Check inArray from a php array

Question

PHP:

$ports = $Db->query('SELECT port FROM servers');

Javascript:

$("#port").on('keyup', function(){
    var port = $("#port").val();
    var portlist = <?php print(json_encode($ports)); ?>;

    if(jQuery.inArray(port, portlist[port])!==-1)
    {
        $("#result").removeClass("label-success");
        $("#result").removeClass("label-warning");
        $("#result").addClass("label label-danger");
        $("#result").html("Port belegt!");
    }
    else
    {
        $("#result").removeClass("label-danger");
        $("#result").removeClass("label-warning");
        $("#result").addClass("label label-success");
        $("#result").html("Port Frei!");
    }



})

Why is this not working?

The array contains 1234 and 7777. So if I type 7777 into the html input field (#port) it should do the first action (if true)

But it always goes into the else action.

Change if(jQuery.inArray(port, portlist[port])!==-1) to if(jQuery.inArray(port, portlist)!==-1). Otherwise you'd need to check if portlist[port] exists and that's not inArray. — Anyone
– Anyone, Commented Jan 30, 2014 at 13:49
Hmm okay. But then the variable portlist only contains [object Object] — Vince
– Vince, Commented Jan 30, 2014 at 14:17

Volkan Ulukut · Accepted Answer · 2014-01-30 13:56:19Z

1

you need to cast integers in array to string by quoting and remove [port] here is a working example of your code http://jsfiddle.net/4X68y/

 var portlist = ["1234","777"];

    if(jQuery.inArray(port, portlist)!==-1)

answered Jan 30, 2014 at 13:56

Volkan Ulukut

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7 Comments

Vince Over a year ago

So the values from php are strings, right? Without the mysql query (var portlist = ["1234","777"];), it works..

Volkan Ulukut Over a year ago

well thats my guess. check the source code if php outputs [1234,7777] or ["1234","7777"]. if its the first, you can always typecast integer values with (string)$intval

Vince Over a year ago

Thats the output from php (print_r): Array ( [0] => Array ( [port] => 7777 ) [1] => Array ( [port] => 1234 ) )

Volkan Ulukut Over a year ago

ports seem to be integer. just iterate the array and parse them to string. foreach($ports as $key=>$port) { $ports[$key] = (string)$port; }

Volkan Ulukut Over a year ago

then try this: foreach($ports as $port) { $portArr[] = $port['port']; }; $portJson = '["' . implode('","',$portArr) . '"]';

|

Lashus · Accepted Answer · 2014-01-31 18:31:38Z

0

jQuery inArray accepts array as second parameter - not specific key.

Try this:

$("#port").on('keyup', function(){
var port = $(this).val();
var portlist = <?php print(json_encode($ports)); ?>;

if(jQuery.inArray(port, portlist)!==-1)
{
    $("#result").removeClass("label-success");
    $("#result").removeClass("label-warning");
    $("#result").addClass("label label-danger");
    $("#result").html("Port belegt!");
}
else
{
    $("#result").removeClass("label-danger");
    $("#result").removeClass("label-warning");
    $("#result").addClass("label label-success");
    $("#result").html("Port Frei!");
}

})

--EDIT--

The following works for me (if the portlist is array not json object. If it isnt maybe it's a good idea to just loop the results and create the array old fashioned way like below)

PHP:

<?php
$q = $db->query("SELECT port FROM ports");
$ports = array();
while($$row = $result->fetch_assoc()){ 
            $ports[] = $row['port'];
        } 
?>

JS CODE:

$("#port").on('keyup', function(){
var port = parseInt($(this).val());
var portlist = <?php print(json_encode($ports)); ?>;

if(jQuery.inArray(port, portlist)===-1)
{
    $("#result").removeClass("label-success");
    $("#result").removeClass("label-warning");
    $("#result").addClass("label label-danger");
    $("#result").html("Port belegt!");
}
else
{
    $("#result").removeClass("label-danger");
    $("#result").removeClass("label-warning");
    $("#result").addClass("label label-success");
    $("#result").html("Port Frei!");
}
})

edited Jan 31, 2014 at 18:31

answered Jan 30, 2014 at 13:55

Lashus

3991 silver badge9 bronze badges

6 Comments

Vince Over a year ago

but without the specified key, I only get [object Object] :S

Lashus Over a year ago

Could you please past what have you got in portlist variable after printing json_encode($ports)?

Vince Over a year ago

Sure. I get [{"port":7777},{"port":1234}]

Lashus Over a year ago

As i thought - u're getting it's parsed as json object that's why you're getting back object - pretty obvious right? Look my edit for answer :) Best regards.

Vince Over a year ago

I'm getting confused :S. Your example does not use the php array, right?

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