Get value from combobox in PHP

Question

I'm trying to get the value from a previous page from a combobox.

[delete.php]

 <form name="delete" action="deleted.php" method="post">
<?php
$connect = mysql_connect("a","b","") or die("Error connecting");
mysql_select_db("c") or die("Error connecting to database");
$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
echo "<select name='forward'>";
while ($row = mysql_fetch_array($result))
  {
 echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";
}
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

[deleted.php]

<?php
$selected = $_POST['forward'];

if ($selected== 'kryptix') 
{
    alert('No one was seleclted');
}else{
alert('Success');
}
?>

but i get the following error:

Notice: Undefined index: forward in C:\xampp\htdocs\folder\deleted.php on line 2

LINE 2 = $selected = $_POST['forward']; *

What am i doing wrong here?

Can you please edit/update your code in your question? (based on the answers you have got) — bestprogrammerintheworld, Apr 03 '13 at 02:17
did you get the warning while page is running or after submit the page? — Maths RkBala, Jul 14 '16 at 10:17

Kai Qing · Answer 1 · 2013-04-03T01:52:34.783

-1

You're not echoing your select. Therefore, $_POST['forward'] will never be set:

"<select name='forward'>";

should be

echo "<select name='forward'>";

You may be better off breaking out of php to write html. It's easier to read and allows better markup writing without escaping and so on.

<select name='forward'>
<?php while ($row = mysql_fetch_array($result)): ?>
    <option class='class' value="<?php echo $row['t']; ?>"><?php echo $row['t']; ?></option>
<?php endwhile; ?>
</select>

Also, as IOIO MAD shows in his markup example, you definitely need this in a form to post anything unless you're using ajax

edited Apr 03 '13 at 01:52

answered Apr 03 '13 at 01:44

Kai Qing

18,793
5
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is alert() a php function you wrote? also, post your form code so we can verify it is constructed correctly – Kai Qing Apr 03 '13 at 01:51
see my updated code. specifically the while loop option markup – Kai Qing Apr 03 '13 at 01:52
right, and now the only thing that looks wrong immediately is that you're using name instead of value in your option tag. That and that I don't know what alert() does. – Kai Qing Apr 03 '13 at 02:04
alert() is javascript? I guess it should be echo ? – bestprogrammerintheworld Apr 03 '13 at 02:08
it could also be a php function if he defined it somewhere. Since we don't see that it is left to assumption. Either way it would spit out an error at least. – Kai Qing Apr 03 '13 at 02:20
okay but what should my GET be? $_GET['forward'] or $_GET[what the value is]? – Xtacy Xtatic Apr 03 '13 at 02:42
it should not be $_GET. it should be $_POST because that's the method your form uses. and it should be $_POST['forward'] because your select box has a name of forward: – Kai Qing Apr 03 '13 at 03:00

internals-in · Answer 2 · 2013-04-05T01:36:45.083

-1

echo "<select name='forward'>"; //cant find a echo there in your code

Add to your php receiver code for extra validation

if(isset($_POST['forward']))
    {

    //your code here  {put [query] here} code
    }

updated: you are missing value attribute in <option> Tag

echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";

the above line in your code doesn't have a value field (ie forward's value is Undefined )? add the value field

echo "<option class='class' name='" .$row['t'] ."' value='".$row['t'] ."'>".$row['t'] ."</option>";

edited Apr 05 '13 at 01:36

answered Apr 03 '13 at 01:44

internals-in

4,798
2
21
38

while file is this from? the first page or second? – Xtacy Xtatic Apr 03 '13 at 01:47
I'm using HTML instead of echo form name .. can that be the problem? – Xtacy Xtatic Apr 03 '13 at 02:05
@XtacyXtatic - no
is outside of the php-scope, so it should be ok.
– bestprogrammerintheworld Apr 03 '13 at 02:19

score -1 · Answer 3 · edited May 23 '17 at 12:26

-1

There is nothing wrong, is just a warning on PHP. If you want to avoid that warning you can use the isset() function or turn off "notice" error reporting in php. See this answer

edited May 23 '17 at 12:26

Community

1
1

answered Apr 03 '13 at 01:46

Jorge Zapata

2,316
1
30
57

so why didn't it echo noting selected? – Xtacy Xtatic Apr 03 '13 at 01:50
It's not good practice to just ignore warnings. In this case the code doesn't even work. – bestprogrammerintheworld Apr 03 '13 at 02:12
Maybe not warnings but I think Notices can be ignored – Jorge Zapata Apr 03 '13 at 17:51

score -1 · Answer 4 · answered Apr 03 '13 at 06:36

You get the undefined index when trying to get value of $_POST['forward'] when having this code:

<form name="delete" action="deleted.php" method="post">
<?php
$connect = mysql_connect("a","b","") or die("Error connecting");
mysql_select_db("c") or die("Error connecting to database");
$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
echo "<select name='forward'>";
while ($row = mysql_fetch_array($result))
  {
 echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";
}
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

When doing a code like this: (I've commented out the database-part)

 <form name="delete" action="deleted.php" method="post">
<?php
//$connect = mysql_connect("a","b","") or die("Error connecting");
//mysql_select_db("c") or die("Error connecting to database");
//$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
$row['t'] = 1; //Dummy
echo "<select name='forward'>";
echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

You will not recieve an undefined index-warning in your deleted.php.

With following code, you will not recieve an index-warning either.

 <form name="delete" action="deleted.php" method="post">
<?php
//$connect = mysql_connect("a","b","") or die("Error connecting");
//mysql_select_db("c") or die("Error connecting to database");
//$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
$row['t'] = 1; //Dummy
$row['u'] = 2; //Dummy
echo "<select name='forward'>";
echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";
echo "<option class='class' name=" .$row['u'] . ">".$row['u'] ."</option>";
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

AND $_POST['forward'] would actually return the name of option when no value is set. (1 or 2). BUT you SHOULD use value="" like this:

 <form name="delete" action="deleted.php" method="post">
<?php
//$connect = mysql_connect("a","b","") or die("Error connecting");
//mysql_select_db("c") or die("Error connecting to database");
//$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
$row['t'] = 1; //Dummy
$row['u'] = 2; //Dummy
echo "<select name='forward'>";
echo "<option class='class' value=" .$row['t'] . ">".$row['t'] ."</option>";
echo "<option class='class' value=" .$row['u'] . ">".$row['u'] ."</option>";
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

However, when you don't have any options in the code (commented out below), you will get an undefined Index warning.

 <form name="delete" action="deleted.php" method="post">
<?php
//$connect = mysql_connect("a","b","") or die("Error connecting");
//mysql_select_db("c") or die("Error connecting to database");
//$result = mysql_query("SELECT * FROM d ORDER BY e ASC");
$row['t'] = 1; //Dummy
$row['u'] = 2; //Dummy
echo "<select name='forward'>";
//echo "<option class='class' value=" .$row['t'] . ">".$row['t'] ."</option>";
//echo "<option class='class' value=" .$row['u'] . ">".$row['u'] ."</option>";
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

This is because the select list contains nothing (no options).

Therefore my question: Do you actually have any options in your select-list?

Try this code, and see what echo print_r($row,true); brings you:

 <form name="delete" action="deleted.php" method="post">
<?php
$connect = mysql_connect("a","b","") or die("Error connecting");
mysql_select_db("c") or die("Error connecting to database");
$result = mysql_query("SELECT * FROM d ORDER BY e ASC");

/*debugging start
$row = mysql_fetch_array($result)
echo print_r($row,true);
debugging end */

echo "<select name='forward'>";
while ($row = mysql_fetch_array($result))
  {
 echo "<option class='class' value=\"" .$row['t'] . "\">".$row['t'] ."</option>";
}
echo "</select><br/><br/><br/><br/>";
?>
<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;<input type="button" id="cls" class="cls" value="Clear">
</form>

score -1 · Answer 5 · answered Jul 24 '13 at 10:15

-1

there is no problem in code.You are using Alert in the PHP.that is why you are getting error. if you want you can isset($_POST(forward)) for checking.

answered Jul 24 '13 at 10:15

NEO

1

score -1 · Answer 6 · edited Jun 13 '16 at 10:56

<form name="delete" action="delete.php" method="post">
<?php
$connect = mysql_connect("localhost","root","menu32") or die("Error connecting");
mysql_select_db("MyProject") or die("Error connecting to database");
$result = mysql_query("SELECT name FROM teacher_detail ORDER BY name ASC limit 5");?>
<select name='forward'>
<option name='' value=''>--select--</option>
<?php 
while ($row = mysql_fetch_array($result))
  {?>
 <option class='class' name="<?= $row['name'] ?>"><?=$row['name'] ?></option>";
<?php } ?>
</select><br/><br/><br/><br/>

<input type="submit" id="thisSbmit" value="Delete Contact" onClick="chck()">&nbsp;&nbsp;
<input type="button" id="cls" class="cls" value="Clear">
</form>

delete.php

<?php

$selected= $_POST['forward'];
if($selected=='Babita')
    echo "matched string";
else
    echo "not found";

score -1 · Answer 7 · answered Jul 14 '16 at 10:15

Try the following:

Your mistake is you are mention name instead of value.

So change the following and test it, let me know.

 echo "<option class='class' name=" .$row['t'] . ">".$row['t'] ."</option>";

Into

echo "<option class='class' value=" .$row['t'] . ">".$row['t'] ."</option>";

Thanks.

Get value from combobox in PHP

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