how to update a database that has a dropdown list in php

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  • leesyaa
    New Member
    • Jan 2014
    • 20

    how to update a database that has a dropdown list in php

    hello,

    I have a html form for user to submit their inquiry. after submitting the form, the data will go to php file and then it will connect to the database. I can do that.

    the problem is, for admin view, it will have a dropdown list (approved/not approved) for admin to validate the data. how to update the database? I managed to do the dropdown, but when I clicked the save button, the database didnt update anything.

    please help me, thanks in advanced !

    ==========form. html===========
    Code:
    <!DOCTYPE html> <head> <meta charset="utf-8"> <title>data</title> <link rel="stylesheet" media="screen" href="http://bytes.com/styles.css" > </head> <body bgcolor="#13b4ff"> <div id="header" style="background-color: #4169E1;"><hr> <form class="form" action="http://bytes.com/submit.php" method="post" name="form" > <ul> <li> <label for="name">Name:</label> <input type="text" name="name" required /> </li> <li> <label for="position:">Jawatan:</label> <input type="text" name="position" /> </li> <li> <label for="unit">Unit:</label> <input type="text" name="unit" required /> </li> <li> <label for="institute">Institute:</label> <input type="text" name="institute" required /> </li> <li> <label for="telefon">No. Telefon:</label> <input type="number" name="telefon" required /> </li> <li> <label for="faks">No. Faks:</label> <input type="number" name="faks"/> </li> <li> <label for="email">E-mail:</label> <input type="email" name="name" placeholder="name@something.com"/ required> <span class="form_hint">proper format<script type="text/javascript">
    /* <![CDATA[ */
    (function(){try{var s,a,i,j,r,c,l,b=document.getElementsByTagName("script");l=b[b.length-1].previousSibling;a=l.getAttribute('data-cfemail');if(a){s='';r=parseInt(a.substr(0,2),16);for(j=2;a.length-j;j+=2){c=parseInt(a.substr(j,2),16)^r;s+=String.fromCharCode(c);}s=document.createTextNode(s);l.parentNode.replaceChild(s,l);}}catch(e){}})();
    /* ]]> */
    </script></span> </li> <li> <label for="data">Data Required:</label> <input type="text" name="data" required/> </li> <li> <label for="purpose">Purpose:</label> <input type="text" name="purpose" required/> </li> <li> <button class="submit" type="submit">Submit</button> </li> </ul> </form> </body> </html>
    ==========submi t.php========== =
    Code:
    <?php
    
    //debug mode
    error_reporting(E_ALL);
    ini_set('display_errors', '1');
    
    //to show some error is smthng went wrong
    $errors = array();
    
    function connect(){
    	$connection = mysql_connect("localhost", "root", "" );
    	$db = mysql_select_db('permohonan_data', $connection);
    	
    if (!$connection || !$db){
    	return false;
    	}
    else{
    	return true;
    	}
    }
    	
    //will run if user did submit the form
    if (!empty($_POST)){
    	
    //connect sql server:
    if (!connect()){
    	$errors[] = "Can't establish link to MySql server";
    	}
    
    	$name = $_POST['name'];
            $position = $_POST['position'];
            $unit = $_POST['unit'];
            $institute = $_POST['institute'];
            $telefon = $_POST['telefon'];
            $faks = $_POST['faks'];
            $email = $_POST['email'];
            $data = $_POST['data'];
            $purpose = $_POST['purpose'];
    	
    //no error til here
    if (empty($error)){
    
    //prevent SQL injection
        $name = mysql_real_escape_string($name);    
        $position = mysql_real_escape_string($position);
        $unit = mysql_real_escape_string($unit);
        $institute = mysql_real_escape_string($institute);
        $telefon = mysql_real_escape_string($telefon);
        $faks = mysql_real_escape_string($faks);
        $email = mysql_real_escape_string($email);
        $data = mysql_real_escape_string($data);
        $purpose = mysql_real_escape_string($purpose);
    
    }
     
    //try insert value
    $query = "INSERT INTO 'user'
    	 (name,position,unit,institute,telefon,faks,email,data,purpose)
            VALUES ('$name', '$position', '$unit', '$institute', '$telefon', '$faks', '$email', '$data', '$purpose')";
    	
    //try
    if (!mysql_query($query)){
    	//
    	//die(mysql_error());
    	$errors[] = "Can't insert the values";
    	}
    else {
    	//on success
    	header("Location:thankyou.php");
    	exit();
    
    }
    
    }	
    
    ?>
    ==========part of admin.php====== =====
    Code:
    <?php
    	$connect = mysql_connect("localhost","root","");
    	if (!$connect){
    		die(mysql_error());
    	}
    	mysql_select_db("permohonan_data");
    	$option = '';
    	$results = mysql_query("SELECT * FROM user ORDER BY id DESC");
    	
    	$option .='<select>';
    	$count=mysql_num_rows($results);
    	
    	if(isset($_POST['submit']))
    {
    	$tindakan = mysql_real_escape_string(stripslashes($_POST['tindakan']));
    	
    	$sql_edit = "UPDATE permohonan_data SET pemohon = tindakan = '".$tindakan."'";
    	if($result_edit = mysql_query($sql_edit))
    	{
    		$status_edit= "Success";
    	}
    	else
    	{
    		$status_edit= "Failed";
    	}
    }
    	
    	while ($row = mysql_fetch_array($results)){
    	
    	?> <tr> <td><?php echo $row['id']?></td> <td><?php echo $row['name']?></td> <td><?php echo $row['unit']?></td> <td><?php echo $row['telefon']?></td> <td><?php echo $row['faks']?></td> <td><?php echo $row['email']?></td> <td><?php echo $row['data']?></td> <td><?php echo $row['purpose']?></td> <td><select> <option <?php if( $row['tindakan'] == '-'){ echo "selected";}?> >-</option> <option <?php if( $row['tindakan'] == 'app'){ echo "selected";}?> >Approved</option> <option <?php if( $row['tindakan'] == 'notapp'){ echo "selected";}?> >Not Approved</option> </select></td> </tr> <?php
    }
    	$option .='</select>';
    ?> </tbody> </table> <input type='submit' value='Save'>

    I need help with my admin.php. I know the code is unorganized. I'm really new to this. :/
    Last edited by Rabbit; Jan 13 '14, 05:33 AM. Reason: Please use [CODE] and [/CODE] tags when posting code or formatted data.
  • Exequiel
    Contributor
    • Jul 2012
    • 288

    #2
    try to determine or put the value for pemohon $sql_edit = "UPDATE permohonan_data SET pemohon = tindakan = '$tindakan'";

    Comment

    • Dormilich
      Recognized Expert Expert
      • Aug 2008
      • 8694

      #3
      Code:
      "UPDATE permohonan_data SET pemohon = tindakan = '".$tindakan."'";
      I doubt this is valid syntax

      Comment

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