How to Insert Value From Radio Button in MySQL in PHP -


i'm making online voting system. system has limit people (population of area) can vote (example-10) & votes add database. want insert values radio buttons in database. tried code. has error.the vote form data doesn't insert db & don't know suitable code case. please if has idea, please let me know. thank-you

<html>      <head>            <title>election</title>            	<link rel="stylesheet" href="bootstrap-3.3.4-dist/css/bootstrap.min.css">          <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>          <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>          <link rel="stylesheet" href="css/presidential.css" type="text/css">  		<link rel="stylesheet" href="css/login.css" type="text/css">          <link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/css/bootstrap.min.css">  		</head>      <body>           <div class="head">               <div class="container-fluid">  		<div class="navbar-header">                     <b> online voting system </b>  	</div>  	      <input type="submit" class="btn btn-primary nextbtn btn-lg pull-right"  name="submit" value="cast vote" />  </div>  </div>                 <div class="container">  <form name="vote" method="post" action="">  <div class="one">       <h2><b>election</b></h2>       <br>  	       <div class="funkyradio">          <div class="funkyradio-default">                <input type="radio" name="radio" id="radio1" value="radio1" />              <label for="radio1"><b>a party</b> <div class="img"><img src="img\a.png" height="75.8em" width="50.3em"  ></img> </div> </label>  		</div>          <div class="funkyradio-default">              <input type="radio" name="radio" id="radio2" value="radio2" />              <label for="radio2"><b>b party</b><div class="img"><img src="img\b.jpg" height="75.8em" width="50.3em"  ></img> </div></label>  		</div>          <div class="funkyradio-default">              <input type="radio" name="radio" id="radio3" />              <label for="radio3"><b>c party</b><div class="img"><img src="img\c.jpg" height="75.8em" width="50.3em"  ></img> </div></label>          </div>          <div class="funkyradio-default">              <input type="radio" name="radio" id="radio4" />              <label for="radio4"><b>d party</b><div class="img"><img src="img\d.jpg" height="75.8em" width="50.3em"  ></img>  </div></label>          </div>          <div class="funkyradio-default">              <input type="radio" name="radio" id="radio5" />              <label for="radio5"><b>e party</b><div class="img"><img src="img\e.jpg" height="75.8em" width="50.3em"  ></img>   </div></label>          </div>          <div class="funkyradio-default">              <input type="radio" name="radio" id="radio6" />              <label for="radio6"><b> f party</b><div class="img"><img src="img\f.jpg" height="75.8em" width="50.3em"  ></img>   </div></label>          </div>      </div>  </div>  </form>  </div>  </body>  </html>    <?php  require'database.php';  if (isset($_post['submit']))  {$radio=$_post['radio'];       if($radio!="")  	 {  	 $query=mysql_query("insert `election`(`candidate_name`) values ('$radio')");  	    if ($query)  		{  			echo"cast vote successfully";  		}  		 else  		 {  			 echo"there problem in database";  		 }  	 }  	   }  ?>

your query won't work because call variable in string. should close string first, call variable , end query in string again.

something this

values ('" . $radio . "');"

the single quotes in query, close query , put php code in, , open string again double quoutes. in string finish query.

look sql injection, because you're vulnerable if use method. question.


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