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newhip

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    Referencing rows

    in PHP Coding Help

    Posted

    this is where you need the SELECT query

     

    blurb.php

    <?php

if (isset($_GET['id'])) {
$id = intval($_GET['id'];
$sql = "SELECT title, path, desct
 FROM media
 WHERE id = $id";

$result = mysql_query($sql);

// process results
}

     

    Ok... but what does that have to do with the blurb.php ?

    Referencing rows

    in PHP Coding Help

    Posted

    Barand is right:

     

    <?php
$conn = mysql_connect("localhost", "root", "");
$select = mysql_select_db("project", $conn);


$title = mysql_real_escape_string($_POST['title'], $conn);
$pic = $_FILES['pic']['tmp_name'];
$pic2 = $_FILES['pic']['name'];
$desc = mysql_real_escape_string($_POST['desc'], $conn);
$path = mysql_real_escape_string("upload/".$pic2, $conn);


$themove = move_uploaded_file($pic,$path);


$sqlqry = mysql_query("INSERT INTO media(title,pic,desct) VALUES ('$title','$path','$desc')", $conn);
$id = mysql_insert_id($conn);

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>New Uploaded Page</title>
</head>

<body>
<h1>
<a href="blurb.php?id=<?php echo $id; ?>">
<?php echo htmlentities($title, ENT_QUOTES, "utf-8"); ?>
</a>
</h1>
<br />
<br />
<img src="<?php echo htmlentities($path, ENT_QUOTES, "utf-8"); ?>" /><br />
<p>
<br />
<?php echo htmlentities($desc, ENT_QUOTES, "utf-8"); ?>
</p>
</body>
</html>

     

    One less query...

     

    Oh ok I think i get it. I didnt know you could do that. But like now how do i go access that page in the url with that info displayed? Like in a www.example.com/upload.php? manner?

    Referencing rows

    in PHP Coding Help

    Posted

    Ok im a super newb and i dont know how to to reference the data in my mysql table. I wrote a code that uploads a file to a server then to a database where its echoed out afterwards. The code also echoes out the id in the table's field so i know the id but i dont know how to afterwards reference it in the url. It's prob really basic but i can't figure it out. Can someone help me out or hint. Would be very appreciated. Here's the code

     

    <?php
$conn=mysql_connect("localhost","root","");
$select=mysql_select_db("project",$conn);
$title=$_POST['title'];
$pic=$_FILES['pic']['tmp_name'];
$pic2=$_FILES['pic']['name'];
$desc=$_POST['desc'];
$path="upload/".$pic2;
$themove=move_uploaded_file($pic,$path);
$sqlqry=mysql_query("INSERT INTO media(title,pic,desct) VALUES ('$title','$path','$desc')");
echo $hmmm=mysql_insert_id($conn);
$searr="SELECT * FROM media WHERE id='$hmmm'";
$seartt=mysql_query($searr);
$burp=mysql_fetch_array($seartt);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>New Uploaded Page</title>
</head>


<body>
<a href="#"><h1><?php  echo $burp['title'];  ?></h1></a><br /><br />
<img src="<?php
  echo $burp['pic'];

?>" /><br /><p><br /><?php
echo $burp['desct'];
?></p>
</body>
</html>

    Request id

    in PHP Coding Help

    Posted

    Ok I'm still having some trouble. Now its echoing out the last thing input in my database and i know what the id is but how do i now reference that in my url? It might be basic stuff but i dont get it. Can anyone help?

    Request id

    in PHP Coding Help

    Posted

    I have a basic understanding of php but there's a lot things I don't get.

    I made an upload form that is processed by an upload.php file. Here's the code.

     

     

    <?php

    $conn=mysql_connect("localhost","root","");

    $select=mysql_select_db("project",$conn);

    $title=$_POST['title'];

    $pic=$_FILES['pic']['tmp_name'];

    $pic2=$_FILES['pic']['name'];

    $desc=$_POST['desc'];

    $path="upload/".$pic2;

    $themove=move_uploaded_file($pic,$path);

    $sqlqry=mysql_query("INSERT INTO media(title,pic,desct) VALUES ('$title','$path','$desc')");

    $searr="SELECT * FROM media ";

    $seartt=mysql_query($searr);

    $burp=mysql_fetch_array($seartt);

    $ab=mysql_query($_REQUEST['desct']);

    ?>

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

    <html xmlns="http://www.w3.org/1999/xhtml">

    <head>

    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

    <title>New Uploaded Page</title>

    </head>

     

    <body>

    <a href="#"><h1><?php echo $burp['title']; ?></h1></a><br /><br />

    <img src="<?php

    echo $burp['pic'];

     

    ?>" /><br /><p><br /><?php

    echo $burp['desct'];

    ?></p>

    </body>

    </html>

     

    I'm trying to request the different pages in the url for the things that are inputted in my table. Can anyone help me understand. or even remotely understand what i'm trying to do.

    I know its dumb

    in PHP Coding Help

    Posted

    I keep getting this error

     

    "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'desc) VALUES ([jskdgkfdgjgfdhfxg],[_1353564571.48585_617.png],[fhvcjhvchjvjhvgjh' at line 1 "

     

     

    I know its just a dumb syntax error but i dont get it. Can someone help? struggling with this part a bit... here's the part of the code

     

     

    $sqlqry=mysql_query("INSERT INTO media(title,pic,desc) VALUES ([$title],[$pic2],[$desc])");

    if (!$sqlqry)

    {

    echo mysql_error();

    }

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