How to get true or false from PHP function using AJAX?

Question

I tested status.php return value with var_dump($result) and alerted it out in check() function like this:

function check() {
    $.ajax({
        url: "status.php"
    }).done(function(data) {
        alert(data);
    });
}

and it did return true or false depending on situation, but when I check if data is true or false inside of check() function it always returns false.

status.php:

<?php 
function status(){
    if(logged() === true) {
        $result = true;
    } else {
        $result = false;
    }
    return $result;
}
status();
?>

check() function: always alerts "false" even though sometimes should be "true"

function check() {
    $.ajax({
        url: "status.php"
    }).done(function(data) {
        if(data === true){
            alert("true");
        } else {
            alert("false");
        }
    });
}

I doubt it return anything because there is no echo in php so nothing will be sendback to browser in ajax request. — Sameer Kumar Jain
– Sameer Kumar Jain, Commented Apr 23, 2016 at 11:56

Barmar · Accepted Answer · 2016-04-23 11:51:24Z

3

You're not sending the return value of the status() function back to PHP. Use:

echo json_encode(status());

And change the AJAX call to expect a JSON response.

function check() {
    $.ajax({
        url: "status.php",
        dataType: 'json'
    }).done(function(data) {
        alert(data);
    });
}

answered Apr 23, 2016 at 11:51

Barmar

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3 Comments

charlietfl Over a year ago

also note that var_dump or json_encode would be string so won't pass === true comparison

charlietfl Over a year ago

"I tested status.php return value with var_dump($result)"

Barmar Over a year ago

I assumed that was just temporary debugging, it's gone now.

JYoThI · Accepted Answer · 2016-04-23 12:01:43Z

2

you just echo the $result like this

ajax not returning value so that we have to echo it.

 <?php function status(){
if(logged() === true) {
    $result = true;
} else {
    $result = false;
}
echo $result;  } status();  ?>

and then should be like this

function check() {
$.ajax({
    url: "status.php"
}).done(function(data) {
    if(data == "true"){
        alert("true");
    } else {
        alert("false");
    }
});   }

edited Apr 23, 2016 at 12:01

answered Apr 23, 2016 at 11:56

JYoThI

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Comments

Anand Jain · Accepted Answer · 2016-04-23 11:56:46Z

1

Use

**dataType: 'json'** 
function check() {
    $.ajax({
        url: "status.php",
        dataType: 'json'
    }).done(function(data) {
        alert(data);
    });
}

and on status.php use

echo json_encode(status());

answered Apr 23, 2016 at 11:56

Anand Jain

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Comments

Tariq · Accepted Answer · 2016-04-23 12:05:11Z

1

You cannot get the response by return method in ajax. to get value "echo" whatever the result in the function , like

function status(){
    if(logged() === true) {
        $result = "1";
    } else {
        $result = "0";
    }
    echo $result;exit;
}

you will get the value 1 or 0 in your ajax success function

answered Apr 23, 2016 at 12:05

Tariq

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Comments

dbf · Accepted Answer · 2016-04-23 13:25:17Z

0

Ajax may not be returning a boolean true or false, rather a string.

So try and put true in double quotes:

if(data=="true")

You can also use the trim function on data to ensure no whitespace is present in the returned data, like so:

if($.trim(data)=="true")

edited Apr 23, 2016 at 13:25

dbf

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answered Apr 23, 2016 at 11:52

Love2Code

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2 Comments

Barmar Over a year ago

The AJAX isn't returning anything, since there's no echo in the PHP.

Love2Code Over a year ago

Thanks for pointing it out. You can use echo $result instead of $return result in status.php.

Shantanu Madane · Accepted Answer · 2016-04-23 13:52:34Z

0

Just Remove Type checking i.e '===' replace with '=='

function check() {
    $.ajax({
        url: "status.php"
    }).done(function(data) {
        if(data == true){
            alert("true");
        } else {
            alert("false");
        }
    });
}

answered Apr 23, 2016 at 13:52

Shantanu Madane

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