javascript - checkbox question with php -

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my question when user click on checkbox, user activated. code working fine. when user again click on uncheckbox, again user de-activated. how that? here working code.

activate user when user click on checkbox

if($_get['doaction'] == 'activate') {     if(!empty($_get['q'])) {         $userid = $_get['q'];         $conn = db_connection();         $query = "update user set activate = '1' userid = '".$userid."' ";         $result=$conn->query($query);     } }

here checkbox

<input type="checkbox" name="app" onchange="calluser(this.value,doaction.value);" value="<?php echo $userid;?>" <?php if($row['approved'] == '1'){ echo "checked=\"true\""; }?>/> <input type="hidden" name="doaction" id="doaction" value="approved" />

thanks much. :-) edit-> here calluser() function

<script type="text/javascript"> function calluser(str,action,third) {  var xmlhttp;     if (str=="") { document.getelementbyid("txthint").innerhtml=""; return; } if (window.xmlhttprequest)   {// code ie7+, firefox, chrome, opera, safari    xmlhttp=new xmlhttprequest(); } else  {// code 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","adminpanel.php?    q="+str+"&doaction="+action+"&app="+third,true);     xmlhttp.send(); }  </script>

here screenshot http://i.stack.imgur.com/mpqfc.png

you should send state of checkbox rather hidden field value.

onchange="calluser(<?php echo $userid;?>,this.value);"

then on php side can a

$userid = $_get['q']; if(!empty($_get['doaction']) {    ... activate ... } else { ... deactivate ... }

also, please aware code sample , anwser both extremely insecure. wide open sql injection attacks , potential permission problems.

edit : fix not sending user id.


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