CODEWITHSUNDEEP
javascriptphpajax
ditto
ditto

Reputation: 6277

JS: Processing form as a POST but retrieving as a GET?

Retrieving the data with PHP, I cannot use $_POST; but $_GET. Why? Am I sending my form data incorrectly?

I'd have thought request.open("POST" would process the form as a POST and not GET? How may I sent it as a POST?

var request = new XMLHttpRequest();
request.open("POST","email.php?text=" + textarea.value + "&email=" + email.value, true);

request.onload = function() {
    if (request.status >= 200 && request.status < 400) {
        var resp = request.responseText;

        console.log(resp);

    }
};

request.send();

Upvotes: 1

Views: 50

Answers (2)

Tushar
Tushar

Reputation: 87203

Because you're adding data inside the URL.

Change your request to:

request.open("POST","email.php", true);
request.setRequestHeader("Content-length", 2); // 2 here is the no. of params to send
....


request.send("text=" + textarea.value + "&email=" + email.value);

Docs: https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest

Upvotes: 3

Faiz Rasool
Faiz Rasool

Reputation: 1379

The reason is you are sending the variables in the url that is why you are getting in get. See this example post

var http = new XMLHttpRequest();
var url = "get_data.php";
var params = "lorem=ipsum&name=binny"; // all prams variable here
http.open("POST", url, true);

//Send the proper header information along with the request
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");

http.onreadystatechange = function() {//Call a function when the state changes.
    if(http.readyState == 4 && http.status == 200) {
        alert(http.responseText);
    }
}
http.send(params);

Upvotes: 2

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