Help a Novice with inserting from a combo box

kposki · #1 (**permalink**) 07-11-08, 12:31 AM

Hi,

I new to PHP and am working on creating an electives database with a web front end using WAMP. As the code below shows, I have created a form that can be used to view data from one table (course) and then insert the user's inputs into another table called suggested. My problem is not with getting values from course table into the combo box....the courseID values are showing in the combo box. My problem is with the inserting of the selected option (courseID) into the second table called suggested...The insert only puts modID, modname and proposer from the input text fields but not courseID from the combo box field. Please is there any help with this...I am so stuck for some days now...Thanks

Code:

PHP Code:


			
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

<html>

<head>

<title>Suggest electives for a selected cohort</title>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

 </head>

<body>



<form method="post">

<table width="600" border="0" cellspacing="1" cellpadding="2">

<tr>

<td width="100">Course Code</td>

<td>

<SELECT NAME='selcosID' ID='selcosID'>

<?php

include 'config.php';

include 'opendb.php';



   $query = "SELECT * FROM course";

$result = mysql_query($query) or die(mysql_error());





while($row = mysql_fetch_array($result)){



    echo "<OPTION VALUE=".$row['selcosID'].">".$row['CourseID']."</OPTION>";

}

?>

</SELECT>

</td>

</tr>



<tr>

<td width="100">Cohort</td>

<td>

<SELECT NAME='selcohort'ID='selcohort'>

<?php

include 'config.php';

include 'opendb.php';



   $query = "SELECT Cohort FROM course";

$result = mysql_query($query) or die(mysql_error());





while($row = mysql_fetch_array($result)){



    echo "<OPTION VALUE=".$row['cohid'].">".$row['Cohort']."</OPTION>";

}

?>

</SELECT>

</td>

</tr>



<tr>

<td width="100">Module Code</td>

<td><input name="modID" type="text" id="modID"></td>

</tr>



<tr>

<td width="100">Module Name</td>

<td><input name="modName" type="text" id="modName"></td>

</tr>



<tr>

<td width="100">Proposer</td>

<td><input name="prop" type="text" id="prop"></td>

</tr>





<tr>

<td width="100">&nbsp;</td>

<td><input name="submit1" type="submit" id="submit1" value="Submit Your Suggestion"></td>

</tr>

</table>

</form>



<?php



if(isset($_POST['submit1']))

{

include 'config.php';

include 'opendb.php';



$selcosID = $_POST['selcosID'];

$selcohort = $_POST['selcohort'];

$modID = $_POST['modID'];

$modName = $_POST['modName'];

$prop = $_POST['prop'];



// prepare the query strings





$query = "INSERT INTO suggested (CourseID, Cohort, ModuleID, ModuleName, Proposer) VALUES ('$selcosID', '$selcohort', '$modID','$modName', '$prop')";



mysql_query($query) or die(mysql_error());





include 'closedb.php';



echo "Your suggestion has been successfully submitted";

}





?>

</body>

</html>

gavintat · #2 (**permalink**) 07-11-08, 12:41 AM

You mean you cannot insert a correct courseID or got sql statement error?

kposki · #3 (**permalink**) 07-11-08, 12:47 AM

HI,
I dont get sql errors. It says insert was successful (suggestion has been submitted successfully) but when i go to view my database table, the columns for others are populated but not for CourseID and Cohort.
Thanks

gavintat · #4 (**permalink**) 07-11-08, 02:08 AM

<SELECT NAME='selcohort'ID='selcohort'>
<?php
include 'config.php';
include 'opendb.php';

$query = "SELECT Cohort FROM course";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result)){

echo "<OPTION VALUE=".$row['cohid'].">".$row['Cohort']."</OPTION>";
}
?>
</SELECT>

replace to
echo "<OPTION VALUE=".$row['Cohort'].">".$row['Cohort']."</OPTION>"; ??

kposki · #5 (**permalink**) 07-11-08, 02:16 AM

Thanks gavinat, It works now..Thanks for your time and may you find help in areas where you need help for yourself.

kposki · #6 (**permalink**) 07-11-08, 03:25 AM

Hi, I am making progress. By your help, what I have achieved is getting the user inputs to be inserted successfully into a table called "suggested" in my database.
Next, what I have just achieved is that another user(this time a coordinator)can view the table called "suggested" with an extra column for checkboxes to allow him approve from the suggestions. This went well.
Now I plan to finally insert the values from this table into a table called approved (meaning only the rows with approved column checked will be inserted into this new table).
Problem is I can view the suggestions now, I can tick the checkboxes but on submit, the database shows no values. No sql errors but form cant seem to release values to approved table in my database. Below is my code. I must thank you for this big help.

PHP Code:


			
<html>

<head>

<title>Approve electives for a selected cohort</title>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

 </head>

<body>



<form method="post">

<table width="600" border="0" cellspacing="1" cellpadding="2">

<tr>

<td width="100">Course Code</td>

<td>

<SELECT NAME='selcosID' ID='selcosID'>

<?php

include 'config.php';

include 'opendb.php';



   $query = "SELECT CourseID FROM course";

$result = mysql_query($query) or die(mysql_error());





while($row = mysql_fetch_array($result)){



    echo "<OPTION VALUE=".$row['CourseID'].">".$row['CourseID']."</OPTION>";

}

?>

</SELECT>

</td>

</tr>



<tr>

<td width="100">Cohort</td>

<td>

<SELECT NAME ='selcohort' ID = 'selcohort'>

<?php

include 'config.php';

include 'opendb.php';



   $query = "SELECT Cohort FROM course";

$result = mysql_query($query) or die(mysql_error());





while($row = mysql_fetch_array($result)){



    echo "<OPTION VALUE=".$row['Cohort'].">".$row['Cohort']."</OPTION>";

}

?>

</SELECT>

</td>

</tr>



<tr>

<td width="100">&nbsp;</td>

<td><input name="submit3" type="submit" id="submit3" value="View Suggested Modules"></td>

</tr>

</table>

</form>



<?php



if(isset($_POST['submit3']))  //if 1

{

include 'config.php';

include 'opendb.php';



$selcosID = $_POST['selcosID'];

$selcohort = $_POST['selcohort'];



$query = "SELECT * FROM suggested WHERE CourseID='$selcosID' AND Cohort = '$selcohort'" ;



$result = mysql_query($query) or die(mysql_error());



if(mysql_num_rows($result) == 0)

{

?>



<p><br><br>Student Table is empty </p>

<?php

}

else

{



echo "<form>";

echo "<table border='1'>

<tr>

<th>Course Code</th>

<th>Cohort</th>

<th>Module ID</th>

<th>Module Name</th>

<th>Proposer</th>

<th>Approved</th>



</tr>";

while($row = mysql_fetch_array($result))

{

echo "<tr>";

  echo "<td>" . $row['CourseID'] . "</td>";

  echo "<td>" . $row['Cohort'] . "</td>";

  echo "<td>" . $row['ModuleID'] . "</td>";

  echo "<td>" . $row['ModuleName'] . "</td>";

  echo "<td>" . $row['Proposer'] . "</td>";

  echo "<td><input type='checkbox' name='approve' value='app'/></td>";



  echo "</tr>";

}

echo "</table>";

echo "</form>";



}

include 'closedb.php';

}



?>



<form method="post">

<table>

<tr>

<td width="100">&nbsp;</td>

<td><input name="submit4" type="submit" id="submit4" value="Submit Approved Electives"></td>

</tr>

</table>

</form>

<?php



if(isset($_POST['submit4']))

{

include 'config.php';

include 'opendb.php'; //HOW DO I GET $MODID AND $MODNAME FROM ANOTHER PHP SCRIPT CALLED suggested or from the php script above....??....or do I use "require".. ???



$selcosID = $_POST['selcosID'];

$selcohort = $_POST['selcohort'];

$modID = $_POST['modID'];

$modName = $_POST['modName'];

$prop = $_POST['prop'];



// prepare the query string

$query = "INSERT INTO approved (CourseID, Cohort, ModuleID, ModuleName) VALUES ('$selcosID', '$selcohort', '$modID','$modName')";



mysql_query($query) or die(mysql_error());





include 'closedb.php';



echo "Your approval has been successful. $modID has been added to approved electives.";

}





?>





</body>

</html>