PHP Forum

**chrisb** · 05-16-2005, 10:23 AM

Hi Guys,

I’m a bit stuck working on a PHP content management system that assigns a photograph with each profile I create.

Now I have my php page set-up so you can enter a variable called $firstname, and this data entered into the textbox is saved in MySQL database in a table called ‘profiles’ when you hit submit - (and it can be retrieved and modified) as shown below:

PHP Code:


			


<?php



// login stuff



include("../ch.php");

include("../cm.php");

checklogin();



$firstname = "";

$image = "";



if(isset($_POST['Submit']))

{



    $firstname = $_POST['firstname'];

    $image = $_POST['image'];

    

    if(!isset($_GET['profilesid']))

    {

        $result = mysql_query("Insert into profiles(firstname,image) values('$firstname','$image')");

        $msg = "New record is saved";

    }

    else

    {

        $result = mysql_query("Update profiles set firstname='$firstname', image='$image' where profilesid=".$_GET['profilesid']);

        $msg = "profiles Record is updated";

    }

}

if(isset($_GET['profilesid']))

{

    $result = mysql_query("Select * From profiles where profilesid=".$_GET['profilesid'],$link);

    $row = mysql_fetch_array($result, MYSQL_BOTH);

    $firstname = $row['firstname'];

    $image = $row['image'];

}

?>

<html>

<head>

<title>Admin</title>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

<link rel="stylesheet" href="../ssheet.css">

</head>



<body>

<div align="center"></div>

<form name="form1" method="post" action="">

  <input name="firstname" type="text" id="firstname" value="<?php echo $firstname?>" size="36">

</form>

</body>

</html>

Now as you can see I also have an $image variable on this page. What I want to do is create an <option></option> drop down menu, which looks at a different table in MySQL database called ‘uploads’ which holds information on .jpg images uploaded to a folder.

Basically each .jpg uploaded has $name and $url data assigned to it in the ‘uploads’ table..

So I want the <option></option> drop down menu to do is:

- Return all $name of uploaded images in the dropdown
- Hit Submit
- The $URL to that image I have selected is held in the ‘profiles’ table
held in the $image variable.

But I have not got a clue how to do it – anybody help?

Thanks

Chris

**OmuCuSucu** · 05-16-2005, 10:29 AM

1. select all image names from uploads
2. in a while create an option for each one. putting the name as value
3. the select will return the name of the image afetr submission.
4. select url from uploads where name = value returned by select
5. update profile with the url returned.

i think this would be one way of doing it....

**chrisb** · 05-16-2005, 10:36 AM

Thanks - but i'm very new to this. Any chance of some code/

I tried the below:

PHP Code:


			


<select name="image"> 

<option value="" selected>select an image...</option> 

<?php 

$query = mysql_query("SELECT * FROM `uploads`"); 

while($image = mysql_fetch_assoc($query)){ 

?> 

<option value="<?php echo $image['url']; ?>"><?php echo $image['name']; ?></option> 

<?php 

} 

?> 

</select>

A while back, which selected the image name, but did not save the $image information in the $image url information in my 'profiles' table?

Any ideads?

**OmuCuSucu** · 05-16-2005, 10:47 AM

that part of code seems corect, i don't know why you add <option value="" selected>select an image...</option>, you can just add it as text before the select box and you woun't have to make sure they don't select it.

any way, back to the problem:
i see you have no submit button, how are you submitting it?

anyway, here is some code you can use as inspiration since i really must leave soon.

PHP Code:


			
<form name="st_stud" method="post" action="index.php?pagina=cont">

  <select name="idpr">

<?

    $query = mysql_query("SELECT `username`, `id_user` FROM `dor_useri` WHERE `tip` = 1");

    while ($row = mysql_fetch_array($query, MYSQL_ASSOC)) {

?>

      <option value="<? echo $row['id_user']; ?>"><? echo $row['username']; ?></option>

<?

    }

?>

  </select>

  <input type="submit" name="stergepr" value="sterge cont profesor">

</form>

PHP Code:


			
if (isset($_POST['stergepr'])) {

    $user = $_POST['idpr'];

    echo $user;

}

the $user in the last part will be your image.

than you do an sql update with it.

SELECT url FROM update WHERE name = $user (your image)

than UPDATE profiles .... with the url.

**chrisb** · 05-16-2005, 11:18 AM

Thanks OmuCuSucu,

Your code seems to have gone over my head for a novice like me, but you seemed to indicate i'm nearly there with:

PHP Code:


			
<select name="image"> 

<option value="" selected>select an image...</option> 

<?php 

$query = mysql_query("SELECT * FROM `uploads`"); 

while($image = mysql_fetch_assoc($query)){ 

?> 

<option value="<?php echo $image['url']; ?>"><?php echo $image['name']; ?></option> 

<?php 

} 

?> 

</select>

This code i've posted. Therefore can anyone please suggest where i'm wrong?

Thanks

**OmuCuSucu** · 11:34 AM

hey, what is the structure of the profiles table?

EDIT:

PHP Code:


			


<form name="form" method="post" action="">

  <select name="image">

<?

    $query = mysql_query("SELECT * FROM uploads");

    while ($image = mysql_fetch_assoc($query)) {

?>

      <option value="<?php echo $image['name']; ?>"><?php echo $image['name']; ?></option> 

<?

    }

?>

  </select>

  <input type="submit" name="select" value="select">

</form> 

<?

if (isset($_POST['select'])) {

    $imagename = $_POST['image'];

    $query = mysql_query("UPDATE profiles SET imageurl = (SELECT url FROM upload WHERE name = $imagename) WHERE [[[[ set condition to identify user here. ]]]]");

}

?>

i made this, but didn't get to test it ... tell me if you don't uderstand something