Run jQuery function if PHP variable exists

Question

I am using a Twitter bootstrap plugin called 'Bootbox' that shows a modal form. I only want the modal form to show up if there is a 'popup' id in a mysql database. Otherwise, I don't want the function to run at all.

Here's what I have:

     var popupid = <?php if ($contact->find_popup()) { echo $contact->popup()->id; } ?>;
 if(popupid) {
     bootbox.dialog({
                    message: "<?php if ($contact->find_popup()) { 
                                    echo $contact->popup()->message;
                        }; 
                    ?>",
                    title: "Contact Pop-Up",
                    buttons: {
                      danger: {
                        label: "Delete...",
                        className: "red",
                        callback: function() {
                           $.ajax({
                                url: "ajax_delete.php?table=popups&id=" + popupid,
                                type: "POST",
                                dataType: 'json',
                                success: function(response) {
                                        //response here if data response
                                    if (response) {
                                            toastr.info('Successfully deleted popup!');
                                            }
                                    }            
                                });
                        }
                      },
                      main: {
                        label: "Ok!",
                        className: "blue",
                        callback: function() {
                        }
                      }
                    }
                });
}

I set a variable called popup that see is there is a popup id present in my DB. My find_popup() method returns true if there is one and false otherwise. If it returns true, the popupid should equal the echoed id I need.

The popup id is then passed into the ajax URL as you can see. I use it to run a delete script that removes the popup if the user selects "Delete...".

Everything works fine right now IF and ONLY IF there is a popup present. If not, my page doesn't work properly. I think it's because the bootbox.dialog is still called.

Maybe I wrote this wrong?

why don't you return some invalid id like -1 when there's no entry in your DB ? Then check for that like if(popupid !== -1) { do your call } — Arkantos
– Arkantos, Commented Mar 5, 2015 at 17:38
your just checking to see if popupid is a variable try something like if(popupid > 0) {, you could also just wrap all of your js in your php if statement and not have your js if — cmorrissey
– cmorrissey, Commented Mar 5, 2015 at 17:39

siliconrockstar · Accepted Answer · 2015-03-05 17:52:45Z

2

So, why render the javascript at all if there is no popup_id in your database?

<?php

if ($contact->findPopup()) {

?>

<!-- javascript/html/whatever goes here -->

<?php

} 

?>

Then your javascript only gets rendered if there is a valid popup_id in the database.

answered Mar 5, 2015 at 17:52

siliconrockstar

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Comments

chris.cavage · Accepted Answer · 2015-03-05 23:04:31Z

var popupid = "<?php if ($contact->find_popup()) { echo $contact->popup()->id; } else { echo false;} ?>";
 if(popupid) {
     bootbox.dialog({
                    message: "<?php if ($contact->find_popup()) { echo $contact->popup()->message; }; ?>",
                    title: "Contact Pop-Up",
                    buttons: {
                      danger: {
                        label: "Delete...",
                        className: "red",
                        callback: function() {
                           $.ajax({
                                url: "ajax_delete.php?table=popups&id=" + popupid,
                                type: "POST",
                                dataType: 'json',
                                success: function(response) {
                                        //response here if data response
                                    if (response) {
                                            toastr.info('Successfully deleted popup!');
                                            }
                                    }            
                                });
                        }
                      },
                      main: {
                        label: "Ok!",
                        className: "blue",
                        callback: function() {
                        }
                      }
                    }
                });
}

Thanks for the suggestions. It wouldn't work properly until I put this line in quotes!

var popupid = "<?php if ($contact->find_popup()) { echo $contact->popup()->id; } else { echo false;} ?>";

Siliconrockstar's suggestion also works, but I am using the above because as it checks for find_popup, the popup id is stored as a variable which I need for the ajax url. Still works though...

Appreciate the help.

score 1 · Accepted Answer · 2015-03-05 17:54:36Z

1

var popupid = <?php if ($contact->find_popup()) { echo $contact->popup()->id; } ?>;

Here, once the PHP is executed, you have :

// If a popup exists
var popupid = 123;

// If there is no popup
var popupid = ;

The second line will make your JS crash.

I suggest:

var popupid = <?php if ($contact->find_popup()) { echo $contact->popup()->id; } else { echo false; } ?>;

Or as Seth suggested:

var popupid = <?php echo (($contact->find_popup()) ? $contact->popup()->id : false); ?>;

edited Mar 5, 2015 at 17:54

answered Mar 5, 2015 at 17:39

user3849602

5 Comments

Seth Over a year ago

Perfect case for Ternary operator. Much more legible that way.

user3849602 Over a year ago

Is the ternary operator better for execution time/resources? (Real question.) Anyway I agree it would have been more readable here, thanks.

user3849602 Over a year ago

Right sorry. I edited my post with that suggestion too.

Seth Over a year ago

No, it doesn't have a performance disadvantage as far as I know. However, you didn't echo the ternary statement in your second example.

user3849602 Over a year ago

@Seth Whoops, fixed! Thank you. :)

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