert(get_char_val("str"));
P.S: Note that making ajax calls synchronous is not advisable.
Since you do an asynchronous ajax call, the response of the server is handled in the "success" function. The return in your success function is useless, you should use the "data" parameter directly in the function
This should work. Use it like this.
function get_char_val(merk) {
var myReturnData= "";
$.ajax({
type: "POST",
url: "char_info2.php",
data: { name: merk },
dataType: "html",
success: function(data){myReturnData = data;}
});
return myReturnData;
}
alert(get_char_val("str"));
Just understand the problem with your piece of code. Use the return statement under the right function.