mysql populated drop down not $_POST ing data

[gesseg]

heres my code not sure why it doesnt work...

 

<form action="insert.php" method="post">
name: <?php
$con = mysql_connect("xxx","xxx","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("xxx", $con);
$query="SELECT name FROM band_data2";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo "<select name='name'> Name</option>";
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=$nt[id]>$nt[name]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>
<br /> 
skill : <select name="skill"> 
<option value="">Do they hold their instuments up the right way?</option> 
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option> 
<option value="4">4</option>
<option value="5">5</option> 
<option value="6">6</option>
<option value="7">7</option> 
<option value="8">8</option>
<option value="9">9</option> 
<option value="10">10</option></select>

<input type="hidden" name="counter" value="1">
<input type="submit" />
</form>

[bugcoder]

try to change the name of drop down name from "name" to something else, works?

[gesseg]

Doesnt it need to be"name" because thats where its getting its data from? i.e. field "name" in database "band_data2". I tried it anyway but it didnt help.

[developerdave]

<form action="insert.php" method="post">
name: <?php
$con = mysql_connect("xxx","xxx","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("xxx", $con);
$query="SELECT name FROM band_data2";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo '<select name="name"> Name</option>';
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=\"{$nt['id']}\">{$nt['name']}</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>
<br /> 
skill : <select name="skill"> 
<option value="">Do they hold their instuments up the right way?</option> 
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option> 
<option value="4">4</option>
<option value="5">5</option> 
<option value="6">6</option>
<option value="7">7</option> 
<option value="8">8</option>
<option value="9">9</option> 
<option value="10">10</option></select>

<input type="hidden" name="counter" value="1">
<input type="submit" />
</form>

 

Sorry, forum killed it before. I've tidied it up a bit, try wrapping variables contained in string with braces. I'm assuming you're running PHP5?

[kenrbnsn]

Since you didn't tell us what was not working, we're guessing here.

 

You are using the "id" from the database records, but you're not selecting it with your query.

 

<?php
mysql_select_db("xxx", $con);
$query="SELECT name, id FROM band_data2";
$result = mysql_query ($query);
echo '<select name="name"> Name</option>';
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
     echo "<option value='{$nt['id']}'>{$nt['name']}</option>\n";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>

 

Ken

[Pikachu2000]

Didn't look at this too deeply, but this line jumps out at me. You're trying to close an <option> tag that hasn't yet been opened.

echo '<select name="name"> Name</option>';

[kenrbnsn]

Oops, missed that one...

 

Ken

[Pikachu2000]

Hope that didn't sound like it was directed at you, kenrbnsn. It wasn't meant to be . . .

[kenrbnsn]

I know. It was directed at the OP.

 

Ken

[gesseg]

Thanks but even with your revisions its still not doing what I want. To be honest Ive only been learning for a week. I can follow most of what the script is doing but havent got a clue how to edit to do what I want.

What I want is a drop down list populated with my "name" field from my database, once the name is selected I want to pass the "name" data on to insert.php as $_POST data. I did have a text box that was working but in order to update the table you have to, obviously, type the name exactly.

[kenrbnsn]

Ok. Please post you current code for both the form and insert.php

 

What is happening now and why is it different from what you want? Have you looked at the generated source to make sure that the generated HTML is correct?

 

Ken

[Pikachu2000]

It also wouldn't hurt to remove the 'Name' string from within the <select></select> tag's structure . . .

 

echo 'Name:<select name="name">';
// instead of //
echo "<select name='name'> Name</option>";

[developerdave]

Oops, missed that one...

 

Ken

 

Me too 

[gesseg]

Heres vote.php. The code is taken from a tutorial on the web,

<form action="insert.php" method="post">
name: <?php
$con = mysql_connect("xxx","xxx","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("a3362005_bands", $con);
$query="SELECT name FROM band_data2";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo 'Name:<select name="name">';
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=\"{$nt['id']}\">{$nt['name']}</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>
<br /> 
skill : <select name="skill">
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option> 
<option value="4">4</option>
<option value="5">5</option> 
<option value="6">6</option>
<option value="7">7</option> 
<option value="8">8</option>
<option value="9">9</option> 
<option value="10">10</option></select>
<input type="hidden" name="counter" value="1">
<input type="submit" />
</form></center>

 

Heres the insert.php

<?php
$con = mysql_connect("xxx","xxx","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("a3362005_bands", $con);
//Add to the table value with form data.
mysql_query("UPDATE band_data2 SET skill = (skill + {$_POST['skill']}) WHERE name='{$_POST['name']}'");
mysql_query("UPDATE band_data2 SET skill_ave = (skill / counter) WHERE name='{$_POST['name']}'");
echo 'vote counted';
mysql_close($con);
print_r ( $_POST );

?>

 

And here is the output from  "print_r ( $_POST );"  "Array ( [name] => [skill] => 10 [counter] => 1 )" i was told to put this in so i could see what was going on. As you can see the "name" from the dropdown isnt making it to the insert.php.

[developerdave]

Since you didn't tell us what was not working, we're guessing here.

 

You are using the "id" from the database records, but you're not selecting it with your query.

 

<?php
mysql_select_db("xxx", $con);
$query="SELECT name, id FROM band_data2";
$result = mysql_query ($query);
echo '<select name="name"> Name</option>';
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
     echo "<option value='{$nt['id']}'>{$nt['name']}</option>\n";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>

 

Ken

 

 

He has  a point, I can't see this amendment in your revised code, perhaps try this? just after the mysql query try

 

echo mysql_error();

 

And see if there's an error preventing the values being returned to the select field if it has no output, chances are it has no value.

[gesseg]

Fixed it sort of. Someone else on another forum had exactly the same problem so copied there revised code. Its still not working properly. It will only return the first word in the "name" field.

 

As above but:

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=$nt[name]>name: $nt[name]}</option>";

 

Now print_r ( $_POST ); gives: Array ( [name] => Coheed [skill] => 10 [counter] => 1 ) I want it to read: ( [name] => Coheed and Cambria [skill] => 10 [counter] => 1 )

[kenrbnsn]

Since you aren't using the "id" field in your query, change your code for vote.php to

<form action="insert.php" method="post">
<?php
$con = mysql_connect("xxx","xxx","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("a3362005_bands", $con);
$query="SELECT name FROM band_data2";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo 'Name:<select name="name">';
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=\"{$nt['name']}\">{$nt['name']}</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box
?>
<br /> 
skill : <select name="skill">
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option> 
<option value="4">4</option>
<option value="5">5</option> 
<option value="6">6</option>
<option value="7">7</option> 
<option value="8">8</option>
<option value="9">9</option> 
<option value="10">10</option></select>
<input type="hidden" name="counter" value="1">
<input type="submit" />
</form></center>

 

Also change the code for insert.php to

<?php
echo '<pre>' . print_r($_POST,true) . '</pre>';
$con = mysql_connect("xxx","xxx","xxx") or die("Could not connect<br>" . mysql_error());
mysql_select_db("a3362005_bands", $con);
//Add to the table value with form data.
$query = "UPDATE band_data2 SET skill = (skill + {$_POST['skill']}), skill_ave = (skill / counter) WHERE name='{$_POST['name']}'";
$rs = mysql_query($query) or die("Problem with the update query: $query<br>" . mysql_error());
echo 'vote counted';
?>

 

You only need one update query.

 

Note: the reason you're new code is only getting one name is that the value is not enclosed in quotes. If you had looked at the generated HTML this would have been apparent.

<?php
echo "<option value=$nt[name]>name: $nt[name]}</option>";
?>

 

See my version above.

 

Ken

[gesseg]

Thats great! Its working now thanks. I just wish I had a better grasp of why it is now working 

 

Dont suppose you want to point me in the right direction to getting the drop down in alphebetical order? 

[kenrbnsn]

To get the list sorted, change the query to

<?php
$query="SELECT name FROM band_data2 ASC";
?>

"ASC" means sort in ascending order, to do a reverse sort use "DESC".

 

Ken

[gesseg]

Thanks mate.