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Having trouble with a Drop Down Menu

perrij3

By perrij3
in PHP Coding Help

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perrij3 Member

perrij3

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Posted

I am new to PHP and I'm having trouble getting my drop down box to work.  I just want a simple drop down box that displays a list of employees from a table.

 

Here is the code that I have been working with, I just don't know where my error is.  ???

 

$query = "SELECT DISTINCT `Name` FROM `employee` ORDER BY `Name`";
$result = mysql_query($query)
or die (mysql_error());

echo "<form action='viewEmp.php' method='POST'>
<select name='EmployeeName'>";


while($row = mysql_fetch_array($result)) // Loop thru recordset and get values
{
echo"<option value='".$row["Name"]."'>".$row["Name"];
}
echo"</select>";
echo"<input type='submit' Value='Submit'>
</form>";

?>

 

I appreciate any help anyone can provide.

 

Thanks

 

Joe

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CroNiX Member

CroNiX

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Posted

1) Are you connected to a DB?  You code doesn't show that part if you are.

2) You shouldn't use single quotes in tag properties, like

<?php echo "<form action='viewEmp.php' method='POST'>

should be:

<?php echo "<form action=\"viewEmp.php\" method=\"POST\">

3) You are not closing your option tags with </option>

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perrij3 Member

perrij3

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I am able to connect to the database and display my data when display it outside of my drop down menu.  I will post my entire code so everyone can view it with the changes mentioned by CroNiX.  When I changed my form tag to what CroNix suggested, I got this error message:

 

Parse error: syntax error, unexpected '/' in C:\wamp\www\hi\findEmp.php on line 41

 

Here is all my code:

 

<?php
echo "<html>
<head>
<title>View All Work Orders Completed by One Employee</title>
<link href='css/gobal.css' rel='stylesheet' type='text/css' />
</head>
<body>";

$host="localhost";
$user="root";
$password="";

$cxn = mysql_connect($host,$user,$password) or die ("Couldn't connect to database.");

$dbname = 'holidayinn';
mysql_select_db($dbname);

$query = "SELECT DISTINCT `Name` FROM `employee` ORDER BY `Name`";
$result = mysql_query($query)
or die (mysql_error());

echo "<div id='header1'><img src='images/hi_logo.gif' alt='Holiday Inn Logo' width='200' height='83' id='logo'/><p align='right'><a href='home.html' target='_self' class='headlink'>Home</a> </p><h1 align='center'>Maintenance Request Tracking</h1></div>
<div>";


while ($row = mysql_fetch_assoc($result)) {
    echo $row["Name"];
}


/*This is the form used to display the drop down box to select an employee. */
echo "<form action=\"viewEmp.php\" method=\"POST\">
<select name='EmployeeName'>";

while($row = mysql_fetch_array($result)) // Loop thru recordset and get values
{
echo"<option value='".$row["Name"]."'>".$row["Name"];
echo"</option>
}
echo"</select>";
echo"<input type='submit' Value='Submit'>
</form>";

?>
</div>

</body>
</html>

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CroNiX Member

CroNiX

Posted
Posted

Wow, you are really mixing a lot of things that shouldn't be.

 

<html>
<head>
<title>View All Work Orders Completed by One Employee</title>
<link href="css/gobal.css" rel="stylesheet" type="text/css" />
</head>
<body>
<?php
$host="localhost";
$user="root";
$password="";

$cxn = mysql_connect($host,$user,$password) or die ("Couldn't connect to database.");

$dbname = 'holidayinn';
mysql_select_db($dbname);

$query = "SELECT DISTINCT `Name` FROM `employee` ORDER BY `Name`";
$result = mysql_query($query)
or die (mysql_error());

echo "<div id=\"header1\"><img src=\"images/hi_logo.gif\" alt=\"Holiday Inn Logo\" width=\"200\" height=\"83\" id=\"logo\"/><p align=\"right\"><a href=\"home.html\" target=\"_self\" class=\"headlink\">Home</a> </p><h1 align=\"center\">Maintenance Request Tracking</h1></div><div>";


while ($row = mysql_fetch_assoc($result)) {
    echo $row["Name"];
}

/*This is the form used to display the drop down box to select an employee. */
echo "<form action=\"viewEmp.php\" method=\"POST\">";
echo "<select name=\"EmployeeName\">";

while($row = mysql_fetch_array($result)) // Loop thru recordset and get values
{
echo "<option value=\"".$row["Name"]."\">".$row["Name"];
echo "</option>";
}
echo"</select>";
echo"<input type=\"submit\" Value=\"Submit\"></form>";

?>
</div>
</body>
</html>

 

In my original post, for #2 I didn't mean just where I showed you, I meant properties in all html tags.

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CroNiX Member

CroNiX

Posted
Posted

No prob, it gave me a nice little break from writing these stupid legal contracts for upcoming client work :)  It's the worst part about being an independent contractor.

 

Did it work?

 

I would also suggest rewriting your code so that you are mixing php and html as little as possible.

This:

1) makes your html much more readable

2) makes it a lot easier to maintain your PHP code

 

Something like this:

<?php include("functions.php"); //include your function(s) ?>
<html>
<head>
<title>View All Work Orders Completed by One Employee</title>
<link href="css/gobal.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="header1">
<img src="images/hi_logo.gif" alt="Holiday Inn Logo" width="200" height="83" id="logo"/>
        <p align="right"><a href="home.html" target="_self" class="headlink">Home</a> </p>
        <h1 align="center">Maintenance Request Tracking</h1>
</div>

<form action="viewEmp.php" method="POST">
<select name="EmployeeName">
        <?php createOptions(); //execute your function defined in 'functions.php' ?>
        </select>
<input type="submit" value="Submit">
</form>

</body>
</html>

 

Then create a 'functions.php' page like:

 

<?php
function createOptions(){
$host="localhost";
$user="root";
$password="";

$cxn = mysql_connect($host,$user,$password) or die ("Couldn't connect to database.");

$dbname = 'holidayinn';
mysql_select_db($dbname);

$query = "SELECT DISTINCT `Name` FROM `employee` ORDER BY `Name`";
$result = mysql_query($query)
	or die (mysql_error());
while ($row = mysql_fetch_assoc($result)) {
	echo $row["Name"];
}

/*This is the form used to display the drop down box to select an employee. */
while($row = mysql_fetch_array($result)) // Loop thru recordset and get values
{
 echo "<option value=\"".$row["Name"]."\">".$row["Name"];
 echo "</option>";
}
} // End createOptions()
?>

 

This is a very basic example and there is really more that should be done, but I hope you see the benefit.

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perrij3 Member

perrij3

Posted
  • Author
Posted

Thanks again for the advice on separating the html code from the php code.  I have learned a lot just from you.  The drop down box is still not being populated though. 

 

When I use this code, it will show me the list of employees so I know that the database is being accessed and the data is being read correctly.

 

while ($row = mysql_fetch_assoc($result)) {
    echo $row["Name"];
}

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