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hey guys

 

im trying to make this form

 

i want it to insert in database on submit..

 

i got

 

if($_POST["submit"] == "Insert"){  

			  $img=$_POST['img'];
			  


		  
			  
mysql_query("UPDATE `_users` SET `logo` = '$img' WHERE `userid` = 2");


     echo "<meta http-equiv=Refresh content=2;url=main.php>";

			   }

?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>

</head>

<body>
<form action="#"  method="post" enctype="multipart/form-data" >
  <p>
    <label>
      <input name="img" type="file" id="img" size="15" />
    </label>
  </p>
  <p>
    <label>
      <input type="submit" name="Insert" value="submit" width="5"/>
    </label>
  </p>
</form>

 

it does not work?

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if($_POST["submit"] == "Insert"){

 

The button's name is Insert, not submit.  And it's value is submit, not Insert.  The if should read :

 

if($_POST["Insert"] == "submit"){

 

Also, there is no INSERT statement, you are using an UPDATE statement, which is not the same as an INSERT.

 

There is no connection to a database, so the mysql_query() call cannot work.

 

 

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https://forums.phpfreaks.com/topic/196061-if-submit/#findComment-1029812
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I took a closer look at your code:

 

<input name="img" type="file" id="img" size="15" />

 

This INPUT element is for a file upload.  You do not get a value in $_POST['img'] from it.  The uploaded file information is stored in the $_FILES array (see the documentation http://us2.php.net/manual/en/reserved.variables.files.php)

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