PHP - Insert Select Options Into This Code

I'm trying to figure out why the options aren't appearing inside the select dropdown. Any ideas why?

Code: [Select]
echo "<label for=" . $row2['fullName'] . ">" . $row2['fullName'] . "</label>";
                       echo "<select name=" . $row2['fullName'] . " id=" . $row2['fullName'] . " class=dropdown title=" . $row2['fullName'] . " />";
                       if ($styleID == 1 || $styleID == 2 || $styleID == 6) {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID = 3
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }   
                       } else {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID IN (1,2,6)
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }
                       }
                       echo "</select>";
                   }

I have set up a form page with a select box of colleges to select. I want the "options" in the select box to be values taken from a field called "name" in a table called "colleges" and they should be ordered alphabetically. I also want the default selected option to be "none."

I have attached a picture to describe what i want. Please be detailed with the code. I am fairly new to php and mysql. Thank you.

Hi, Im not sure whether this is a PHP or MySQL question, but im trying to get a select menu to only display the number of options that are defined in the mysql database.   For example   The code I have that retrieves the $quantity from the database.   Lets say that   $quantity = 3   Now I would like the option menu to only display three options, like this:

<select name="quantity">
<option value="<?php echo $quantity ?>"><?php echo 'Maximum of ' . $quantity . ' available'?></option>
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option>
</select>

but for example, if the $quantity was equal to 10 then I would like the menu to be displayed as such

<select name="quantity">
<option value="<?php echo $quantity ?>"><?php echo 'Maximum of ' . $quantity . ' available'?></option>
<option value="1">1</option> 
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
<option value="10">10</option>
</select>

How is it possible to do this? I am completely at a loss here.   Many Thanks   aquaman

Hey Guys,

Here is my issue. I have a MySQL database with a table of products. I use this table to generate a mutliselect form element that is used int a larger form. The larger form contains other information relevant to an issue that we are having with that product (such as an outage). Anyway, the form works fine and the form validation works fine and I write it to the database fine. The issue I'm having is how to save the mutliselect options most efficiently.

This is my current process:

1. User creates a new entry from the form and clicks save
2. The form is validated
3. All entries are written to the database in a table called "incidents" (content inclues, timestamps, description of the problem, the impact, who to contact, etc)
4. Because the products list can vary from 1 product to all (~30) products I broke it out into a separate table called "impacted_products". I then loop through each of the selected products and write one row for each product using the id from the row in the "incidents" table to define the tables relationship.

Example of DB results:

"incidents"

+-----------------+------+----------+---------------------+---------+---------+
| notification_id | what | impact   | time_start          | summary | contact |
+-----------------+------+----------+---------------------+---------+---------+
| 235235          | Test | None.... | 2011-02-08 13:41:00 | Test    |       3 |
+-----------------+------+----------+---------------------+---------+---------+

"impacted_products"

+-----+-----------------+------------+---------------------+---------+
| id  | notification_id | product_id | timestamp           | deleted |
+-----+-----------------+------------+---------------------+---------+
| 202 | 235235          |          7 | 2011-02-10 14:30:42 |       0 |
| 203 | 235235          |         37 | 2011-02-10 14:30:42 |       0 |
| 204 | 235235          |         23 | 2011-02-10 14:30:42 |       0 |
+-----+-----------------+------------+---------------------+---------+

5. Now the users wants to go back and make some updates.
6. They click the edit button from the menu on screen and the form is brought back up and all options/fields are filled out from that which is stored in the database.
7. They make their edits and save again.
8. Now this time it simple updates the "incidents" table but for the products table I do a:

"UPDATE impacted_products SET deleted=1 WHERE notification_id='$ID'"

and "delete" all the old impacted_products for this particular incident and then loop through all the products that the user still had selected and write them to the database again.
9. And repeat

So as you can see this could end up with a lot of "useless" entries in the database since my scripts only pay attention to those "impacted_products" whose deleted value is set to 0.

Example:

+-----+-----------------+------------+---------------------+---------+
| id  | notification_id | product_id | timestamp           | deleted |
+-----+-----------------+------------+---------------------+---------+
| 176 | 235235          |         37 | 2011-02-08 15:26:25 |       1 |
| 177 | 235235          |         43 | 2011-02-08 15:26:25 |       1 |
| 178 | 235235          |         37 | 2011-02-08 15:37:58 |       1 |
| 179 | 235235          |         43 | 2011-02-08 15:37:58 |       1 |
| 180 | 235235          |          1 | 2011-02-08 15:39:49 |       1 |
| 181 | 235235          |          7 | 2011-02-08 15:39:49 |       1 |
| 182 | 235235          |         37 | 2011-02-08 15:39:49 |       1 |
| 183 | 235235          |         43 | 2011-02-08 15:39:49 |       1 |
| 184 | 235235          |          1 | 2011-02-08 15:40:53 |       1 |
| 185 | 235235          |          7 | 2011-02-08 15:40:53 |       1 |
| 186 | 235235          |         37 | 2011-02-08 15:40:53 |       1 |
| 187 | 235235          |         43 | 2011-02-08 15:40:53 |       1 |
| 188 | 235235          |         37 | 2011-02-10 10:00:47 |       1 |
| 189 | 235235          |          1 | 2011-02-10 12:17:05 |       1 |
| 190 | 235235          |          7 | 2011-02-10 12:17:05 |       1 |
| 191 | 235235          |         13 | 2011-02-10 12:17:05 |       1 |
| 192 | 235235          |         37 | 2011-02-10 12:17:05 |       1 |
| 193 | 235235          |         23 | 2011-02-10 12:17:05 |       1 |
| 194 | 235235          |          1 | 2011-02-10 12:21:52 |       1 |
| 195 | 235235          |          7 | 2011-02-10 12:21:52 |       1 |
| 196 | 235235          |         13 | 2011-02-10 12:21:52 |       1 |
| 197 | 235235          |         37 | 2011-02-10 12:21:52 |       1 |
| 198 | 235235          |         23 | 2011-02-10 12:21:52 |       1 |
| 199 | 235235          |          7 | 2011-02-10 12:22:26 |       1 |
| 200 | 235235          |         37 | 2011-02-10 12:22:26 |       1 |
| 201 | 235235          |         23 | 2011-02-10 12:22:26 |       1 |
| 202 | 235235          |          7 | 2011-02-10 14:30:42 |       0 |
| 203 | 235235          |         37 | 2011-02-10 14:30:42 |       0 |
| 204 | 235235          |         23 | 2011-02-10 14:30:42 |       0 |
+-----+-----------------+------------+---------------------+---------+

I did it this way beacuase it seems the easiest and fastest way. Otherwise I would have to lookup all the impacted_products from the table that match that notification_id and check

1) Was there a new product selected? If so, add it.
2) Was a product that was selected no longer selected? If so, delete it.
3) Was a product that was selected before still selected now? If so, leave it alone.

This seemed like a lot of extra looping and a lot of extra DB queries to essentially end up at the same place.

However, I feel that there has still got to be a more efficient way of doing this where I won't have all the extra entries in the impacted_products table. 

Any ideas?

Thanks in advance!

This seems to be a bit of a challenge but I am creating a multiple page form and on one of the pages I have a select field. I would like the user to have the ability to select multiple options but I am using some functions to move the data in hidden fields from page to page. I don't think my functions are jiving with my my foreach loop cause I keep getting an invalid argument error.

Thanks in advance for any help.

Here is my function:

    function setSelected($fieldName, $fieldValue) {
        if(isset($_POST[$fieldName]) && $_POST[$fieldName] == $fieldValue) {
            echo 'selected="selected"';
        }
    }

And here is my loop:

    $selections = "";
   
    if(isset($_POST["selections"])) {
        foreach ($_POST["selections"] as $selection) {
            $selections .= $selection . ", ";
        }
    }

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=357712.0

Hi,

I would like to do the following in PHP using a MySQL database, but not sure how.

select a, b,c etc.. from mailbox where id = X
then
insert id,a,b,c etc.. into problem

When I select, I will be calling the id

So I would like it to get data from mailbox and then insert it into problem.
How would I go about doing this?

Thanks for any help you can give me

Hello.

I have  a simple enough code that takes information from one table and drops it into another.  This is great, but I have 2a new complexities that I have been unable to code correctly.

A.  'lastname', 'firstname' on table1 need to be combined into 'name'

How & Where do I combine these strings and then pass them?

Code: [Select]
<?php
include('dbconfig.php');


// Make a MySQL Connection
mysql_connect("localhost", "$user", "$password") or die(mysql_error());
mysql_select_db("$database") or die(mysql_error());

$result = mysql_query(
"INSERT INTO table2 (lastname, firstname, email)
SELECT lastname, firstname, email
FROM table1
WHERE email='someemail@gmail.com' ") or die(mysql_error());

?>

Well, I made this yesterday then realised, I need to check is it exists...
I got this but when I go to accept the application and the member exists in the table it enters it anyway...

Code: [Select]
<?php
$member=$_POST['memberid'];
$status=$_POST['Status'];

$con = mysql_connect("host","user","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("a2186214_hbclan",$con);
$sql="UPDATE application SET Status = '$status' WHERE ID = '$member'";
$sql1="INSERT INTO table_members(name) SELECT application.Name FROM application WHERE application.ID = '$member'";
if ($status == 'ACCEPTED')
{
if(mysql_num_rows(mysql_query("SELECT name FROM table_members WHERE name = '$member'")))
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
if(mysql_query($sql1, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
}
else
{
die('Could not submit: ' . mysql_error());
}
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
mysql_close($con);

?>

I have tried several attempts and can get data to select and using echo display it, however, I need to take this data and insert it into the database in a separate table.   I have the following which does hafl the job, can I get some pointers on the rest.  I have looked everywhere and not found a solution, at all.


// Selects the data I need
<?php
mysql_connect("PRIVATE INFO","PRIVATE INFO","PRIVATE INFO") or
    die("Could not connect: " . mysql_error());

mysql_select_db("wpdb");

$result = mysql_query("SELECT ID FROM wp_posts

WHERE post_title LIKE '%future%'
AND post_status = 'publish'
OR post_title LIKE '%option%'
AND post_content LIKE '%fundamental%'
AND post_status = 'publish'

   ORDER BY post_date DESC");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{   echo $row['ID'];
   echo "<br />";};

?>



Thought maybe something like the following would work, but am at a loss:
INSERT INTO wp_term_relationships (object_id, term_taxonomy_id, term_order)
       SELECT ID FROM wp_posts

            WHERE post_title LIKE '%future%'
            AND post_status = 'publish'
            OR post_title LIKE '%option%'
            AND post_content LIKE '%fundamental%'
            AND post_status = 'publish'

            ORDER BY post_date DESC

            while ($row = mysql_fetch_array($result, MYSQL_ASSOC))

VALUES
($row['ID'], '25', '0')

Hi, I need to insert some code into my current form code which will check to see if a username exist and if so will display an echo message. If it does not exist will post the form (assuming everything else is filled in correctly). I have tried some code in a few places but it doesn't work correctly as I get the username message exist no matter what. I think I am inserting the code into the wrong area, so need assistance as to how to incorporate the username check code.

$sql="select * from Profile where username = '$username';
$result = mysql_query( $sql, $conn )
      or die( "ERR: SQL 1" );
if(mysql_num_rows($result)!=0)
{
process form
}
else
{
echo "That username already exist!";
}


the current code of the form


<?PHP
//session_start();
require_once "formvalidator.php";
$show_form=true;

if (!isset($_POST['Submit'])) {



$human_number1 = rand(1, 12);



$human_number2 = rand(1, 38);



$human_answer = $human_number1 + $human_number2;



$_SESSION['check_answer'] = $human_answer;
}

if(isset($_POST['Submit']))
{





if (!isset($_SESSION['check_answer'])) {
echo "<p>Error: Answer session not set</p>";
}


if($_POST['math'] != $_SESSION['check_answer']) {
echo "<p>You did not pass the human check.</p>";
exit();
}


   $validator = new FormValidator();
    $validator->addValidation("FirstName","req","Please fill in FirstName");







$validator->addValidation("LastName","req","Please fill in LastName");
$validator->addValidation("UserName","req","Please fill in UserName");
$validator->addValidation("Password","req","Please fill in a Password");
$validator->addValidation("Password2","req","Please re-enter your password");
$validator->addValidation("Password2","eqelmnt=Password","Your passwords do not match!");
$validator->addValidation("email","email","The input for Email should be a valid email value");
$validator->addValidation("email","req","Please fill in Email");
$validator->addValidation("Zip","req","Please fill in your Zip Code");
$validator->addValidation("Security","req","Please fill in your Security Question");
$validator->addValidation("Security2","req","Please fill in your Security Answer");
    if($validator->ValidateForm())
    {
        $con = mysql_connect("localhost","uname","pw") or die('Could not connect: ' . mysql_error());
        mysql_select_db("beatthis_beatthis") or die(mysql_error());
$FirstName=mysql_real_escape_string($_POST['FirstName']); //This value has to be the same as in the HTML form file
$LastName=mysql_real_escape_string($_POST['LastName']); //This value has to be the same as in the HTML form file
$UserName=mysql_real_escape_string($_POST['UserName']); //This value has to be the same as in the HTML form file
$Password= md5($_POST['Password']); //This value has to be the same as in the HTML form file
$Password2= md5($_POST['Password2']); //This value has to be the same as in the HTML form file
$email=mysql_real_escape_string($_POST['email']); //This value has to be the same as in the HTML form file
$Zip=mysql_real_escape_string($_POST['Zip']); //This value has to be the same as in the HTML form file
$Birthday=mysql_real_escape_string($_POST['Birthday']); //This value has to be the same as in the HTML form file
$Security=mysql_real_escape_string($_POST['Security']); //This value has to be the same as in the HTML form file
$Security2=mysql_real_escape_string($_POST['Security2']); //This value has to be the same as in the HTML form file



$sql="INSERT INTO Profile (`FirstName`,`LastName`,`Username`,`Password`,`Password2`,`email`,`Zip`,`Birthday`,`Security`,`Security2`) VALUES ('$FirstName','$LastName','$UserName','$Password','$Password2','$email','$Zip','$Birthday','$Security','$Security2')"; 
//echo $sql;
if (!mysql_query($sql,$con)) {

 die('Error: ' . mysql_error());

} else{



mail('email@gmail.com','A profile has been submitted!',$FirstName.' has submitted their profile',$body);

echo "<h3>Your profile information has been submitted successfully.</h3>";
}

mysql_close($con);
        $show_form=false;
    }
    else
    {
        echo "<h3 class='ErrorTitle'>Validation Errors:</h3>";

        $error_hash = $validator->GetErrors();
        foreach($error_hash as $inpname => $inp_err)
        {
            echo "<p class='errors'>$inpname : $inp_err</p>\n";
        }        
    }
}

if(true == $show_form)
{
?>

I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue...

Here is my select box code which is working perfect:

<select name="city">
<?php
$sql = "SELECT id, city_name FROM cities ".
"ORDER BY city_name";
$results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!");
while($row = mysqli_fetch_array($results_set)) {
echo "<option value=$row[id]>$row[city_name]</option>";
}
?>
</select>

Are any variables defined in a while loop global to the while loop only?

Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form.

Any ideas for a workaround to get this value passing as normal?

if (isset($_POST['addPosting'])) {
      $query = "INSERT INTO Postings (id, city_id, title, description) VALUES
('','$row[id]','$_POST[title]','$_POST[description]')";

could anyone please help me with the code which is i have already displayed data as a multi select list but now i need to select one or more from them and insert into another database table. would be appreciate your help. thanx

I use this type of a code to send automatic emails from my website:

Code: [Select]
$headers = ;
$headers .= ;
$to = ;

Click here to go to Google.

", $headers);

I am having hard time figuring out how to do hyperlink on words (like here). If I do something like this:

Code: [Select]
<a href='http://www.google.com'>here</a>
it spits out that exact thing out.

Thanks you for your input

hi, this piece of code is not working, is it ok?

Code: [Select]
$query1 = "SELECT * FROM members_copy WHERE RSUSER = '".$RSUSER."' AND RSPASS = '".$RSPASS."'";
$result1 = mysql_query($query1);
$row1 = mysql_fetch_array($result1);

$_SESSION['USERID']=$row1['USERID'];
$_SESSION['rsTown']=$row1['rsTown'];
$_SESSION['RSEMAIL']=$row1['RSEMAIL'];
$_SESSION['RSUSER']=$row1['RSUSER'];

$pSQL = "INSERT INTO favepub_copy (USERID,PUBID)";
$pSQL = $pSQL."SELECT ".$_SESSION['USERID'].", PUBID ";
$pSQL = $pSQL."FROM pubs ";
$pSQL = $pSQL."WHERE rsTown = '".$rsTown;
//echo "hello world";
//echo $pSQL;
//exit();
mysql_query($pSQL);