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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
    2,413
    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.
    Yes, I know I'm about as subtle as being hit by a bus..(\\.\ Aug08)
    Yep... I say it like I see it, even if it is like a baseball bat in the nutz... (\\.\ Aug08)
    I want to leave this world the same way I came into it, Screaming, Incontinent & No memory!
    I laughed that hard I burst my colostomy bag... (\\.\ May03)
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  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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