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  1. #1
    New Coder
    Join Date
    Aug 2011
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    Exclamation MYSQL/PHP problem

    Hi,

    Creating a website where users can upload files then download them on another page. I've made it so users can upload their files now and I am currently working on the downloading portion. Here is my code and the problem:

    PHP Code:
    <?php

    $Server
    ="xxxx";
    $User="xxxx";
    $Password="xxxx";
    $Database="xxxx";

    $con mysql_connect($Server,$User,$Password);

    if(!
    $con){
        
        die(
    "Couldn't Connect " mysql_error());
        
    }

    mysql_select_db($Database,$con);

    $sql "SELECT * FROM IB2FILES";

    $ctq mysql_query($sql,$con);
    if (!
    $ctq)
    {
        die(
    "SQL Error! Query is $query<br />Error is ".mysql_error());


    echo 
    "Complete";
    echo 
    "<br />";
    echo 
    $ctq;


    ?>
    After running through this code I get this on the next page:

    Complete
    Resource id #3

    The problem here is that I have 3 files on the IB2FILES table and they have 5 fields (id(int),name(varchar),type(varchar),size(int),content(mediumBLOB).

    I have no idea what Resource id #3 means but I keep getting that... Help Please!!!


    Thanks in advance!


    I don't know if it is relevant but here is the user input and upload handling code:

    Form Code:

    Code:
     <form action="show.php" method="post" enctype="multipart/form-data">
             
    <!--         <p style="font-size:20px">Upload a file:</p> -->
             
             <tr><td>  IB2: 
    	<input type=radio name="grade" value="IB2" />
    	<br />
    	IB1: 
    	<input type=radio name="grade" value="IB1" />
    	<br />
    	FY: 
    	<input type=radio name="grade" value="FY" />
    	<br />
    	Y2: 
    	<input type=radio name="grade" value="Y2" />
    	<br />
    	Y1: 
    	<input type=radio name="grade" value="Y1" /> <br /> </td></tr>
             	<tr><td><input type="file" method="post" name="file" enctype="multipart/form-data" /> 
    	<input type="submit" name="submit" value="Upload File" />
    Upload Code:

    PHP Code:
    <?php

    //if (isset($_POST['submitb'])) {

    $TheGrade $_POST['grade'];
                
    $Server="xxxx";
    $User="xxxx";
    $Password="xxxx";
    $Database="xxxx";

    $name $_FILES['file']['name'];
    $type $_FILES['file']['type'];
    $size $_FILES['file']['size'];
    $file $_FILES['file']['tmp_name'];

    if (((
    $_FILES["file"]["type"] == "image/gif")
    || (
    $_FILES["file"]["type"] == "image/jpeg")
    || (
    $_FILES["file"]["type"] == "image/png")
    || (
    $_FILES["file"]["type"] == "application/msword")
    || (
    $_FILES["file"]["type"] == "application/pdf")
    || (
    $_FILES["file"]["type"] == "text/plain")
    || (
    $_FILES["file"]["type"] == "image/pjpeg")))

      {
          
        if(
    $_FILES["file"]["size"] < 5242880){

    if(
    $TheGrade==IB2){

    $gradetable="IB2FILES";
        
    } elseif(
    $TheGrade==IB1){

    $gradetable="IB1FILES";
        
    } elseif(
    $TheGrade==FY){

    $gradetable="FYFILES";
        
    } elseif(
    $TheGrade==Y2){

    $gradetable="Y2FILES";
        
    } elseif(
    $TheGrade==Y1){

    $gradetable="Y1FILES";
        
    } else {
        
        echo 
    "Connection error: No grade submission";
        
    $gradetable=null;
        echo 
    $gradetable;
        
    }

    $connection mysql_connect($Server$User$Password);
    if(!
    $connection){
        die(
    "Couldn't Connect" mysql_error());
    }

    $size "'" $size "'";
    $type "'" $type "'";    
    $name "'" $name "'";    

    mysql_select_db($Database$connection);

    $query "INSERT INTO " $gradetable " (name,type,size) VALUES (" $name "," $type "," $size ")";

    $table mysql_query($query,$connection);
    if (!
    $table)
    {
        die(
    "SQL Error! Query is $query<br />Error is ".mysql_error());
    }   


        } else {
            
            echo 
    "File size is too large";
            
        }

    } else {
                       
               
    $fte "'" "filelist.html" "'" "," "'" "File type not accepted. Please review the page below." "'";    
            
          echo 
    "<script type=text/javascript>
                window.open(" 
    $fte ")
                </script>"
    ;    
           
      }

    echo 
    "File Uploaded!";
        
    ?>

  • #2
    Supreme Master coder! Old Pedant's Avatar
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    The result of calling mysql_query(...) will be *either* a boolean false value (which you are testing for when you do if (!$ctq) ... ) *OR* a resource identifier.

    When you simply do echo $ctg; PHP can't output anything more sensible than the nature of the resource identifier.

    There is NO VALUE OR POINT WHATSOEVER in doing echo $ctg;

    Instead, if you want to *use* the results of you mysql_query(), you must now call mysql_fetch_assoc($ctg) or mysql_fetch_array($ctg) (the first of those is generally easier to work with).

    Go read the PHP documentation:
    http://www.php.net/manual/en/function.mysql-query.php
    http://www.php.net/manual/en/functio...etch-assoc.php
    http://www.php.net/manual/en/functio...etch-array.php
    An optimist sees the glass as half full.
    A pessimist sees the glass as half empty.
    A realist drinks it no matter how much there is.


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