php - Storing database value into variable -


my table category has these columns:

 idcategory categorysubject users_idusers 

i have form simple radio buttons , textbox. have select statement category , need idcategory stored variable ($getcatid) can use statement:

$sql="insert topic(subject, topicdate, users_idusers,     category_idcategory, category_users_idusers) values('($_post[topic])', '$date', '$_session[userid]', '$getcatid', '$_session[userid]');";  

what best way , store categoryid?

 if($_server['request_method'] != 'post') //show form if not posted     {             $sql = "select * category;";       $result = mysqli_query($conn,$sql);           ?>     <form method="post" action="createtopic.php">         choose category:         </br>         </br>          <?php          while ($row = mysqli_fetch_assoc($result)) {             echo "<div class= 'choice'><input type='radio' name='category' value='". $row['idcategory'] . "'>" . $row['categorysubject'] ."</div></br>";          }                 echo 'topic: <input type="text" name="topic" minlength="3" required>              </br></br>              <input type="submit" value="add topic" required>              </form>';              }      if ($_post){        if(!isset($_session['signedin']) && $_session['signedin'] == false) {      echo 'you must signed in contribute'; }        else{          $sql="insert topic(subject, topicdate, users_idusers, category_idcategory, category_users_idusers) values('($_post[topic])', '$date', '$_session[userid]', '$getcatid', '$_session[userid]');";               $result = mysqli_query($conn,$sql);             echo "added!"; 

if understand question correctly, you'll have $getcatid (id of category) in $_post['category'] (after sending form) in case


Comments

Popular posts from this blog

java - SSE Emitter : Manage timeouts and complete() -

jquery - uncaught exception: DataTables Editor - remote hosting of code not allowed -

java - How to resolve error - package com.squareup.okhttp3 doesn't exist? -