demuro1 Posted December 2, 2012 Share Posted December 2, 2012 Hello, I'm trying to save some keystrokes and I have assigned document.getElementById('pageNumber').value; to a variable. The first section of code is the function that I am having trouble with. The second section is all the code I am currently using. In the if statement the last alert will give the correct value but the if statement won't populate on the web page. If I copy the document.getElementById('pageNumber').value statement and replace it for the variable p in the if statement p=1, then the value is reassigned and populated properly on the page. Any ideas? Thanks function page(input, form) { alert(input.name); alert(form.pageNumber.value); var p = document.getElementById('pageNumber').value; if(p < 1) { p = 1; } alert(p); } <html><head> <title>test</title> <script type='text/javascript'>/*make some magic*/ function page(input, form) { alert(input.name); alert(form.pageNumber.value); var p = document.getElementById('pageNumber').value; if(p < 1) { p = 1; } alert(p); } function keystrokeSubmit(textbox,e,form) { e = e? e : window.event; var prskeyrtrn = e.keyCode? e.keyCode : e.which? e.which : e.charCode? e.charCode :null; if (prskeyrtrn == 13 && textbox.id == 'pageField') { //alert(textbox.id); page(textbox,form); } } function buttonSubmit(button,form) { if (button.id == 'previousPage' || button.id == 'nextPage') { //alert(button.id); page(button,form); } } </script></head> <body> <form name="headForm" id="headForm" method="POST" action="<?php echo ($_SERVER['PHP_SELF']); ?>"> <input type="text" name="pageField" id="pageField" placeholder="enter a page number" onkeypress="keystrokeSubmit(this,event,form)"> <br> <input type="text" style="visibility:visible" name="pageNumber" id="pageNumber" value="0"> <br> <input type="button" name="nextPage" id="nextPage" value="nextPage" onclick="buttonSubmit(this,form);"> <input type="button" name="previousPage" id="previousPage" value="previousPage" onclick="buttonSubmit(this,form);"> </form> </body> </html> Link to comment Share on other sites More sharing options...
Don E Posted December 2, 2012 Share Posted December 2, 2012 var p just contains the value to document.getElementById('pageNumber').value, not a reference to it. That's why when you did p = 1, it doesn't get populated on the webpage. Link to comment Share on other sites More sharing options...
demuro1 Posted December 2, 2012 Author Share Posted December 2, 2012 ahhh that makes sense. thanks Link to comment Share on other sites More sharing options...
davej Posted December 3, 2012 Share Posted December 3, 2012 function keystrokeSubmit(textbox, e, form) { ..... } </script></head> <body> <form ....> <input type="text" name="..." id="..." placeholder="..." onkeypress="keystrokeSubmit(this, event, form)"> .... </form> Interesting code. Link to comment Share on other sites More sharing options...
demuro1 Posted December 3, 2012 Author Share Posted December 3, 2012 (edited) how so is the code interesting? and do you have any suggestions to make this code better? I'd love any feedback you have Here's what it ended up as: <html><?php$maxPage = 9;$pageNumber = $_POST['pageNumber'];if(!isset($pageNumber)){$pageNumber=1;}?> <head> <title>page number</title> <script type='text/javascript'> function page(input,form) { //alert(input.id); //alert(document.getElementById('pageNumber').value); var p = parseInt(document.getElementById('pageNumber').value); //input.id is numberField if ((input.id == 'pageField') && ( (isNaN(document.getElementById('pageField').value)) || (parseFloat(document.getElementById('pageField').value) % 1 !== 0) || (parseInt(document.getElementById('pageField').value) < 1) || (parseInt(document.getElementById('pageField').value) > <?php echo $maxPage; ?>) )) { alert('I\'m Sorry, ' + document.getElementById('pageField').value + '\ is not a valid page number. \nYou will be returned to page ' + document.getElementById('pageNumber').value); } else if (input.id == 'pageField') { document.getElementById('pageNumber').value = parseInt(document.getElementById('pageField').value); }//input.id is previousPage else if (input.id == 'previousPage' && p <= 1) { alert('You are already on page ' + p); } else if (input.id == 'previousPage') { p -= 1; document.getElementById('pageNumber').value = p; }//input.id is nextPage else if (input.id == 'nextPage' && p >= <?php echo $maxPage; ?>) { alert('You are on the last page.'); } else if (input.id == 'nextPage') { p += 1; document.getElementById('pageNumber').value = p; } else {alert('pageNumber Error');} form.submit(); } function keystrokeSubmit(textbox,e,form) { e = e? e : window.event; var prskeyrtrn = e.keyCode? e.keyCode : e.which? e.which : e.charCode? e.charCode :null; if (prskeyrtrn == 13 && textbox.id == 'pageField') { //alert(textbox.id); page(textbox,form); } } function buttonSubmit(button,form) { if (button.id == 'previousPage' || button.id == 'nextPage') { //alert(button.id); page(button,form); } } </script></head> <body> <form method="POST" action="<?php echo ($_SERVER['PHP_SELF']); ?>"> <input type="text" name="pageField" id="pageField" placeholder="Jump to Page" onkeypress="keystrokeSubmit(this,event,form);"><br> <input type="text" name="pageNumber" id="pageNumber" value="<?php echo $pageNumber; ?>"><br> <input type="button" name="previousPage" id="previousPage" value="previous page" onclick="buttonSubmit(this,form);"> <input type="button" name="nextPage" id="nextPage" value="next page" onclick="buttonSubmit(this,form);"> </form> <?php print("The current page number is ". $pageNumber); print("<br>");?> </body> </html> Edited December 3, 2012 by demuro1 Link to comment Share on other sites More sharing options...
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