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POHow do i select buttons using php and enter into database?
 

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    <p>Here is my current code..</p> <pre><code>&lt;form action="js.php" method="post"&gt; &lt;input type="button" id='approve' value="Yes" onclick="a()" class="approve" /&gt; &lt;input type="button" id='reject' value="No" onclick="r()" class="reject"/&gt; &lt;input type="submit" name="submitted" value="TRUE" /&gt; &lt;/form&gt; </code></pre> <p>This is my form and the page title is js.php..</p> <p>I want that if the user clicks on any of the button the name of the button should bge entered into the database into a table called response under the heading response.</p> <p>How Do i do that?Thanks..</p> <p>I have done some PHP part...</p> <pre><code>&lt;?php require_once('connect.php'); if(isset($_POST['submitted'])) { //if on button click mysqli_select_db($connect,"check"); //entering query } ?&gt; </code></pre> <p>here is what Im using to enter the data</p> <pre><code>&lt;?php require_once('connect.php'); //if on button click mysqli_select_db($connect,"pubd"); //entering query if ($_POST['action'] == 'approve') { // user clicked Yes $yes = "insert into response (button) values ('approve')"; $yesr = mysqli_query($connect,$yes) or die(mysqli_error($connect)); } else if ($_POST['action'] == 'reject') { $no = "insert into response (button) values ('reject')"; $nor = mysqli_query($connect,$no) or die(mysqli_error($connect)); } ?&gt; &lt;form action="js.php" method="post"&gt; &lt;input type="button" id='approve' name="action" value="Approve" onclick="a()" class="approve" /&gt; &lt;input type="button" id='reject' value="Reject" name="action" onclick="r()" class="reject"/&gt; &lt;/form&gt; </code></pre>
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