How to get a variable from a php file with javascript?

Question

What I'm trying to do is query a variable in my check.php file and pull that variable to my index.php file, compare it to a different variable, and reload the page if they don't match. This is what I have so far:

check.php:

some query
echo json_encode($checknew);

index.php:

var checkold = "<?php echo $old['ertek'];?>"; 

function checkupdates() {
  setInterval(function() {
    $.ajax({
        type: 'POST',
        url: 'check.php',
        dataType: 'json'});

        if (checkold != checknew){
            location.reload(); 
       }  
    , 3000)}
};

$( document ).ready( checkupdates );

How do I "catch" the $checknew variable from php and turn it into a variable for my function?

You are not using the ajax call correctly. Need to use .done(function() {} to get the echo value from check.php — Gary Holiday
– Gary Holiday, Commented Aug 4, 2016 at 16:16
success: function() { and here you can do whatever you want with your variables that you got from backend } — impregnable fiend
– impregnable fiend, Commented Aug 4, 2016 at 16:17
Also, var checkold = "<?php echo $old['ertek'];?>"; is not doing what you think it is. checkold will contain the literal string <?php echo $old['ertek'];?> not the value inside $old['ertek']. Please check out Ajax Documentation — Gary Holiday
– Gary Holiday, Commented Aug 4, 2016 at 16:18

milan kyada · Accepted Answer · 2016-08-04 16:19:57Z

2

In your check.php you are echoing json_encode($checknew). Be specific what you are writing if it is only one value then you can simply write echo $checknew;.
And your JAVASCRIPT should be like this

var checkold = "<?php echo $old['ertek'];?>"; 

function checkupdates() {
  setInterval(function() {
    $.ajax({
        type: 'POST',
        url: 'check.php', 
        success:function(checknew){
          if (checkold != checknew){
            location.reload(); 
          }  
        }
    });

     },3000);
}

$( document ).ready( checkupdates() );

answered Aug 4, 2016 at 16:19

milan kyada

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8 Comments

phpnoob Over a year ago

yea, checknew is only one value, not an array

milan kyada Over a year ago

write only echo $checknew; :)

phpnoob Over a year ago

ok so I did what milan suggested and is working now, the only problem is the page reloads every time, even if I set $checkold="1" and var checknew="1"

milan kyada Over a year ago

Why did you set Interval? That is executing every 3sec.

t.niese Over a year ago

Even if $checknew is just a primitive value and not an array it is still a good idea to write echo json_encode($checknew);, because then it can be ensured, that it is always properly encode no matter what the content of $checknew is and in the code it is clear that the data should be returned as json. If it is $checknew is the number 1 then output of json_encode will also be 1. @phpnoob

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