drop down php/mysql

**Markus** · Jan 21 '09, 12:10 PM

Ok, the MySQL will need to be an UPDATE query. You will also need to know how to get the selected value of the dropdown (I will explain that below). However, currently, looking at your form, there is a problem: you <form> has no action or method. Not a huge issue, but it is good practice to tell your form what you want it to do. I expect the form is posting to the same page, but I don't know whether you plan on using GET or POST (or if you even know about those?). I'll assume you will use POST. Make your <form> into this:

Code:

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">

Now when the form is submitted, all the values of the form will be in the POST array, which we can access by using the name attribute of the form elements as the array key, like so:

Code:

<?php

echo $_POST['element_name'];

?>

Overlooking your code again, I see your <select> has no name attribute - this is necessary. Otherwise, the element's value won't be available in the POST array. So add name="user" to your <select>. Note: you will have to have a name attribute for every form element.

Check this tutorial out, for help with forms & php: PHP Tutorial - Forms

**Dormilich** · Jan 21 '09, 12:18 PM

Originally posted by Markus

However, currently, looking at your form, there is a problem: you <form> has no action or method. Not a huge issue, but it is good practice to tell your form what you want it to do.

it's a POST to "drop.php", line 28. there's just one <form> too many.

btw. "action" is a required attribute

**brendanmcdonagh** · Jan 21 '09, 01:48 PM

so is the code so bad that it can't be made to work?

**Dormilich** · Jan 21 '09, 01:50 PM

if you apply Markus' improvements, then certainly not.

**brendanmcdonagh** · Jan 21 '09, 01:54 PM

Thanks Dormilich for swift reply

Here is code with suggestions added:

Code:

<html>
<?php
$connection = mysql_connect("localhost", "root", "");
if(!$connection)
{
die("database failed " . mysql_error());
}
$db_select = mysql_select_db("staff_status", $connection);
if(!$db_select)
{
die("database selection failed " .mysql_error());
}
echo $db_select;
?>
 
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<select name="user">
<?php 
$sql="SELECT id,staff_name FROM status_staff";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result))
{
echo ("<option value=".$data['id'].">". $data['staff_name']."</option>"); 
?>
<?php } ?>
<?php 
  
echo $_POST['staff_name']; 
  
?>
</select>
</form> 
<form action="drop.php" method="post">
 
    
    <br>
Location: 
<input type="text" name="location">
<br>
Time Leaving: 
<input type="text" name="time">
<br>
<input type="Submit">

</html>

I know I'm out of my depth here but if I was to get it working I could understand/learn from reading the working code.

Any further instructions please?

**Dormilich** · Jan 21 '09, 02:07 PM

there's still work to do...
minor details:
- line 23, you should html-quote the attribute values

Code:

echo '... value="' . $var . '" ...';
// or
echo "... value=\"" . $var . "\" ...";
// giving
... value="content_of_$var" ...

- line 24–26, replace by } (the <?php and ?> remove themselves)

major details:
- the values from the first form are never sent (there's no submit button for that form). you should make only one <form> which will process the results. I guess it's the second one. remember, only the values from inside the <form> are submitted (you may have multiple form elements on a side, but only the one whose submit is pressed actually sends data).

**brendanmcdonagh** · Jan 21 '09, 02:30 PM

thanks again but just don't understand what you are saying to do.

I have never asked anyone on the site to show me with my variables, but I just don't think ill get it any other way.

If you can't do that for me, thanks again, I'll just stick to java!

**Dormilich** · Jan 21 '09, 02:51 PM

in the end, the html code should look like this (more or less):

Code:

<html>
  
<form action="drop.php" method="post">
   <select name="user">
     <option value="[I]ID1[/I]">[I]NAME1[/I]</option>
     <option value="[I]ID2[/I]">[I]NAME2[/I]</option>
     <option value="[I]ID3[/I]">[I]NAME3[/I]</option>
 // ...
  </select>
  <br>
Location: 
   <input type="text" name="location">
   <br>
Time Leaving: 
   <input type="text" name="time">
   <br>
   <input type="Submit" value="send">
</form>

</html>

in the final version, drop lines 13 and 28 (this is nothing for the eyes of the user, only for debugging)

the file "drop.php" has to handle the database update.

EDIT: the problem of your code does not come from the PHP side, it's the HTML code that's not right.

**brendanmcdonagh** · Jan 21 '09, 03:07 PM

Dormilich, thank you! I now know what the meaning of the word post means!

So after a bit of copying and pasting (and learning) I have 2 files which look like they should do what im trying.

here's insert1.php

Code:

<html>
<?php
$connection = mysql_connect("localhost", "root", "");
if(!$connection)
{
die("database failed " . mysql_error());
}
$db_select = mysql_select_db("staff_status", $connection);
if(!$db_select)
{
die("database selection failed " .mysql_error());
}
echo $db_select;
?>
 
<form action="insert2.php" method="post">
<select name="user">
<?php 
$sql="SELECT id,staff_name FROM status_staff";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result))
{
echo ("<option value=".$data['id'].">". $data['staff_name']."</option>");
?>
</select>
<br>
Location: <input type="text" name="location">
Time Leaving: <input type="text" name="time">
 
<input type="Submit">
 
 
 
</form> 
</html>

I have tried to stay with importing the names from database to show in dropdown box as this was working and has been my only success today:(

Here's insert2.php

Code:

 
<?php
 
$connection = mysql_connect("localhost", "root", "");
if(!$connection)
{
die("database failed " . mysql_error());
}
$db_select = mysql_select_db("staff_status", $connection);
if(!$db_select)
{
die("database selection failed " .mysql_error());
}
echo $db_select;
 
$name = $_POST['staff_name']; 
$location = $_POST['location'];
$time = $_POST['time'];
$query2 = "INSERT INTO status_staff VALUES ('$name','$location','$time')";
mysql_query($query2);
mysql_close();
?>

Am i getting closer to the concept of html/php/mysql?

Edit:

**Dormilich** · Jan 21 '09, 03:21 PM

Originally posted by brendanmcdonagh

Am i getting closer to the concept of html/php/mysql?

yes

Originally posted by brendanmcdonagh

Code:

echo ("<option value=".$data['id'].">". $data['staff_name']."</option>");

this could be improved to

Code:

echo "<option value=\"", $data['id'], "\">", $data['staff_name'], "</option>";

(printing " to the html too and executing a bit faster)

Originally posted by brendanmcdonagh

Here im getting a parse error on line 30 which no matter what i do won't go away but I think it's a syntax error more than me not having a clue !!

which is line 30, the display is currently a bit strange here....
EDIT: fixed now, I'll have a second look

**brendanmcdonagh** · Jan 21 '09, 03:23 PM

sorry, ignore line 30 error, i was looking at wrong page, doh!

**Dormilich** · Jan 21 '09, 03:25 PM

Originally posted by brendanmcdonagh

Code:

$name = $_POST['staff_name'];

should be

Code:

$name = $_POST['user'];

second, but nonetheless important: never trust user input! never! you can sanitize the input using mysql_real_esca pe_string()

**brendanmcdonagh** · Jan 21 '09, 03:41 PM

I can't thank you enough, I have it doing what it should!!!!!!!!! !

**Markus** · Jan 21 '09, 03:50 PM

I go out to walk the dog, and I come back to another happy customer.

Brendan, just a note, you should really check out these sites (they helped me from n00b to being able to basic-intermediate stuff).

Introduction to PHP

PHP Tutorial - Introduction

drop down php/mysql

drop down php/mysql

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