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RayMatrix

01-01-2003 12:42 PM

PHP-Beginner - Probs with Dropdown

Hi,

is there a way to show the results of an 'select' event in dropdownmenues ?

All Hints or Codesamples are welcome

Thanks for your help !!

Regards

GSHelpBoy

01-01-2003 04:36 PM

Sure.

PHP Code:


		
			
<?php

// connect

// select db

// sql query

$query = mysql_query($sql);

$i = 0;

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

    print('<option value="' . $row[$i] . "'>' . $row[$i] . '</option>');

    $i++;

}

?>

RayMatrix

01-01-2003 06:23 PM

But it wont work at my side.

This is my Code:

PHP Code:


		
			
<?

include ("config.php"); // connect to database

$result=mysql_query("SELECT mitarbeiter_betreuer FROM mitarbeiter");

$i = 0;

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

{

print('<option value='" . $row[$i] . "'>' . $row[$i] . '</option>');

 $i++;

}

mysql_close();

?>

But i get an Parse error at the line:

PHP Code:


		
			
 print('<option value='" . $row[$i] . "'>' . $row[$i] . '</option>');

any Idea ?

RayMatrix

01-01-2003 06:51 PM

. . .in that line, but it wont work also.

This is the hole Code now:

PHP Code:


		
			
<!-insert.php-> // Name of File

<html>

<body>

<?

if ($button==" OK ") {

include ("config.php"); // connect to the Database

mysql_query("INSERT INTO mitarbeiter(mitarbeiter_vorname,mitarbeiter_name,mitarbeiter_nummer,mitarbeiter_betreuer)VALUES('$vorname','$name','$nummer','$betreuer')");



mysql_close();

?>

<p>Die Daten wurden eingetragen<br>

  <a href="insert.php">neuer Eintrag</a><br>

  <a href="fetcharr.php">zur ?bersicht</a> 

  <?

}

else {

?><form method="post" action="insert.php">

  <input type="text" name="vorname" value="vorname">

  <br>

  <input type="text" name="name" value="name">

  <br>

  <input type="text" name="nummer" value="nummer">

  <br>

  <?

include ("config.php"); // connect to the Database

$result=mysql_query("SELECT mitarbeiter_betreuer FROM mitarbeiter");

$i = 0;

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

{

print('<option value=' . "$row[$i]" . '>' . $row[$i] . '</option>');

 $i++;

}

mysql_close();

?>

<br><input type="submit" name="Submit" value="Abschicken">

</form>

</body>

</html>

But now it tells me an Parseerror in the last line. . .

RayMatrix

01-01-2003 09:22 PM

. . .my mistake.

The Code do not bring up any more errors but now there are no entries in the dropdown :(

here the Code again:

PHP Code:


		
			
<!-insert.php->

<html>

<body>

<?

if ($button==" OK ") {

include ("config.php");

mysql_query("INSERT INTO mitarbeiter(mitarbeiter_vorname,mitarbeiter_name,mitarbeiter_nummer,mitarbeiter_betreuer)VALUES('$vorname','$name','$nummer','$betreuer')");



mysql_close();

?>

<p>Die Daten wurden eingetragen<br>

  <a href="insert.php">neuer Eintrag</a><br>

  <a href="fetcharr.php">zur ?bersicht</a> 

  <?

}

else {

?><form method="post" action="insert.php">

  <input type="text" name="vorname" value="vorname">

  <br>

  <input type="text" name="name" value="name">

  <br>

  <input type="text" name="nummer" value="nummer">

  <br>

  <select name="select">

  <?

include ("config.php");

$result=mysql_query("SELECT mitarbeiter_betreuer FROM mitarbeiter");

echo $result;

$i = 0;

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

{

print('<option value="' . $row[$i] . '">' . $row[$i] . '</option>');

$i++;

}

mysql_close();

?>

</select>

<br><input type="submit" name="Submit" value="Abschicken">

</form>

<?

}

?>

<form name="form1" method="post" action="">

  <select name="select">

    <option value="1">test</option>

  </select>

</form>

</body>

</html>

GSHelpBoy

01-02-2003 12:57 AM

Does config.php connect to the database, because otherwises, you will not get any result.

RayMatrix

01-02-2003 02:35 PM

. . .i got it now to work.

There were some little Errors in your Code ;)
Here is the one wich works fine for me:

PHP Code:


		
			
<!-insert.php->

<html>

<body>

<?

if ($button==" OK ") {

include ("config.php");

mysql_query("INSERT INTO mitarbeiter(mitarbeiter_vorname,mitarbeiter_name,mitarbeiter_nummer,mitarbeiter_betreuer)VALUES('$vorname','$name','$nummer','$betreuer')");

$result=mysql_query("SELECT * FROM mitarbeiter");

//echo  mysql_num_rows($result);

mysql_close();

?>

<p>Die Daten wurden eingetragen<br>

  <a href="insert.php">neuer Eintrag</a><br>

  <a href="fetcharr.php">zur ?bersicht</a> 

  <?

}

else {

?>

<form method="post" action="insert.php">

  <input type="text" name="vorname" value="vorname">

  <br>

  <input type="text" name="name" value="name">

  <br>

  <input type="text" name="nummer" value="nummer">

  <br>

  <select name="betreuer">

<?

include ("config.php");

$result=mysql_query("SELECT mitarbeiter_name FROM mitarbeiter");

// echo  mysql_num_rows($result);

$i = 0;

while ($row = mysql_fetch_array($result))

{

print('<option value="' . $row[$i] . '">' . $row[$i] . '</option>');

// $i++;

}

mysql_close();

?>

</select>

<br><input type="submit" name="button" value=" OK ">

</form>

<?

}

?>

</body>

</html>

Big Thanks again, your Hints were great !!

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