Php Select Results from Menu

Cork Skate · 26-04-2005 6:17pm #1

Hi there, i have what probably is a very basic question ....

I have a drop down menu and it is populated from a SELECT statement and the user clicks on Submit .... i want to use the option that the user picks to run another SELECT and show the results on another page !!

How can i run the query based on that ..... i dont know how to carry out the select and format it into columns etc. and send this all to another page. I want to use the other page for loads of different types of searches ... all with very different results.

Here is my code so far .....


<html>

<body>

<?php

$username = "username";
$password = "password";
$hostname = "IP address or localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
print "Connected to MySQL<br>";
$selected = mysql_select_db("databasename",$dbh) 
	or die("Could not select my database");
print "Connected to database<br>";


function generate_box() { 

    $sql = "SELECT ClubNo, ClubName FROM club"; 
    $result = mysql_query($sql) or die(mysql_error());
    $entries = mysql_num_rows($result); 
    $options = '<option value="-1"></option>'; 

    while($row = mysql_fetch_assoc($result)) { 
            $options .= "<option value=\"{$row['ClubNo']}\">{$row['ClubName']}</option>"; 
     } 

    return $options; 

} 

$options = generate_box();

?>


<table>
<form method="get" name="forum_select" action="search.php"><table cellspacing="0" cellpadding="0" border="0">

	<tr>

                <td><span>Search By: 

<select name="by">

<? echo "$options"; ?>

<input type="submit" value="Submit"></span></td>

	</tr>

</table></form>

</body>

</html>

hostyle · 26-04-2005 7:45pm

if ($_REQUEST['submit'] == 'submit') {
  do_something_with($_REQUEST['by']);
}

Cork Skate · 26-04-2005 8:07pm

hostyle wrote:

if ($_REQUEST['submit'] == 'submit') {
  do_something_with($_REQUEST['by']);
}

Could i do something like this ??

if ($_REQUEST['submit'] == 'submit') {
[COLOR=DarkSlateBlue]mysql_query("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, ManagerNo, CoachNo FROM club");
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
print "Club Name: ".$row{'ClubName'}."<br> League: ".$row{'DomesticLeague'}."<br> Telephone: ".$row{'TelePhoneNo'}."<br> Fax No.: ".$row{'FaxNo'}."<br> Email: ".$row{'EmailAddress'}."<br> Website: ".$row{'Website'}."<br> GroundNo: ".$row{'GroundNo'}."<br> Established: ".$row{'YearEstablished'}."<br> Status: ".$row{'Status'}."<br> ManagerNo: ".$row{'ManagerNo'}."<br> CoachNo: ".$row{'CoachNo'}."<br>"; 
}([/COLOR]$_REQUEST['by']);
}

Cork Skate · 27-04-2005 11:01am

I am getting an error ...

Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in \\fullpath\club.php on line 70

Here is the code ....

<?Php

[B]if ($_REQUEST['submit'] == 'submit') {[/B]
$clubID = $_POST['Clubno']; 
$result=mysql_query("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, ManagerNo, CoachNo FROM club WHERE Clubno=$clubID"); 
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
echo "Club Name: $row['ClubName'] <br> League: $row['DomesticLeague'] <br> Telephone: $row['TelePhoneNo'] <br> Fax No.: $row['FaxNo'] <br> Email: $row['EmailAddress'] <br> Website: $row['Website']<br> GroundNo: $row['GroundNo'] <br> Established: $row['YearEstablished'] <br> Status: $row['Status'] <br> ManagerNo: $row['ManagerNo'] <br> CoachNo: $row['CoachNo'] <br>"; 
}($_REQUEST['by']);
}

?>

The IF statement above is line 70

Any one got any ideas ??

Cork Skate · 27-04-2005 1:09pm

Have anyone got any ideas ..... i cant figure this one out ..... it is expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' but i cant understand why ??

hostyle · 27-04-2005 1:09pm

Remove this line:

}($_REQUEST['by']);

Cork Skate · 27-04-2005 1:17pm

hostyle wrote:
Remove this line:
}($_REQUEST['by']);

I just tried it but still the same ...... there could be a problem with that but i dont think there is right now, the error is on line 70 (so it is either that line or before it, never after) ....... then again, i have been known to be wrong

hostyle · 27-04-2005 2:50pm

When you get errors break it down into smaller pieces:

<?php

if ($_REQUEST['submit'] == 'submit') {
	$clubID = $_POST['Clubno']; 
	$result=mysql_query("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, ManagerNo, CoachNo FROM club WHERE Clubno=$clubID"); 
	while ($row = mysql_fetch_array($result)) {
		echo "Club Name: " . $row['ClubName'] . "<br> League: " . $row['DomesticLeague'] . "<br>";
//		echo "Telephone: $row['TelePhoneNo'] <br> Fax No.: $row['FaxNo'] <br>";
//		echo "Email: $row['EmailAddress'] <br> Website: $row['Website']<br>";
//		echo "GroundNo: $row['GroundNo'] <br> Established: $row['YearEstablished'] <br>";
//		echo "Status: $row['Status'] <br> ManagerNo: $row['ManagerNo'] <br> CoachNo: $row['CoachNo'] <br>"; 
	}
}

?>

Change the rest of the lines to match.

Cork Skate · 27-04-2005 3:58pm

I have tried this ......

<?Php

if ($_REQUEST['submit'] == 'Submit') {
$clubID = $_POST['Clubno']; 
$result=mysql_query("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, ManagerNo, CoachNo FROM club WHERE Clubno=$clubID"); 
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
echo "Club Name: {$row['ClubName']} <br> League: {$row['DomesticLeague']} <br> Telephone: {$row['TelePhoneNo']} <br> Fax No.: {$row['FaxNo']} <br> Email: {$row['EmailAddress']} <br> Website: {$row['Website']} <br> GroundNo: {$row['GroundNo']} <br> Established: {$row['YearEstablished']} <br> Status: {$row['Status']} <br> ManagerNo: {$row['ManagerNo']} <br> CoachNo: {$row['CoachNo']} <br>"; 
}
}

?>

And it works .... well .... it doesn't give an error ... but it doesn't show the result of the query.

The form tag is <form method="post" name="forum_select" action="result.php">

And it brings me to that but it is meant to display the results on that page .... but it aint !!

Any ideas ??

hostyle · 27-04-2005 6:43pm

PHP is case sensitive.

if ($_REQUEST['submit'] == '[u]S[/u]ubmit') {

Is your submit buttons value submit or Submit?

Cork Skate · 27-04-2005 7:09pm

hostyle wrote:
PHP is case sensitive.
if ($_REQUEST['submit'] == '[u]S[/u]ubmit') {
Is your submit buttons value submit or Submit?

<input type="submit" value="Submit">

I doubt it is that .... i get to result.php alright so it is working ... it is just not bringing the results with it

I dont know if anything has to go onto result.php to indicate which cell the data will appear ... i tried something like this ....

<?php echo $_POST; ?>

But that is not giving me anything ?? It just prints the word 'array', i tried it on another page and the same so it does nothing.

I dont know what should go in there to be honest !!

hostyle · 28-04-2005 9:59am

Honestly this is going nowhere.

1. What are you trying to do?
2. Post your code somewhere (if its over 100 lines link to it externally - if less post it here)
3. Look up some PHP tutorials. This is very, very basic stuff. The reason I've been giving short answers is to try and give you hints rather than do it for you.

Cork Skate · 28-04-2005 12:21pm

hostyle wrote:

Honestly this is going nowhere.

Tell me about it !!

3. Look up some PHP tutorials. This is very, very basic stuff. The reason I've been giving short answers is to try and give you hints rather than do it for you.

I have .... i have tried to avoid asking question, searching through previous posts, looking for similar. I haven't found a tutorial that really covers what i need and i wouldn't mind but i would have thought that it is not that convoluted really. ITs nuts and bolts stuff.

I know its basic, but i am a total newbie .... i am struggling to be honest but what can i do, i need to get this done.

hostyle · 28-04-2005 1:13pm

I've PMed you my email address. Send on the code you have so far and I'll get the basics working for you.

Ph3n0m · 28-04-2005 1:33pm

Cork Skate wrote:

I know its basic, but i am a total newbie .... i am struggling to be honest but what can i do, i need to get this done.

searching in past posts here isnt going to help you.

You need to goto places like

www.php.net
www.phpbuilder.com/board/search.php

and other places that specilise in php, and its tutorials

Cork Skate · 28-04-2005 8:38pm

Problem solved !!

Just wanna post up the correct code ......

This is to populate the drop down menu ....


<?php
$username = "username";
$password = "ipassword";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) 
	or die("Could not select my database");

function generate_box() { 
    $sql = "SELECT ClubNo, ClubName FROM club"; 
    $result = mysql_query($sql) or die(mysql_error());
    $entries = mysql_num_rows($result); 
    $options = '<option value="-1"></option>'; 
    while($row = mysql_fetch_assoc($result)) { 
            $options .= "<option value=\"{$row['ClubNo']}\">{$row['ClubName']}</option>"; 
     } 
    return $options; 
} 

$options = generate_box();

?>

And this is the form ...... and it brings you to result.php


<form method="post" name="forum_select" action="result.php"> <table cellspacing="0" cellpadding="0" border="0">
	<tr><td border="1"><span>Select a club from the drop down list: <br>

<select name="by">

<? echo "$options"; ?>

</select>

<input type="submit" name="submit "value="Submit"></span>

And this is the code on result.php ..... in the cell where everything appears.


<?php
$username = "username";
$password = "ipassword";
$hostname = "localhost";		
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) or die("Could not select my database");
$clubID = $_POST['by']; 
$query = ("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, 

ManagerNo, CoachNo FROM club WHERE ClubNo=$clubID");
$result = mysql_query($query) or die(mysql_error()); 
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
extract($row);
echo "Club Name: $row[ClubName] <br> League: $row[DomesticLeague] <br> Telephone: $row[TelePhoneNo] <br> Fax No.: $row[FaxNo] <br> Email: $row[EmailAddress] <br> Website: $row[Website] <br> GroundNo: $row[GroundNo] <br> Established: $row[YearEstablished] <br> Status: $row[Status] <br> ManagerNo: $row[ManagerNo] <br> CoachNo: $row[CoachNo] <br>"; 
}

?>

Hope this helps someone else !!

hostyle · 28-04-2005 8:45pm

You should change the last one to make it secure from QUERY_STRING poisoning


<?php
$username = "username";
$password = "ipassword";
$hostname = "localhost";		
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) or die("Could not select my database");

[b]if ($_POST['by'] == intval($_POST['by'])) {
  $clubID = $_POST['by'];
}

if ($clubID) {[/b]
$query = ("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, 

ManagerNo, CoachNo FROM club WHERE ClubNo=$clubID");
$result = mysql_query($query) or die(mysql_error()); 
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
extract($row);
echo "Club Name: $row[ClubName] <br> League: $row[DomesticLeague] <br> Telephone: $row[TelePhoneNo] <br> Fax No.: $row[FaxNo] <br> Email: $row[EmailAddress] <br> Website: $row[Website] <br> GroundNo: $row[GroundNo] <br> Established: $row[YearEstablished] <br> Status: $row[Status] <br> ManagerNo: $row[ManagerNo] <br> CoachNo: $row[CoachNo] <br>"; 
}
[b]} else {
  echo "<p>Invalid selection</p>";
}[/b]
?>

Php Select Results from Menu

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