kamsmartx Posted April 25, 2008 Share Posted April 25, 2008 Welcome Question I want to change the value of text field when choosing from a list value of the solution Be interpreted as follows I'm in a field next to the texts and also list and is required when selecting the value of the list and let 1 box sent to neighbouring value figure 1 in the database and let Name Quote Link to comment Share on other sites More sharing options...
Goose Posted April 25, 2008 Share Posted April 25, 2008 kamsmartx I don't quite follow what you are asking here. If you have a JavaScript object, lets cal it obj. // this changes the value of the object obj.value = 'a new value'; You can also but the above code into some sort of function and add an event listener to your list. Something like this: <input id="list" onchange="yourFunction();" /> Hopefully this helps, if not, please try and re-ask your question. Or maybe someone else gets it Quote Link to comment Share on other sites More sharing options...
kamsmartx Posted April 25, 2008 Author Share Posted April 25, 2008 take this code <select name="FileNo" id="select2" onChange="change_title();"> <? $query="SELECT * FROM pationts"; $result=mysql_query($query) or die("error: " . mysql_error()); while($row=mysql_fetch_array($result)) { echo "bla"; ?> <option value="<?php echo $row['FileNo'] ; ?>"><?php echo $row['FileNo'] ; ?></option> <? } ?> </select> the above code is for list i need to select value from it and when i select value we must change vale on the below code <input name="PationtName" type="text" id="title"/> Quote Link to comment Share on other sites More sharing options...
Goose Posted April 25, 2008 Share Posted April 25, 2008 So in the change_title() function you want to say something like document.getElementById('title').value = document.getElementById('select2').value; Quote Link to comment Share on other sites More sharing options...
kamsmartx Posted April 25, 2008 Author Share Posted April 25, 2008 no your code print Number of Pationt but i need when i select Number of Pationt the change on next text filed print name of pationt Quote Link to comment Share on other sites More sharing options...
kamsmartx Posted April 25, 2008 Author Share Posted April 25, 2008 ??? Quote Link to comment Share on other sites More sharing options...
kamsmartx Posted April 25, 2008 Author Share Posted April 25, 2008 i think put pationname with list am i Quote Link to comment Share on other sites More sharing options...
kamsmartx Posted April 25, 2008 Author Share Posted April 25, 2008 i think u have to display corresponding PationtName value should be displayed for selected FileNo am i right???????? Quote Link to comment Share on other sites More sharing options...
zenag Posted April 25, 2008 Share Posted April 25, 2008 <? $con=mysql_connect("localhost","root",""); mysql_select_db("dbname",$con); ?><html> <head> <script type="text/javascript"> function chang() { } var xmlHttp function showCustomer(str) { xmlHttp=GetXmlHttpObject(); if (xmlHttp==null) { alert ("Your browser does not support AJAX!"); return; } var url="chan.php"; url=url+"?id="+str; alert(url); xmlHttp.onreadystatechange=stateChanged; xmlHttp.open("GET",url,true); xmlHttp.send(null); } function stateChanged() { if (xmlHttp.readyState==4) { //alert(xmlHttp.responseText); document.myForm.patname.value=xmlHttp.responseText; } } function GetXmlHttpObject() { var xmlHttp=null; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { // Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } } return xmlHttp; } </script> </head><body> <form name="myForm"> <select onChange="showCustomer(this.value)"name="FileNo" id="select2" onChange="change_title();"> <? $query="SELECT * FROM table"; $result=mysql_query($query) or die("error: " . mysql_error()); while($row=mysql_fetch_array($result)) { ?> <option value="<?php echo $row['field'] ; ?>"><?php echo $row['field'] ; ?></option> <? } ?> <input type="text" name="patname" /> </form> <!--<div id="Hint"></div>--> </select> chan.php <? $con=mysql_connect("localhost","root",""); mysql_select_db("dbname",$con); $id=$_GET["id"]; $sql=mysql_query("select * from table where fieldname='$id'"); while($rows=mysql_fetch_array($sql)) { echo $rows["fieldname"]; } ?> Quote Link to comment Share on other sites More sharing options...
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