Javascript - Adding variable to a string of code

Question

I realized how bad I am with javascript :(

I have this:

function getUrlVars() {
var vars = {};
var parts = window.location.href.replace(/[?&]+([^=&]+)=([^&]*)/gi,     
function(m,key,value) {
    vars[key] = value;
});
return vars;
}

var theidis = getUrlVars()["id"];

and

$.tzPOST = function(action,data,callback){
$.post('php/ajax.php?id='+theidis,'&action='+action,data,callback,'json');
}

$.tzGET = function(action,data,callback){
$.get('php/ajax.php?id='+theidis,'&action='+action,data,callback,'json');
}

The first gets the ?id=value parameter properly. I tested with an alert(theidis).

I am looking to do a simple thing - sorry if this is stupid -.- I have worked as this for so long that I am beginning to oversee things.

I want an ?id=value to be added to the php/ajax.php. I tried with the above, but it doesn't seem to work. I also tried by just adding it.

Thank you in advance. :-)

David Hedlund · Accepted Answer · 2012-10-03 13:41:40Z

1

Your $.get call specifies

'php/ajax.php?id='+theidis, '&action='+action, data ...

It looks like you're passing data twice. Either &action='+action should be your data parameter, or data. Did you mean to concatenate the &action part into the URL?

answered Oct 3, 2012 at 13:41

David Hedlund

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8 Comments

Christian Over a year ago

Hmm, they are three seperate functions. I know the script works with "theidis" removed. I am looking to - when calling the file - adding another parameter called ID to ID things in the database when running the script. Is it even possible? Or have I slammed my head voluntarily into a wall -.-

David Hedlund Over a year ago

@Christian: $.get has the signature get( url [, data] [, success(data, textStatus, jqXHR)] [, dataType] ). So your first parameter is always your URL, which may contain additional data, by passing querystrings. If you want to add more data, you can do this in the second parameter. This could be a string or an object. You're currently passing '&action='+action as your second parameter. Then you're passing data which doesn't make sense when you've already passed string data. Your third parameter is supposed to be the callback.

Christian Over a year ago

So everything could be combined into +action?

David Hedlund Over a year ago

@Christian: we don't know enough of your context here. Data is being passed to the server in the form "variable=value&variable2=value" and these can be sent either as form data or querystrings. By appending values to the URL you will be sending some querystring data, regardless of whether or not you specify data. But you could equally well have passed 'id='+theids+'&action='+action as your data parameter, and left your url clean. Which approach you want to use depends on how your server is retreiving the values, what control you have over that, and what other data you want to send.

Christian Over a year ago

I see it. Hmm. Well I got some help with the script. But I know it works - it sends data from a form to a database with the help of those PHP-files. However, the only - and last thing - I need it to do to work properly is that is must choose a variable database. I have set it up so that $id specifies which database in ajax.php. In ajax.php some files are required - and those files handles data, action etc. properly. I just need to add the id parameter to make e.g. the query DELETE FROM ".$id."_chatusers work. Did that clarify? :-)

|

Mortalus · Accepted Answer · 2012-10-03 13:55:33Z

1

you probably meant:

$.tzGET = function(action,data,callback){
    $.get('php/ajax.php?id='+theidis + '&action='+action,data,callback,'json');
}

Change the , sign to + sign.

NOTE!: data should be null if you are cocncating string to the URL's query string.

answered Oct 3, 2012 at 13:55

Mortalus

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1 Comment

Christian Over a year ago

I don't have the code here - but this should call in the ID? Data should be null? Sorry if that's a dumb question. :-)

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