String Variables

[watagal]

Greetings-My Code:

var validation = ajaxCall('Html_php_code/validateSignIn2.php',"POST",'u='+userName+'&p='+password, false);  // ajaxCall() returns xmlHttp.responseTextif (validation.substr(0,18) == "Invalid User Name:") {}

The error I get via firebug is:validation has no propertiesHow can force the ajaxCall() return as a string object?thanks,-Gal

[Jesdisciple]

I would first try console.log()'ing validation to make sure you have what you think you have. If that works, try adding an empty string to it in the return statement.

[watagal]

Thanks Jesdisciple. In Firefly, under the console tab, then under the Response tab - it returns the string "Successful", which is the only thing echoed in my PHP file.

	var validation = ''+ajaxCall('Html_php_code/validateSignIn.php',"POST",'u='+userName+'&p='+password, false);if (validation.substr(0,18) == "Invalid User Name:") {	document.getElementById('divInvalidUserName').style.display = "block";	document.getElementById('txtUserName').value = "";	} else if (validation.substr(0,17) == "Invalid Password:") {	document.getElementById('divInvalidPassword').style.display = "block";	document.getElementById('txtPassword').value = "";	} else if (validation == "Successful") {	window.location="index.php?p="+FRM_SIGNIN;	} else {	document.write(validation);}

it should satisfy the third condition above, but it goes the the 4th condition and writes "undefined" in the browser.I also added an empty string to validation - no change.Thanks, Gal

[Jesdisciple]

I think we need the definition of ajaxCall...

[watagal]

Yes, you're right. I think I found the strange problem/solution: I was returning the "xmlHttp.responseText" directly (commented out now), now I assign it to a variable first and return the variable -- IT WORKS!

// AJAX callfunction ajaxCall(dataSource, method, encode, async) {	xmlHttp = createAjaxObject();	if (xmlHttp != null) {		xmlHttp.onreadystatechange = function() {			if ((xmlHttp.readyState == 4 || xmlHttp.readyState == "complete" ) && xmlHttp.status == 200) {				// return xmlHttp.responseText;				var response = xmlHttp.responseText;				return response;			}		}										if (method == "POST") {  xmlHttp.open("POST", dataSource, async);											 xmlHttp.setRequestHeader('Content-Type','application/x-www-form-urlencoded');											 xmlHttp.send(encode+"&sid="+Math.random());				 } else {						 xmlHttp.open("GET", dataSource+"&sid="+Math.random(), async);											  xmlHttp.send(null);		 }	} else		 alert ("Browser does not support HTTP Request");}

Thanks, if you or anybody have any insight on this, please don't hesitate to shed some light. I feel like i'm in a dark room, feeling around for the instruction manual.--Gal

[watagal]

Sorry, I was wrong - it still does not work.help? still in the dark

[justsomeguy]

Use console.log like he mentioned, it will write something to the console in Firebug that you can inspect. e.g.var validation = ajaxCall('Html_php_code/validateSignIn2.php',"POST",'u='+userName+'&p='+password, false);console.log(validation);It will write the object where you can click on it and look through it to see all of the properties and methods. If you want to convert something to a string you can do this:var validation = new String(ajaxCall('Html_php_code/validateSignIn2.php',"POST",'u='+userName+'&p='+password, false));

[watagal]

I went back to a post by Aspnetguy in another thread and implemented a callback function (validateSignIn2) and I think it works now. I'm still in the dark reading the manual!Thanks all,

// AJAX callfunction ajaxCall(dataSource, method, encode, async, callback) {	xmlHttp = createAjaxObject();	if (xmlHttp != null) {		xmlHttp.onreadystatechange = function() {			if ((xmlHttp.readyState == 4 || xmlHttp.readyState == "complete" ) && xmlHttp.status == 200) {				if (callback) { callback(xmlHttp.responseText); }			}		}											if (method == "POST") {		xmlHttp.open("POST", dataSource, async);									xmlHttp.setRequestHeader('Content-Type','application/x-www-form-urlencoded');									xmlHttp.send(encode+"&sid="+Math.random());											} else {					xmlHttp.open("GET", dataSource+"&sid="+Math.random(), async);									xmlHttp.send(null);		}	} else		alert ("Browser does not support HTTP Request");}//////////////////function validateSignIn() {	var userName = document.frmSignIn.txtUserName.value.toLowerCase();	var password = document.frmSignIn.txtPassword.value;	if (userName == '') {document.getElementById('divInvalidUserName').style.display = "block"; return;}	if (password == '') {document.getElementById('divInvalidPassword').style.display = "block"; return;}		ajaxCall('Html_php_code/validateSignIn.php',"POST",'u='+userName+'&p='+password, false, validateSignIn2);}function validateSignIn2(responseText) {  // callback function 	document.getElementById('divInvalidUserName').style.display = "none";	document.getElementById('divInvalidPassword').style.display = "none";		var validation = responseText;		if (validation.substr(0,18) == "Invalid User Name:") {		document.getElementById('divInvalidUserName').style.display = "block";		document.getElementById('txtUserName').value = "";			} else if (validation.substr(0,17) == "Invalid Password:") {		document.getElementById('divInvalidPassword').style.display = "block";		document.getElementById('txtPassword').value = "";			} else if (validation == "Successful") {		window.location="index.php?p="+FRM_SIGNIN;			} else {		document.write(validation);	}}

[watagal]

Use console.log like he mentioned, it will write something to the console in Firebug that you can inspect. e.g.var validation = ajaxCall('Html_php_code/validateSignIn2.php',"POST",'u='+userName+'&p='+password, false);console.log(validation);It will write the object where you can click on it and look through it to see all of the properties and methods. If you want to convert something to a string you can do this:var validation = new String(ajaxCall('Html_php_code/validateSignIn2.php',"POST",'u='+userName+'&p='+password, false));

Good to know, I'll use this alot! Thanks.