CODEWITHSUNDEEP
phpjqueryajax
Brian Powell
Brian Powell

Reputation: 3411

How to determine what a PHP page posts back to an AJAX call

I've used ajax quite a bit to send data off to a php page and get something back, but I'm not sure how to determine what specifically gets sent back. I understand what most of the fields here do, but I have a few questions as well.

My current code looks like this:

$.ajax({
              url: "search.php", 
              type: "post",
              dataType: "json",
              data: {partNumber: q , newType:newType},
              success: function(data) {
                  console.log(data);
              }
          });

One thing I don't fully understand is how the success: line works. When I say success: function(data), it's my understanding that I'm saying "upon success (of something? I'm not sure what), the information that is returned to this ajax function will be called data," and then inside the brackets, I'm choosing to console.log that data.

Please correct me if I'm wrong in that assumption.

Ultimately, my php page will do a ton of work and return an array with data in it. If the last line of my php code is:

$myArray = array("bob","dan","carl","mike");

how do I choose for that to be returned to ajax? The php page is working fine, but right now my console.log(data) line of code is not returning anything.

Upvotes: 1

Views: 168

Answers (2)

Demodave
Demodave

Reputation: 6652

1) Your on "success" is basically saying when there is not an error that occurs.

Here is testing error

$.ajax({
  url: 'search.php',
  success: function(){
    alert('success');
  },
  error: function(){
    alert('failure');
  }
});

2) For your "data" what you echo on the php side gets returned into the variable defined in your 

function(results) {

for example if you want to return an array you may want to return "echo" json

your array should include a key and a value

$myArray = array("key" => "value");

echo json_encode($myArray);

and on the jquery side parse the json object returned.

function(data) {
    var obj = jQuery.parseJSON(data);

    alert(obj.key);

3) Pass JSON Objects

var JSONObject= {"partNumber":q, "newType":newType};
var jsonData = JSON.parse( JSONObject );  

var request = $.ajax({
  url: "search.php",
  type: "POST",
  data: jsonData,
  dataType: "json"
});

Upvotes: 1

developerwjk
developerwjk

Reputation: 8659

You are not supposed to return anything from the php page called by your ajax, you're supposed to echo it (i.e. write it to the response). 

echo json_encode($myArray);

And if what you're echoing is not JSon, then take out dataType: "json". The success function means that the request was successful, i.e. a response was received back from the server with status 200. There is also an error method you can use for when you don't get a successful response back.

          success: function(data) {
              console.log(data);
          },
          error: function(data) {
              console.log("Error: "+data);
          }

Upvotes: 4

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