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azra

what is wrong with the code?

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1. HTML FORM

#for user to enter the data

<html>
<title>reg</title>
<style type="text/css">
body {
background-color: rgb(200,200,200);
color: white;
padding: 20px;
font-family: Arial, Verdana, sans-serif;}
h4 {
background-color: DarkCyan;
padding: inherit;}
h3 {
background-color: #ee3e80;
padding: inherit;}
p {
background-color: white;
color: rgb(100,100,90);
padding: inherit;}
</style>

<form method="POST" action="login_back.php" enctype="multipart/form-data"></br>
&nbsp<font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username">
</br></br>
&nbsp<font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/>
</br></br>
&nbsp<font color="DarkCyan"> Last name:</font><input type="text" name="lastname"/>
</br></br>
&nbsp<font color="DarkCyan"> File: <input type="file" name="image"></font>
</br></br>
<input type="submit" value="Save and Proceed">
</form>
</html>


----------
2 STORING IN DATABASE
#backend processing to store and retrieve data from db

<?php
#echo "<body style='background-color:rgb(200,200,200)'>";
session_start();
if( isset($_POST['username']) && isset($_FILES['image']) )
{
$_SESSION['username']=$_POST['username'];
$_SESSION['firstname']=$_POST['firstname'];
$lastname=$_POST['lastname'];
$file=$_FILES['image']['tmp_name'];
$image_size=getimagesize($_FILES['image']['tmp_name']);
if(!isset($file))
echo"please select an image";
else
{
$image_name=$_FILES['image']['name']; //grabing image name
$image_size=getimagesize($_FILES['image']['tmp_name']); //getting image size
}
echo "</br>";


#connection to db
mysql_connect("localhost","root","")or die(mysql_error());
mysql_select_db("wordgraphic")or die(mysql_error());



#checking the available username
$query = mysql_query("SELECT * FROM userdata WHERE username = '" . $_SESSION['username'] . "'" );
$ans=mysql_num_rows($query);
if ($ans > 0)
{
echo "Username already in use please try another.";

}

else if($image_size==FALSE)
{
echo"That's not an image.";

}

else
{
#Insert data into mysql

#1.Inserting user name & image into db
$sql="INSERT INTO userdata(username, firstname, lastname, image)VALUES('" . $_SESSION['username'] . "', '" . $_SESSION['firstname'] . "', '$lastname','$image')";
$result1=mysql_query($sql);

if($result1)
{
echo "</br>";
echo "Registration successful";
echo "</br>";

//displaying image
$lastid=mysql_insert_id();//get the id of the last record
echo "uploaded image is :";
echo "<img src='get.php?id=".$lastid."'>"; > this command has some mistake

}#if insertion into db successful
else
{
echo "Problem in database operation";
}
}# else block of unique username n img
}#end of isset
?>


----------

3. GET.PHP
#additional file that retrieve image from database

<?php
#connection to db
mysql_connect("localhost","root","")or die(mysql_error());
mysql_select_db("wordgraphic")or die(mysql_error());
if(isset($_REQUEST['id']) ) > this block of code is not runninng
{
$mid=(int)($_REQUEST['id']);

$image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error());
$image=mysql_fetch_assoc($image);
$image=$image['image'];
header("Content-type: image/jpeg");
echo $image;
}
else
{
echo"error";
}
?>

 

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You should really elaborate on what you need help with. What exactly isn't right in your mind?

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Having now seen your duplicate posting (!!!) I will offer this..

 

1 - as was already mentioned, turn on error checking. (See my signature)

2 - Where you try to insert a record you reference the $image var. That has not been defined. And what exactly are you trying to store in your db - an image file name? Did you ever save the uploaded file someplace? I don't see any code for that. And since you never assigned a value to $image, you're probably getting an error but not seeing it displayed.

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