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Thread: How to get a link from an uploaded file?

  1. #1
    Join Date
    Dec 2013
    Posts
    12

    Thumbs up How to get a link from an uploaded file?

    PHP Code:
    @move_uploaded_file$upload_Temp $uploadFile); 
    chmod($uploadFile0644);
    $upload_URL "uploads/".$upload_Name ;
    $uploadFile "uploads/".$upload_Name ;
    $id=$_POST['id'];
     
    $name=$_POST['name']; 
     
    $upload_Name=($_FILES['FILE']['name']);
     
    //Writes the information to the database
     
    mysql_connect("localhost""freak_admin""what") or die(mysql_error()); 
     
    mysql_select_db("freak_doc") or die(mysql_error()); 
     
    mysql_query("INSERT INTO `users_cv` VALUES ('$id', '$name', '$upload_Name')"); 

    hello Developers

    i have tried to find my way around this but no show.from the php admin everything works fine including the name of the file visible.



    the above code works when i view the info on my view.php(the field for name shows)

    however the field for upload_name is blank because the idea I had isnt working below

    <td>&nbsp;echo "<a href=" ./ $upload_URL . basename($row_users['file']) . ">
    {$row_users['upload_Name']}</a>";</td>



    Can someone help me with what i am doing wrong ?



    thanks
    A newbie in php

  2. #2
    <td>&nbsp;echo "<a href=" ./ $upload_URL . basename($row_users['file']) . ">
    {$row_users['upload_Name']}</a>";</td>
    ...
    You can try with :
    <td>&nbsp;echo "<a href=./uploads/" . basename($row_users['file']) . ">
    {$row_users['upload_Name']}</a>";</td>

  3. #3
    Join Date
    Dec 2013
    Posts
    12
    Hello Hong

    I this i messed up previously of the output i sent earlier on

    This is the code i used <?php echo $row_users['upload_Name']; ?> but it only reveals the file name and not the link to it on the server ho to i echo it to linked to the file on the server?

  4. #4
    Function basename(path,suffix)
    <?php
    $path = "/testweb/home.php";
    //Show filename with file extension
    echo basename($path) ."<br/>";
    //Show filename without file extension
    echo basename($path,".php");
    ?>

    So your link is : ./ $upload_URL . basename($row_users['file']) .
    you check it again and change it.
    The output of the code above will be:
    home.php
    home

  5. #5
    Join Date
    Mar 2007
    Location
    localhost
    Posts
    2,591
    You have to escape the strings that get inserted in to a database.

    mysql_ commands are going to be retired very soon, so switching to the alternative mysqli_ or PDO methods of accessing a database is going to be an important change that you should consider doing now otherwise you will be rewriting your database routines very soon.
    If your post falls off the page, bump it. ...
    Please remember to wrap any code you have in forum tags:-

    [CODE]...[/CODE] [HTML]...[/HTML] [PHP]...[/PHP]

    If you can't think outside the box, you will be trapped forever with no escape...

  6. #6
    Join Date
    Oct 2013
    Posts
    29
    You didn't insert the filename in the database.
    You only inserted id,name, and upload_Name.
    mysql_query("INSERT INTO `users_cv` VALUES ('$id', '$name', '$upload_Name')");

  7. #7
    Join Date
    Dec 2013
    Posts
    12
    thanks you all i finally got my database working but there is an issue here,
    I have about 118 variables to insert. each time i fill the forms it says submityed but when i chack the database table it is'nt there!

    When I fill part of the form and submit behold i see it all in the table.
    is there a limit to the number of variables a form can write to a database?

    sperox says thanks for quick answer

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