Cannot store multiple form selections to a mysql d

scifo · August 3rd, 2003, 10:22 AM

I have a form document (below) and would like to store that data in a mysql database.
Form.php:
<html>
<head>
<title>Form.php</title>
</head>
<body>
<?
echo â
<form method=\"post\" action=\"handleform.php \">
<table border=\"1\">
<tr><td>Doctor: </td><td>
<select name = \"doctornames\">
<option value=\"blank\" selected=\"selected\"></option>
<option value=\"McEnroe\">Dr John McEnroe</option>
<option value=\"Becker\">Dr Boris Becker</option>
<option value=\"Borg\">Dr Bjorn Borg</option>
<option value=\"Navratilova\">Dr Martina Navratilova</option>
<option value=\"Williams\">Dr Serena Williams</option>
</select></tr>
</table>
";
echo "
<p><input type=\"submit\" name=\"mybutton\" value=\"Submit\" />
<input type=\"reset\" value=\"Reset Form\" />
</p>
</form>
";
?>
</body>
</html>

I am using PHP 4.0 thatâs why I am using $HTTP_POST_VARS instead of _POST
handleform.php :

<body> <?php
global $HTTP_POST_VARS; global $doctor;
for ($x=0; $x < count($HTTP_POST_VARS['doctornames']); $x++) {
$doctor = $HTTP_POST_VARS['doctornames[$x]'];

if (isset($HTTP_POST_VARS['mybutton'])) {
echo "1.You clicked the button\n";
echo "2.You typed ". $doctor . " in the textbox.<br>\n";
$Host = "***";
$DBName = "***";
$TableName = "***";
$Link = mysql_connect ($Host, '****', '****'); if ($Link==false) { echo mysql_errno().": ".mysql_error()."<BR>\n";
}
mysql_select_db ($DBName, $Link);
$Query = "INSERT INTO $TableName VALUES ('{$doctor}')";

if (mysql_query ($Query, $Link)) {
print "The query was successfully executed!<BR>\n";
}
else {
print "The query could not be executed!<BR>\n";
echo mysql_errno().": ".mysql_error()."<BR>\n";
}
mysql_close($Link);
}
else {
echo "you must have come here from somewhere else.\n";
}
?>
</body>
I also used switch()
switch ($HTTP_POST_VARS['doctornames']) {
case "blank":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
case "McEnroe":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
case "Becker":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
case "Borg":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
case "Navratilova":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
case "Williams":
$doctor = $HTTP_POST_VARS['doctornames'];
break;
}

instead of lines
for ($x=0; $x < count($HTTP_POST_VARS['doctornames']); $x++) {
$doctor = $HTTP_POST_VARS['doctornames[$x]'];
but it is not storing the data. What is the problem? Could u help me?

richard.york · August 3rd, 2003, 12:45 PM

Well,
Unless you are needing multiple selections from the select field, it is uneccessary to loop it. Use the loop if you are using this attribute:

<select name = \"doctornames\" multiple=\"multiple\" size=\"4\">

Doing it this way would allow you to toggle multiple selections, thereby making it neccessary to have a loop, which you also had some errors in:

Code:

foreach ($HTTP_POST_VARS['doctornames'] as $key => $value) {

    $doctor = $HTTP_POST_VARS['doctornames'][$key];

} // snip snip

The counter variable goes in its own array, and use use the $key, $value to keep in line with the selections made by the user.

Code:

<?php
global $HTTP_POST_VARS; global $doctor;


Host = "***";
$DBName = "***";
$TableName = "***";
$Link = mysql_connect ($Host, '****', '****'); 

/*
Making a database connection is resource intensive, you should do this only once at very beginning of a script, and definitely not within a loop
*/

if ($Link==false) { 

    echo mysql_errno().": ".mysql_error()."<BR>\n";

}

mysql_select_db ($DBName, $Link);

    foreach ($HTTP_POST_VARS['doctornames'] as $key => $value) {

        # I don't understand the point of this:
        # $doctor = $HTTP_POST_VARS['doctornames[$x]'];

        if (isset($HTTP_POST_VARS['mybutton'])) {

            echo "1.You clicked the button\n";
            echo "2.You selected ". $HTTP_POST_VARS['doctornames'][$key] . " from the select field.<br>\n";

            $Query = "INSERT INTO $TableName VALUES ('{$HTTP_POST_VARS['doctornames'][$key]}')";

            if (mysql_query ($Query, $Link)) {

                print "The query was successfully executed!<BR>\n";

            } else {

                print "The query could not be executed!<BR>\n";
                echo mysql_errno().": ".mysql_error()."<BR>\n";

            }

                mysql_close($Link);
        } else {

            echo "you must have come here from somewhere else.\n";

        }

?>

If you did in fact want multiple <select> choices that's how it would be done. You were trying to assign the $HTTP_POST_VARS value to a regular global, which is fine, but you didn't treat the $doctors variable as an array. And not to mention were trying to insert the counter in the string.

For a regular <select> field, without the multiple="multiple" attribute it would be exactly the same as accessing data from a text field. Just a single name, value pair - no sub array.

hth,
: )
Rich

:::::::::::::::::::::::::::::::::
Smiling Souls
http://www.smilingsouls.net
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richard.york · August 3rd, 2003, 04:16 PM

I noticed some more errors which I did not notice before:

The call to:
mysql_close($Link);
Should be done after the loop is closed.

Otherwise you're closing the database connection after the first selection is inserted. Which will cause some errors.

I also left out the closing '}' on the foreach loop.

: )
Rich

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