mrhussein Posted October 8, 2012 Share Posted October 8, 2012 Hello, regarding to this example in your website: this example display data from mysql database\table with parameter (dropdown list),and i do input the choices to this list,what if i want same this but the choices in the (dropdown list) come from database\table as well ? anyone can help me plz? this example containing two files as the following: 1. HTML File: <html><head><script type="text/javascript">function showUser(str){if (str=="") { document.getElementById("txtHint").innerHTML=""; return; }if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); }else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); }xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } }xmlhttp.open("GET","getuser.php?q="+str,true);xmlhttp.send();}</script></head><body><form><select name="users" onchange="showUser(this.value)"><option value="">Select a person:</option><option value="1">Peter Griffin</option><option value="2">Lois Griffin</option><option value="3">Glenn Quagmire</option><option value="4">Joseph Swanson</option></select></form><br /><div id="txtHint"><b>Person info will be listed here.</b></div></body></html> 2. PHP File: <?php$q=$_GET["q"];$con = mysql_connect('localhost', 'peter', 'abc123');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("ajax_demo", $con);$sql="SELECT * FROM user WHERE id = '".$q."'";$result = mysql_query($sql);echo "<table border='1'><tr><th>Firstname</th><th>Lastname</th><th>Age</th><th>Hometown</th><th>Job</th></tr>";while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName'] . "</td>"; echo "<td>" . $row['Age'] . "</td>"; echo "<td>" . $row['Hometown'] . "</td>"; echo "<td>" . $row['Job'] . "</td>"; echo "</tr>"; }echo "</table>";mysql_close($con);?> Thank you very much... Regards, Link to comment Share on other sites More sharing options...
php_developer Posted October 8, 2012 Share Posted October 8, 2012 save the html file with .php extntion first: Then for select box do as follows:<?php $query=mysql_query("select * from table-name where ..."); ?><select name="users" onchange="showUser(this.value)"><option value="">Select a person:</option><?php while($data=mysql_fetch_array($query)){<option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?></select>don't forget to create the connection with database... Link to comment Share on other sites More sharing options...
mrhussein Posted October 8, 2012 Author Share Posted October 8, 2012 Thank You Very Very Much your helpfull Link to comment Share on other sites More sharing options...
mrhussein Posted October 8, 2012 Author Share Posted October 8, 2012 actully i changed the extntion to .php and did the following but the page dont want to apear: <html><head><script type="text/javascript">function showUser(str){if (str=="") { document.getElementById("txtHint").innerHTML=""; return; }if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); }else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); }xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } }xmlhttp.open("GET","data.php?q="+str,true);xmlhttp.send();}</script></head><body><form><?php $con = mysql_connect('localhost', 'wordpressuser461', 'r_uw7x.-EJh&');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("wordpress461", $con);$query=mysql_query("select id from marks"); ?><select name="users" onchange="showUser(this.value)"><option value="">Select a person:</option><?php while($data=mysql_fetch_array($query)){<option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?></select></form><br /><div id="txtHint"><b>Person info will be listed here.</b></div></body></html> it looks there is some error in the code Link to comment Share on other sites More sharing options...
Ingolme Posted October 8, 2012 Share Posted October 8, 2012 Here's the first thing I spotted. <?php while($data=mysql_fetch_array($query)){<option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?> You should read the error messages. They're very descriptive and they tell you exactly where the error is occurring so that you can find it easily. Link to comment Share on other sites More sharing options...
mrhussein Posted October 8, 2012 Author Share Posted October 8, 2012 (edited) thank u i found it and this is the correction : <?php while($data=mysql_fetch_array($query)){ ?><option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?> Edited October 8, 2012 by mrhussein Link to comment Share on other sites More sharing options...
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