[SOLVED] Displaying information in a update form

[dark22horse]

I still havent been able to find any thing.  Does anyone have a link, to exokain this to me? or tell me another way

[dark22horse]

I am giving up trying to put the file into the file type.  I am trying to write an if statement. 

It hasnt worked, but this is what I am trying to do.

 

If $photo = error4 (no file has been uploaded)

Run the script that doesnt include the upload files

 

else Run the script that includes the upload file. 

 

I have attempted the script below, but it doesnt work. 

 

 

 

<?
require_once("mysql_config.php");

$target_path = $target_path . basename( $_FILES['photo']['name']); 
$_FILES['photo']['tmp_name'];

$target_path = "uploads/";

$target_path = $target_path . basename( $_FILES['photo']['name']); 


if(move_uploaded_file($_FILES['photo']['tmp_name'], $target_path)) {

$id = $_POST['id'];
$price = $_POST['price'];
$description = $_POST['description'];
$photo = $_FILES['photo']['name'];
$photo1 = $_FILES['photo1']['name'];
$photo2 = $_FILES['photo2']['name'];
$photo3 = $_FILES['photo3']['name'];
$car_of_the_week = $_POST['car_of_the_week'];
$status1 = $_POST['status1'];

$id = mysql_real_escape_string($id);
$price = mysql_real_escape_string($price);
$description = mysql_real_escape_string($description);
$car_of_the_week = mysql_real_escape_string(car_of_the_week);
$status1 = mysql_real_escape_string($status1);


if($photo=""){
      $sql_string = "UPDATE car SET price='$price', description='$description', car_of_the_week='$car_of_the_week', status1='$status1'"." WHERE id='$id'";

} else {
      $sql_string = "UPDATE car SET price='$price', description='$description', photo='$photo', photo1='$photo1', photo2='$photo2', photo3='$photo3', car_of_the_week='$car_of_the_week', status1='$status1'"." WHERE id='$id'";
}
mysql_query($sql_string) OR die(mysql_error());

echo "<br/>SQL STRING WE WANT TO RUN: ".$sql_string;

echo "db connection: ".$dbconn;
echo "table connection: ".$table;

$ok = mysql_query($sql_string);

echo "OK: ".$ok;


if($ok){
	echo "<h2>all is good, upload another</h2>";
} else {
	echo "<h2>check the string".$sql_string."</h2>";
}
} else{

echo "There was an error uploading the file, please try again!";

}




?>