newphpcoder Posted May 3, 2012 Share Posted May 3, 2012 Hi....I have form which I put save function on the last textbox:here is my code: <?php error_reporting(0); date_default_timezone_set("Asia/Singapore"); //set the time zone $con = mysql_connect('localhost', 'root','');if (!$con) { echo 'failed'; die();}mysql_select_db("mes", $con);?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html><head><script type="text/javascript" > var input_size = 1;function checkTextBox(bc){ var barcode_ = bc.tabIndex; if ( bc.value.length > input_size ) { for(i=0; i<document.barcode.elements.length; i++) { if( document.barcode.elements[i].tabIndex == (barcode_+1) ) { document.barcode.elements[i].focus(); break; } } } }function postSet() { if (window.event.keyCode==13 || window.event.keyCode==10) { document.getElementById('code_read_box6').disabled = true; save(); alert('code_read_box6'); }} </script><script type="text/javascript">var ajaxTimeOut = null;var ajaxTimeOutOperator = null;var responsePHP; // = "no_reply"var responsePHPOperator;var changeFocus; //= false;var transactionWasSaved;function remoteRequestObject() { var ajaxRequest = false; try { ajaxRequest = new XMLHttpRequest(); } catch(err) { try{ ajaxRequest = new ActiveXObject("MSxml2.XMLHTTP"); } catch(err) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch(err){ // --> change to DOM alert("Not Supported Browser") + err.description; notify('Not Supported Browser.'); return false; } } } return ajaxRequest;}var ajaxRequest; // = remoteRequestObject();var ajaxRequestOperator;</script><script type="text/javascript">function save() { ajaxRequest.onreadystatechange = function () { if (ajaxRequest.readyState==4 && ajaxRequest.status==200) { var result = ajaxRequest.responseText; alert (result); if (result == "failed") { document.getElementById('code_read_box6').disabled = false; document.getElementById('code_read_box6').value = ""; document.getElementById('code_read_box6').focus(); notify("Please scan again."); } if (result == "saved") { alert(result); notify("Transaction has been saved."); reset(); } } } var url = "save_barcode.php"; ajaxRequest.open("POST", url, true);ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");ajaxRequest.setRequestHeader("Content-length", parameters.length);ajaxRequest.setRequestHeader("Connection", "close");ajaxRequest.send(parameters);} </script> </head><body onLoad="document.barcode.code_read_box1.focus();"><form name="barcode" ><input type="text" tabindex="1" id="code_read_box1" value="" onkeyup="checkTextBox(this);"/><br/><input type="text" tabindex="2" id="code_read_box2" value="" onkeyup="checkTextBox(this);"/><br/><input type="text" tabindex="3" id="code_read_box3" value="" onkeyup="checkTextBox(this);"/><br/><input type="text" tabindex="4" id="code_read_box4" value="" onkeyup="checkTextBox(this);"/><br/><input type="text" tabindex="5" id="code_read_box5" value="" onkeyup="checkTextBox(this);"/><br/><input type="text" tabindex="6" id="code_read_box6" value="" onkeyup="checkTextBox(this);" onkeypress="postSet()"/><br/></form></body> </html> I got an error:'ajaxRequest' is null or not an object on line 72It display the error when I press enter on the last textbox.Sorry, I'm not familiar in ajax..I hope somebody can help meThank you Link to comment Share on other sites More sharing options...
justsomeguy Posted May 3, 2012 Share Posted May 3, 2012 The line where you're supposed to set the ajaxRequest object is commented out. It also looks like you're trying to use a function called notify that is never defined. Link to comment Share on other sites More sharing options...
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