Unable to display and open file in php and html

Question

Sophia_1 0 Junior Poster in Training

9 Years Ago

Hi everyone, am trying to save excel/docs filename in mysql and then display the filename and open the file to edit. The filename can be save in mysql but unable to display the filename and open the file. Appreciate if you could advise. Thanks a lot.

    <div id="tabs-1"> <?php
    error_reporting(E_ALL ^ E_NOTICE);
    $con = mysql_connect("localhost","user","");
    if (!$con){
    die("Can not connect: " . mysql_error());
    }
    mysql_select_db("pq",$con);

    $Picid = $_GET['Picid'];
    $nwQty = "SELECT * FROM progress WHERE Picid = '$Picid'";
    // check if the progress is already
    $solution = mysql_query($nwQty);
    if(isset($_POST['submit'])){
    if (mysql_num_rows($solution) == 0){
        // put the progress in table
    $sql = "INSERT INTO progress(Attachment1,Attachment11,
    Picid) VALUES ('" . $_POST["Attachment1"] . "','" . $_POST["Attachment11"] . "','" . $Picid . "')"; 
    $result = mysql_query($sql);
    } else {
        // update progress in table
    $sql =("UPDATE progress SET  Attachment1='" . $_POST["Attachment1"] . "',
    Attachment11='" . $_POST["Attachment11"] . "'    WHERE Picid='" . $Picid . "'");
    $result = mysql_query($sql);
    echo('Record Updated');
    }
    $result = mysql_query("SELECT * FROM progress WHERE  Picid='$Picid' ");
    $row= mysql_fetch_array($result); 
    }
    ?> <form name="progress" id="progress" method="post" action=""> <p><ENCTYPE="multipart/form-data">Forms: <input type="file" name="Attachment1" id="Attachment1" MAXLENGTH=50  ALLOW="text/*" value="<?php echo $row['Attachment1']; ?>"></p> <p><ENCTYPE="multipart/form-data">Department Forms: <input type="file" name="Attachment11" id="Attachment11" MAXLENGTH=50  ALLOW="text/*" value="<?php echo $row['Attachment11']; ?>"></p> <input type="hidden" name="Picid" id="Picid" value="<?php echo $row['Picid']; ?>" > <td colspan="2"><input type="submit" name="submit" id="submit" value="Save" class="btnSubmit"></td> <?php
    mysql_close($con);  
    ?> </div> </form>

display html-css php

2 Contributors
4 Replies
362 Views
3 Days Discussion Span
Latest Post 9 Years Ago Latest Post by Sophia_1

diafol

9 Years Ago

1) Don't use deprecated code (mysql_* functions)
2) Avoid mixing PHP and HTML wherever possible. Only include echo / loops / conditionals if able. Bulk PHP shuld go above the DTD or in an include file.
3) You are wide open to SQL injections - THIS IS IMPT!!!
4) You are trying to programmatically put the filename into a file input. This can't be done.
5) The form tag looks messed up - form attributes (enctype) are placed as some sort of tag. Wrong. No close form tags. The HTML is a mess, start again.

Please read my tutorial ( DW Tutorial: Common Issues with MySQL and PHP ) on the use of PDO or mysqli. This code is downright dangerous.

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Sophia_1 0 Junior Poster in Training · Answer 1 · 2015-10-08T01:13:40+00:00

Hi Diafol, thanks a lot for your advise. So for item 2), should i separate the php coding for file display and open in another file and the html in another file? Thanks.

Sophia_1 0 Junior Poster in Training · Answer 2 · 2015-10-08T02:36:00+00:00

Hi, i was able to save the file to server in folder14 using the coding below. Appreciate if you could advise how to display the file name in the form and view the file from the form. Thanks a lot.

            <?php
              if ( isset($_POST['myupload'])){

                if($_FILES['myfile']['size'] > 0){
                    $path = $_SERVER['DOCUMENT_ROOT']."\folder14";
                    $targetfolder = "\\upload\\"; 
                    $targetfolder = $path . $targetfolder . basename( $_FILES['myfile']['name']) ;          

                    $result = move_uploaded_file($_FILES["myfile"]["tmp_name"], $targetfolder);
                    if($result) {
                        echo "The file ". basename( $_FILES['file']['name']). " is uploaded<br>";
                    } else {
                        echo "Problem uploading file<br>";
                    } 
                }else{       
                    echo "Error on uploading file";
                }
            }
            ?>
             </table>
             <html>
             <head>
             <title>A simple example of uploading a file using PHP script</title>
             </head>
             <body>
             <b>Simple example of uploading a file</b><br/>
             Choose a file to be uploaded<br />
             <form action="upload.php" method="post" enctype="multipart/form-data">
             <input type="file" name="myfile" size="50" />
             <br />
             <input type="submit" name="myupload" value="Upload" />
             </form>
             </body>
             </html>

Sophia_1 0 Junior Poster in Training · Answer 3 · 2015-10-09T03:45:01+00:00

Hi, i am now able to upload,display and view the file with the coding below. However, i still cannot save the filename in mysql. Kindly please advise. Thanks a lot.

        <?php 
        $con = mysql_connect("localhost","user","");
        if (!$con){
        die("Can not connect: " . mysql_error());
        }
        mysql_select_db("pq",$con);

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

            if($_FILES['Attachment1']['size'] > 0){
                $path = $_SERVER['DOCUMENT_ROOT']."\folder14";
                $targetfolder = "\\upload\\"; 
                $targetfolder = $path . $targetfolder . basename( $_FILES['Attachment1']['name']) ;             

                $result = move_uploaded_file($_FILES["Attachment1"]["tmp_name"], $targetfolder);
                $sql="INSERT INTO progress(Attachment1) VALUES('$targetfolder')";
         mysql_query($sql); 

                if($result) {
                    echo "The file ". basename( $_FILES['file']['name']). " is uploaded<br>";

                } else {
                    echo "Problem uploading file<br>";
                } 
            }else{       
                echo "Error on uploading file";
            }
        }

        ?>
         </table>
         <html>
         <head>
         <title>A simple example of uploading a file using PHP script</title>
         </head>
         <body>
         <b>Simple example of uploading a file</b><br/>
         Choose a file to be uploaded<br />
         <form action="upload.php" method="post" enctype="multipart/form-data">
         <input type="file" name="Attachment1" size="50" />
         <br />
         <input type="submit" name="myupload" value="Upload" />
         </form>
         </body>
         </html>
         <?php
         $handle = opendir('upload');
        if($handle){
            while(($entry = readdir($handle)) !== false){
                if($entry != '.' && $entry != '..'){
                    echo "<a href=\"upload/$entry\">$entry</a><br>";
                }
            }
            closedir($handle);
        } 
        ?>