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Thread: parameter is not entering in xmlhttp.open( "POST", 'show.php', true);

  1. #1
    Join Date
    Jan 2014
    Posts
    2

    parameter is not entering in xmlhttp.open( "POST", 'show.php', true);

    my code is uploading anything what i want. Problem is that it is not entering in the page show.php. Although i did debug and failed to understand why this is not going to other page.

    **show.php**

    <?php
    echo"heloo00";
    $con=mysql_connect("localhost","root",'');
    mysql_select_db("project",$con) or die("error db");

    $sql="select * from upload";
    $query=mysql_query($sql);
    while($row=mysql_fetch_array($query))
    {
    $image=$row ['name'];

    echo '<img src="data:image/png;base64,' . base64_encode( $row['content'] ) . '" />';
    }
    ?>

    **test.php**

    <html>
    <head>
    <style>
    #boox{
    overflow:auto;
    width:600px;
    height:400px;
    }
    </style>

    <script>
    alert("helo");
    </script>
    <script>

    function pict()
    { alert("hel");
    xmlhttp = new XMLHttpRequest();
    var picInput = document.getElementById('userfile').value;
    var uploadpic = document.getElementById('upload').value;
    document.getElementById('usf').innerHTML = picInput;
    document.getElementById('upic').innerHTML = uploadpic;
    xmlhttp.onreadystatechange = function () {

    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
    var resp = xmlhttp.responseText;
    alert(resp);
    } alert("a");
    xmlhttp.open( "POST", 'show.php', true); //POST Because you use $_POST in php
    xmlhttp.send();
    }

    }
    </script>
    </head>
    <div id="boox">
    <?php
    if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
    {
    $fileName = $_FILES['userfile']['name'];
    $tmpName = $_FILES['userfile']['tmp_name'];
    $fileSize = $_FILES['userfile']['size'];
    $fileType = $_FILES['userfile']['type'];
    $fp = fopen($tmpName, 'r');
    $content = fread($fp, filesize($tmpName));
    $content = addslashes($content);
    fclose($fp);
    if(!get_magic_quotes_gpc())
    {
    $fileName = addslashes($fileName);
    }
    $con=mysql_connect("localhost","root",'');
    mysql_select_db("project",$con) or die("error db");

    $query = "INSERT INTO upload (name, size, type, content ) ".
    "VALUES ('$fileName', '$fileSize', '$fileType', '$content')";
    mysql_query($query) or die('Error, query failed');
    mysql_close($con);
    echo "<br>File $fileName uploaded<br>";

    }
    ?>
    </div>
    <body>
    <form method="post" enctype="multipart/form-data">
    <input type="hidden" name="MAX_FILE_SIZE" value="2000000">
    <input name="userfile" type="file" id="userfile">
    <input name="upload" type="submit" onClick="pict()" class="box" id="upload" value=" Upload ">
    </form>
    </body>
    </html>

  2. #2
    Join Date
    Jan 2014
    Posts
    2
    no one????

  3. #3
    Join Date
    May 2006
    Location
    Somewhere behind your screen
    Posts
    1,686
    what you mean by xmlhttp.open( "POST", 'show.php', true); //POST Because you use $_POST in php ?
    there is nothing dealing with $_POST in your show.php code

    and please use the [code]code goes here[/code] tags when posting code
    Last edited by Padonak; 01-14-2014 at 11:28 AM.

  4. #4
    Join Date
    Mar 2007
    Location
    localhost
    Posts
    2,585
    Also... storing images inside a database has a toll on the server turn around, get allot of requests and the server gives up.

    Databases are meant to serve data, just because someone has found a way to stuff a database with data and files doesn't mean that you should.
    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...

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