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photo edit can't

Hello.my Friend,Today i found some problem when edit my text and photo update
I have this dir

upload
list.php
edit.php

list.php

PHP Code:


		
			
<!DOCTYPE html>

<html>

    <head>

        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">

        <title></title>

    </head>

    <body>

        <table border="1">

        <tr > <td>id </td>

              <td>Name </td>

               <td>email</td>

                <td>photo</td>

                <td>Edit</td>

        </tr>

        <?php

        $host= "localhost";

        $name="root";

        $pass="";

        $db="test";

        $con=mysql_connect($host,$name,$pass);

        $seldb=mysql_select_db($db,$con);

        $query=mysql_query("SELECT * FROM user");

        while($row=mysql_fetch_array($query)){

        ?>

        <tr>

            <td><?php echo $row['id']; ?></td>

            <td><?php echo $row['name']; ?></td>

            <td><?php echo $row['email']; ?></td>

            <td><img src="upload/<?php echo $row['photo']; ?>" width="100" height="100"/></td>

      <td>  <a href="edit.php?id=<?php echo $row[0]; ?>">Edit</a></td>

        </tr>

        <?

        }

        ?>

        </table>

    </body>

</html>

edit.php

Code:

<?php



 $host= "localhost";

        $name="root";

        $pass="";

        $db="test";

        $con=mysql_connect($host,$name,$pass);

        $seldb=mysql_select_db($db,$con);

        $query=mysql_query("SELECT * FROM user");

        if(isset($_GET['id'])){

            $id=$_GET['id'];

             $query=mysql_query("SELECT * FROM user");

             while($row=  mysql_fetch_assoc($query)){

       

?>

<form action="edit.php" method="post" enctype='multipart/form-data'>

Name: <input type="text" name="name" value="<?php echo $row['name']; ?>"/>

Email : <input type="text" name="email" value="<?php echo $row['email']; ?>"/>

Photo: <input type="file" name="photo" value="upload/<?php echo $row['photo']; ?>"/>

<?php 

        } }

        ?>

<input type="hidden" name="hid" value="<?php echo $row['id']; ?>"/>

<input type="submit" name="save" value="Update"/>

</form>

<?php

if(isset($_POST['save'])){

    $hid=$_POST['hid'];

    $name=$_POST['name'];

    $email=$_POST['email'];

    $photo=$_FILES['photo'];

    

   $update= mysql_query("UPDATE user SET name='$name',email='$email',photo='$photo' WHERE id='$hid'");

         move_uploaded_file($_FILES['photo']['tmp_name'], "upload/".$photo);

    if($update){

      echo "<meta http-equiv=refresh content=0,url=list.php>";

    }

}

?>

i just want to change text when i edit the text but my problem is,when
i edit the text if i don't choose the image,which is empty
so,How can i solve the problem ?
Thanks

Don't let them change the photo image using that particular edit script.
If they need to change the photo, create a separate edit script for that.
Use the script you're showing us for only changing textbox information
such as name and email.

Quote:

Originally Posted by mlseim (Post 1304263)

So,when i only edit the photo ,how to recall the photo from the list.php

i call with photo id value like this value="<?php row['photo']; ?>

and i don't get the photo value in edit.php

Sorry, I don't follow you ...

There is one form to edit the photo.
You show them the photo that is currently there,
and then when they pick a new one from their PC
and submit, the current photo is overwritten by
the new one.

Quote:

Originally Posted by mlseim (Post 1304363)

yes, i pick the picture from my pc the photo is overwritten on existinig photo

So, i just edit the name and don't edit photo ,i want to still the photo
last existance photo don't pick the photo again ,
if update the photo is empty or my script is not work if i don't choose the photo
Do you know to solve this?
Thanks for your suggestion

You can just change the photo filename name if you wish, the filename and not the actual photo.
I just don't know why you would want to do that.

When a photo is uploaded, it puts the photo(s) into a directory somewhere and
the photo filename is written into your database table? I assume that is what you're doing.

So your column name "photo" is the filename of the photo itself.

Is that how you have it set up?