Getting results from Select Multiple HTML tag

dpjoff · January 30th, 2004, 05:43 PM

;)Can anyone tell me what I'm doing wrong, or not doing?
I have tried $_POST['names'], and foreach($names as $allnames);
and I can not get the values to pass to the submit.php form.
Thanks :-)

Code:

<TABLE border="1">
<TH> This is the HEADER</TH>
<tr>
<form method="POST" action="submit.test2.php">
<tr><td><select name="names[]" multiple size="5">
  <option value="Tammy Jones">Tammy Jones</option>
  <option value="Tommy Jones">Tommy Jones</option>
  <option value="Fred Flintstone">Fred Flintstone</option>
  <option value="Heather Locklear">Heather Locklear</option>
</td><tr><td align=center>
 </select><input type="submit" value="Submit" >
</form> </td></tr>
</BODY>
</HTML>

;);)

Snib · January 30th, 2004, 06:45 PM

I don't know if this will help with your PHP, but I think your HTML is faulty...

It seems you end the row and column that contain the select menu before you close the select menu itself.

Try this (remove the crossed-out red and insert the bold red):

Code:

<TABLE border="1">
<TH> This is the HEADER</TH>
<tr>
<form method="POST" action="submit.test2.php">
<tr><td><select name="names[]" multiple size="5">
  <option value="Tammy Jones">Tammy Jones</option>
  <option value="Tommy Jones">Tommy Jones</option>
  <option value="Fred Flintstone">Fred Flintstone</option>
  <option value="Heather Locklear">Heather Locklear</option>
</select></td><tr><td align=center>
 <s></select></s><input type="submit" value="Submit" >
</form> </td></tr>
</BODY>
</HTML>

----------
---Snib---
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Snib · January 30th, 2004, 06:47 PM

Another thing: you don't close the table. In this sample you would close it after the last <tr>.

HTH

----------
---Snib---
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dpjoff · January 30th, 2004, 06:59 PM

Thanks,
I changed that. makes the code follow nicer but still doesn't tell the array to save the clicked options. I think I'll have to use some javascript...

Don

richard.york · January 30th, 2004, 09:04 PM

Quote:

quote:
I have tried $_POST['names'], and foreach($names as $allnames);
and I can not get the values to pass to the submit.php form.
Thanks :-)

Don't forget that you're creating a multidimensional array...

$_POST['names'] becomes an array itself. Also foreach is a control structure, its purpose is to loop, in javascript you'd use for/in (not an exact equivalent but the closest).. but you still have to create the statement as a control structure, or it doesn't do anything useful.

foreach($_POST['names'] as $name)
{
echo $name;
}

The syntax is:

foreach ($array as $key => $value) ...

Have a look at:
http://www.php.net/manual/en/control-structures.php

: )
Rich

:::::::::::::::::::::::::::::::::
Smiling Souls
http://www.smilingsouls.net
:::::::::::::::::::::::::::::::::

richard.york · January 30th, 2004, 09:09 PM

Also, some helpful debugging functions are..

print_r and var_dump

If you can't figure out where your POST data is going, then dump it out..

echo '<pre>';
print_r($_POST);
echo '</pre>';

http://www.php.net/print_r
http://www.php.net/var_dump

print_r provides information about a variable, var_dump provides a slightly more complex analysis.

: )
Rich

:::::::::::::::::::::::::::::::::
Smiling Souls
http://www.smilingsouls.net
:::::::::::::::::::::::::::::::::

nikolai · January 30th, 2004, 10:17 PM

Hi guys,

Quote:

quote:Originally posted by quesadilla5
Don't forget that you're creating a multidimensional array...

$_POST['names'] becomes an array itself. Also foreach is a control structure, its purpose is to loop, in javascript you'd use for/in (not an exact equivalent but the closest).. but you still have to create the statement as a control structure, or it doesn't do anything useful.

foreach($_POST['names'] as $name)
{
echo $name;
}

The syntax is:

foreach ($array as $key => $value) ...

Well, the fact that $_POST['names'] is an array is why you can use it in a foreach loop in the first place. The only difference between this:
  foreach($_POST['names'] as $name)

and this:
  foreach($_POST['names'] as $key => $val)

is that the 2nd approach allows you to also get the key of each element in the array, should you need it.

Since the HTML form doesn't explicitly SET the array keys (they're numeric keys created on the fly via the empty bracket notation, i.e. "names[]"), you don't necessarily NEED the actual key value, so omitting it from the foreach() loop is okay.

In fact, it's a wise decision because you don't introduce the overhead of creating a variable to store a value that you won't end up using.

All that being said, there should really be no reason why your code doesn't work -- that is, it works for me when I use this simple test script (basically a copy/paste of your original code with only slight modifications and a lot of beautification/reformatting):

<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<table border="1">
  <tr>
    <th> This is the HEADER</th>
  </tr>
  <tr>
    <td>
      <select name="names[]" multiple size="5">
        <option value="Tammy Jones">Tammy Jones</option>
        <option value="Tommy Jones">Tommy Jones</option>
        <option value="Fred Flintstone">Fred Flintstone</option>
        <option value="Heather Locklear">Heather Locklear</option>
      </select>
    </td>
  </tr>
  <tr>
    <td align=center>
     <input type="submit" value="Submit" >
    </td>
  </tr>
</table>
</form>
<pre>
<?php
if (isset($_POST['names']))
{
    foreach($_POST['names'] as $name)
    {
        echo "Name is $name\n";
    }
}

?>

Also, notice how much easier proper indentation and formatting makes it to verify that all your HTML tags are properly nested -- that is, your tags don't overlap boundaries.

In your original code, you started the <select> tag in one table cell, and closed it in a completely separate row. That's really really really bad style, and should be avoided at all costs.

Take care,

Nik
http://www.bigaction.org/

dpjoff · February 2nd, 2004, 10:30 AM

Thanks for all your comments.
I am really looking to see why when the "Submit" button is pushed it's not passing the "Value of the Array" to the submit.php form action like it should. I'm only getting one word "Array" as the value of the names Array.

Code:

  <TABLE border="1">
   <TH> This is the HEADER</TH>
    <tr>
     <tr><td><form method="POST" action="submit.test2.php"><select name="names[]" multiple size="5">
        <option value="Tammy Fay">Tammy Fay</option>
        <option value="Angy Everheart">Angy Everheart</option>
        <option value="Bridget Hall">Bridget Hall</option>
        <option value="Christy Turlington">Christy Turlington</option>
        <option value="Claudia Shiffer">Claudia Shiffer</option>
        <option value="Estella Warren">Estella Warren</option>
        <option value="Naomi Campbell">Naomi Campbell</option>
        <option value="Stephanie Seymore">Stephanie Seymore</option>
        <option value="Sybil Bucks">Sybil Bucks</option>
        <option value="Tyra Banks">Tyra Banks</option>
        <option value="Yasmeen Ghauri">Yasmeen Ghauri</option>
     </select></td><tr>
    <td align=center><input type="submit" value="Submit" ></form></td>
   </tr>
  </TABLE>

dpjoff · February 2nd, 2004, 12:25 PM

WOWI Finally got it to work.
It required me to implode the value after the post and I was able to set it and insert into the DB.
Thanks for ya'lls help.

Code:

<html>
 <body>
 <?php
  require("log.inc");

  $dojo = implode(",", $names);
  //echo ("<p>");
  print ("INSERT INTO logtest set SSYS = '$dojo' ");

  ?>

 <META HTTP-EQUIV="REFRESH" CONTENT = "4; URL=http://hs01sd12/hs/thankyou.html"> ?>

 </body>
</html>

nikolai · February 2nd, 2004, 04:03 PM

Okay, well, I'm glad you got it working.

Bear in mind, of course, that the question you originally asked had nothing to do with what your problem really was. It would've helped a lot to have a more detailed description of your problem and/or some source code to show what you were really trying to do with it.

If you have an array variable, you can't just convert it to string like this:

$names = array("Naomi Campbell", "Stephanie Seymore", "Tyra Banks");

echo "My names are $names."

This will echo "My names are Array." since PHP can't convert arrays to strings on-the-fly.

So the problem you were REALLY having wasn't so much that you weren't receiving array input from the user, it was that you didn't know how to create a SQL query string from an array of values.

Take care,

Nik
http://www.bigaction.org/