Notification when new record is added in database php-jquery

Question

I need advice about the title above. I have the following piece of code :

PHP :

$sql = "SELECT * FROM i_case";
$qry = pg_query($connection, $sql);
$res = pg_num_rows($qry);   

$sql_1 = "SELECT * FROM i_case";
$qry_1 = pg_query($connection, $sql_1);
$res_1 = pg_num_rows($qry_1);

do {
    echo "new record on i_case";
} while($res !=  $res_1 and !(usleep(10000)));

JQUERY :

setInterval(function(){    
    $.ajax({
        type : "POST",
        url : "file.php",
        success : function(response){
            alert(response);
        }
    });
},1000);

It's not work that I expecting.

I don't know where is the matter from the code?

but there is no new record on table... no error! i put error: function(){ alert("ERROR"); } — Jin Sun
– Jin Sun, Commented Jan 27, 2012 at 7:48

xdazz · Accepted Answer · 2012-01-27 08:17:48Z

6

Your loop even won't run once because $res is same with $res_1 in almost all cases. If the loop runs, it's a horrible infinite loop.

What you should do is like below.

The javascript:

var old_count = 0;

setInterval(function(){    
    $.ajax({
        type : "POST",
        url : "file.php",
        success : function(data){
            if (data > old_count) {
                alert('new record on i_case');
                old_count = data;
            }
        }
    });
},1000);

And the PHP code:

$sql = "SELECT count(*) as count FROM i_case";
$qry = pg_query($connection, $sql);
$row = pg_fetch_assoc($qry);
echo $row['count'];

edited Jan 27, 2012 at 8:17

answered Jan 27, 2012 at 7:50

xdazz

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2 Comments

Jin Sun Over a year ago

but, when i refresh the page, last notification is shown again and again, any advice?

xdazz Over a year ago

@jin you can load the initial record count first not just set to 0.

Siddharth Mahajan · Accepted Answer · 2016-07-07 09:33:51Z

5

A little tweak to xdazz answer. To avoid alert for first time when you load the page.

JS Code:

var old_count = 0;
var i = 0;
setInterval(function(){    
$.ajax({
    type : "POST",
    url : "notify.php",
    success : function(data){
        if (data > old_count) { if (i == 0){old_count = data;} 
            else{
            alert('New Enquiry!');
            old_count = data;}
        } i=1;
    }
});
},1000);

PHP Code:

include("config.php");   
$sql = "SELECT count(*) as count FROM Enquiry";
$qry = mysqli_query($db, $sql);
$row = mysqli_fetch_assoc($qry);
echo $row['count'];

P.S. My first Stackoverflow answer.

answered Jul 7, 2016 at 9:33

Siddharth Mahajan

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1 Comment

usertest Over a year ago

Can you explain the the code { if (i == 0){old_count = data;} else{ alert('New Enquiry!'); old_count = data;} } i=1;

DerVO · Accepted Answer · 2012-01-27 08:37:09Z

2

I think your whole approach will not work. I suggest, PHP to only generate the count and JS to check if it has changed.

$sql = "SELECT * FROM i_case";
$qry = pg_query($connection, $sql);
$res = pg_num_rows($qry);   
echo $res;

and in JS, you could do somesthing like this:

var count_cases = -1;
setInterval(function(){    
    $.ajax({
        type : "POST",
        url : "file.php",
        success : function(response){
            if (count_cases != -1 && count_cases != response) alert('new record on i_case');
            count_cases = response;
        }
    });
},1000);

edited Jan 27, 2012 at 8:37

answered Jan 27, 2012 at 7:50

DerVO

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4 Comments

Jin Sun Over a year ago

so it call count_cases from -1, what if cases count already 50 record?

DerVO Over a year ago

-1 is just an init value, to prevent the alert in the first run of the function. After the 1st run, count_cases is initiated with -let's say- 50 and the function will trigger the alert, as soon as the PHP returns a number != 50 (likely > 50?)

DerVO Over a year ago

btw: A full SELECT to just count the rows is very inefficient. Have a look on the COUNT statement in xdazz's answer

Jin Sun Over a year ago

okay, i've tried to add new record, but notification is not appearing... why?

vkGunasekaran · Accepted Answer · 2012-01-27 07:50:19Z

1

Have you checked the you're connected to db?

i mean $connection variable returns true?, because i didn't found any code in php that connects to database.

If you could tell the what response you're getting from the server end, then i can help you bit more.

answered Jan 27, 2012 at 7:50

vkGunasekaran

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