PHP- Unable to Access Data from DB in dropdown menu

Question

I am trying to create a drop down menu that will display Project Names off all Projects present in the table Project in the database Testing... The drop down is created but it is not accessing the database and retrieving the required data... The code is as follows:

<html>
<head>
<title>Dynamic Drop Down List</title>
</head>

    <form id="form1" name="form1" method="post" action="<?php echo $PHP_SELF; ?>">
        Project List :
        <select Project Name='NEW'>
        <option value="">--- Select ---</option>
        <?
            $serverName = "Swagatha-PC"; 
            $connectionInfo = array( "Database"=>"Testing");
            $conn = sqlsrv_connect( $serverName, $connectionInfo);

            if( $conn ) 
            {
            echo "Connection established.<br />";
            }
            else
            {
            echo "Connection could not be established.<br />";
            die( print_r( sqlsrv_errors(), true));
            }
            if (isset ($select)&&$select!="")
            {
            $select=$_POST ['NEW'];
        }
        ?>
        <?
            $query=mysql_query("select ProjectName from dbo.Project");
            $menu=" ";
            while($row = sqlsrv_fetch_array($query))
            {
               $menu ="<option>" . $row["ProjectName"] . "</option>";
            }


            ?>
        </select>
        <input type="submit" name="Next" value="Select" />
    </form>
</body>
</html>

Dropdown Error

What Do i do to fix this??

mysql_query and sqlsrv_fetch_array? hmm.. Not even a Can I mix MySQL APIs in PHP?. But for real, you're probably looking for sqlsrv_query — FirstOne
– FirstOne, Commented Jun 7, 2017 at 16:50

u_mulder · Accepted Answer · 2017-06-07 17:06:07Z

1

The solution is

while($row = sqlsrv_fetch_array($query))
{
   echo "<option>" . $row["ProjectName"] . "</option>";
}

See? You output data with echo instead assinging data to variable.

And as @FirstOne noticed using mysql_query with sqlsrv_connect is an error too. I hope it's just your typo:

$query=sqlsrv_query($conn, "select ProjectName from dbo.Project");

As another sidenote

if (isset ($select)&&$select!="")
{
    $select=$_POST ['NEW'];
}

this if will always be false, because you first check $select and then define it. It definitely should be:

if (isset($_POST['NEW']) && $_POST['NEW'] != "")
{
    $select = $_POST['NEW'];
}

edited Jun 7, 2017 at 17:06

answered Jun 7, 2017 at 16:51

u_mulder

54.8k5 gold badges54 silver badges72 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

FirstOne Over a year ago

Besides from connecting with sqlsrv_connect and running mysql_query xD

u_mulder Over a year ago

Yeah, mentioned it, I suppose now we can close this as dupe(

user7390100 Over a year ago

Nope its still not working..Have made both the changes

FirstOne Over a year ago

I feel like commenting here because this will become more complete, instead of a bunch of answers with each a little info. Sorry if it's trouble for ya ^^. 1: the op doesn't have anything setting $select before if (isset ($select)&&$select!=""). 2: <select Project Name='NEW'> what's Project supposed to mean? 3: $select=$_POST ['NEW']; seems like a self fulfilling prophecy - though it will never be fulfilled xD since to select something, you'll need to have selected something (new is the name of the select itself).

user7390100 Over a year ago

@FirstOne It doesn't seem to be working still..Not gttng wer i hv gone wrong

Farhad Hossen · Accepted Answer · 2017-06-07 17:01:07Z

0

Here is you fixed code. You can try it

<html>
<head>
<title>Dynamic Drop Down List</title>
</head>

    <form id="form1" name="form1" method="post" action="<?php echo $PHP_SELF; ?>">
        Project List :
        <select Project Name='NEW'>
        <option value="">--- Select ---</option>
        <?php
            $serverName = "Swagatha-PC"; 
            $connectionInfo = array( "Database"=>"Testing");
            $conn = sqlsrv_connect( $serverName, $connectionInfo);

            if( $conn ) 
            {
            echo "Connection established.<br />";
            }
            else
            {
            echo "Connection could not be established.<br />";
            die( print_r( sqlsrv_errors(), true));
            }
            if (isset ($select)&&$select!="")
            {
            $select=$_POST ['NEW'];
        }


            $query=sqlsrv_query("select ProjectName from dbo.Project");
            $menu=" ";
            while($row = sqlsrv_fetch_array($query))
            {
               echo "<option>" . $row["ProjectName"] . "</option>";
            }


            ?>
        </select>
        <input type="submit" name="Next" value="Select" />
    </form>
</body>
</html>

edited Jun 7, 2017 at 17:01

answered Jun 7, 2017 at 17:00

Farhad Hossen

2731 silver badge6 bronze badges

2 Comments

u_mulder Over a year ago

Remove = after echo

Farhad Hossen Over a year ago

Thanks for suggested me @u_mulder

Collectives™ on Stack Overflow

PHP- Unable to Access Data from DB in dropdown menu

2 Answers 2

5 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related