wright67uk Posted January 15, 2013 Share Posted January 15, 2013 In the coe below the status of the drop down list is passed to the $q variable and used in an sql query. In what way would I change the code if I wanted to have a second drop down? xmlhttp.open("GET","getuser.php?q="+str,true); xmlhttp.send(); <html> <head> <script> 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> Quote Link to comment Share on other sites More sharing options...
stijn0713 Posted January 15, 2013 Share Posted January 15, 2013 what do you mean? you want the returned data from the ajax call to be in a dropdown? in that case just create a dropdown, like so: $personID = $_GET['q']; //use this $personID in your query $rs = mysql_query($sql); $content = '<form><select name="secondDropdown" onchange="anotherFunction(this.value)">'; while ($row = mysql_fetch_assoc($rs)) { $content .= '<option value="'.$row['ID'].'">'.$row['name'].'</option>'; } $content .= /// close form echo $content; Quote Link to comment Share on other sites More sharing options...
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