[SOLVED] Setting ID

[dark22horse]

Hi guys

 

I have succesfully selected pictures and text from a database and displayed them in a table.  But now I need to be able to click on the picture and then it opens a page with all the details from the databse about it.  I know I need to have some code that selects the ID from the databse and then shows the pictures and text.

 

The problem I am having is getting the picture to have the unique ID.  any help??

[MadTechie]

see here (same idea)

 

if you want someone you do the work for you see here

[Mutley]

In your database make a new field called "id" set it as a MEDIUMINT (type) and auto_increment (other), as a Primary Key.

 

Then when you click an image, you can do an 'id' (I'm guessing it lists all the images, so add the 'id' to the query) and display it in the URL of the link that will goto the new page with all the picture information on, like:

 

<a href="info.php?id=<?=$id?> image here </a>

 

On the info.php, you get the 'id' by doing:

 

$id = $_GET['id'];

 

Then make a new query to SELECT all the info WHERE id = '$id'

 

I hope that helps!

[dark22horse]

Cheers for the help, a bit late doing this.  I have an error,

 

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\Program Files\xampp\htdocs\site\Cars2.php on line 76

 

This is the link part, unsure what I have done wrong.  That is as far as I can get, I have posted the info.php as well.

 

<?
//connect to mysql
//change user and password to your mySQL name and password
mysql_connect("localhost","root","");

//select which database you want to edit
mysql_select_db("car"); 

//select the table
$result = mysql_query("select * from car");

//grab all the content
while($r=mysql_fetch_array($result))
{	
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $price=$r["price"];
   $description=$r["description"];
   $photo=$r["photo"];
   $id=$r["id"];
  
   //display the row
   echo "$price";

   echo	"$description";

   echo "<a href="info.php?id=<?=$id?> image here </a>";


}
?>

 

 

info.php

<?

$id = $_GET['id'];

mysql_connect("localhost","root","");

mysql_select_db("car"); 

$result = mysql_query("select * from car WHERE id = '$id'");

while($r=mysql_fetch_array($result))
{
  
   $price=$r["price"];
   $description=$r["description"];
   $photo=$r["photo"];
   $id=$r["id"];

?>

[per1os]

echo "<a href="info.php?id=<?=$id?> image here </a>";

should be

echo '<a href="info.php?id=' . $id . '"> image here </a>';

[dark22horse]

Cheers for that, I know have it linking to the the info.php

 

But now that isnt working. 

 

<?

$id = $_GET['id'];

mysql_connect("localhost","root","");

mysql_select_db("car"); 

$result = mysql_query("select * from car WHERE id = '$id'");

while($r=mysql_fetch_array($result))
{
  
   $price=$r["price"];
   $description=$r["description"];
   $photo=$r["photo"];
   $id=$r["id"];
}
?>

 

This time I dont get an error, but it doesnt show anythign.

[per1os]

Time for some debugging.

 

$result = mysql_query("select * from car WHERE id = '$id'") or DIE(mysql_error());

 

Also I do not see where you are echoing anything to the screen, am I blind?

[MadTechie]

you see nothing cos you printed nothing..

[dark22horse]

Yeah sorry guys, didn't paste the whole echo.  It works now, cheers for all the help.  This forum is quality!!!!