php $_REQUEST[ ] dosnt work

Question

i have form in html file , it's file body

<form action="" method="post">       
    <input id="name" type="text" name="username" placeholder="username" required></br></br>
    <input id="password" type="password" name="password" placeholder="password" required></br></br>
    <p id="warn"></p></br></br>
    <input type="submit" value="Login" onclick="f()"></br>
</form>

<script>
function f() {
    var name = document.getElementById("name").value ;
    var password = document.getElementById("password").value ;
    var xttp = new XMLHttpRequest();
    xttp.onreadystatechange = function () {
        if (this.readyState == 4 && this.status == 200) {
            document.getElementById("warn").innerHTML = this.responseText ;             
        }
    };

    xttp.open("GET", "tst1.php?name="+name+"& password"+password, true);
    xttp.send();
}
</script>

and i have php file callled tst1.php

<?php 
    echo $_REQUEST["name"]+"...."+$_REQUEST["password"] 
?>

im trying to use ajax to show the entered value below them in p tag with id="warn" but it dose not show it

You are submitting your form at the same time you do your Ajax call. Your page refreshes before you can see if it actually works or not. — Ivar, Sep 02 '17 at 17:55
Question is not clear enough to give a good answer. What have you done to debug? What are you seeing passed back in your console net tab? — Brian Gottier, Sep 02 '17 at 17:55
Possible duplicate of [Submit and onclick not working together](https://stackoverflow.com/questions/13839459/submit-and-onclick-not-working-together) — Ivar, Sep 02 '17 at 18:01

score 3 · Accepted Answer · answered Sep 02 '17 at 18:01

3

There are few things you need to change in your code, such as:

echo $_REQUEST["name"]+"...."+$_REQUEST["password"], + is not a concatenation operator, . is. So this statement should be like this:
```
echo $_REQUEST["name"] . "...." . $_REQUEST["password"]; 
```
Missing = sign after password in this statement, xttp.open("GET", "tst1.php?name="+name+"& password"+password, true);. This statement should be like this:
```
xttp.open("GET", "post.php?name="+name+"&password="+password, true);
```
Return false immediately after calling f() function, otherwise the page would refresh and you won't be able to see the output value here <p id="warn"></p>.
```
<input ... onclick="f(); return false;">
```

answered Sep 02 '17 at 18:01

Rajdeep Paul

16,887
3
18
37

thanks for solving problem and sorry for stupid mistakes , could you plain explain me what that return false do ? dose it stop the form to be submitted or ... ? @Rajdeep Paul – mike Sep 02 '17 at 18:49
@mamad What `return false;` will do is, it will prevent the form from being submitted in the first place. That way you'll be able to see the output here, `
`. Also, consider *accepting* the answer if it resolved your issue. [How to accept answer on Stack Overflow?](https://meta.stackexchange.com/a/5235) – Rajdeep Paul Sep 02 '17 at 18:55
sorry im new to web developing , what is common way to have a login page ? i want to check the entered values in my database if there is user show his homepage if not show error ? is it ajax good for this purpose ?can you plz give me some advice ? @ Rajdeep Paul – mike Sep 02 '17 at 19:31
@mamad You can start off with some [AJAX tutorials](https://developer.mozilla.org/en-US/docs/AJAX/Getting_Started). And on top of that, you would fine plenty of tutorials on Google on [how to implement a simple login page using JavaScript/jQuery and AJAX](https://www.google.co.in/search?q=implement+login+page+using+ajax&oq=implement+login+page+using+ajax&gs_l=psy-ab.3...2327.2327.0.2555.1.1.0.0.0.0.119.119.0j1.1.0....0...1.1.64.psy-ab..0.0.0.e7qLEaW9eqI&gws_rd=cr&dcr=0&ei=_AerWYGNK8GLvQSdm7bAAg). – Rajdeep Paul Sep 02 '17 at 19:35

score 0 · Answer 2 · answered Sep 02 '17 at 17:55

0

First you have forgot = equal sign for password here

xttp.open("GET", "tst1.php?name="+name+"& password="+password, true);
xttp.send();

Also you should first check request data by isset like this

$name = isset($_REQUEST["name"]): $_REQUEST["name"] : "";

Also change submit button as button from submit to make ajax work

answered Sep 02 '17 at 17:55

Niklesh Raut

34,013
16
75
109

thanks , but why should i change submit button , i left action empty , what's the difference ? @user2486 – mike Sep 02 '17 at 18:46
It will submit the same page – Niklesh Raut Sep 02 '17 at 19:04

php $_REQUEST[ ] dosnt work

2 Answers2