dropdown table

[phpjay]

can you please help me  i want to display all my table in mysql in dropdown list and after that if the user select one table

there is another dropdown list to display the column

[phpSensei]

Can you show some code on what you have done so far?

 

 

[phpjay]

i dont have code yet thats why i post  that message dont have any idea to do this

[trq]

PHP is generally used to output html. All you need to do is make that html create a drop down.

[phpjay]

i have this one

 

<?php

include('connection.php');

$con = mysql_connect("$host","$username","$password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$db", $con);

$sql = "SHOW TABLES FROM $db";
$result = mysql_query($sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

while ($row = mysql_fetch_row($result)) {
    echo "List of Food: {$row[0]}<br>";

}

mysql_free_result($result);

mysql_close($con);


?>

need to display in dropdown list

 

tablename1

tablename2

tablename3

 

this are my sample table in my database.

[trq]

Do you know how to create a drop down in html?

[phpjay]

yes like this

 

<select name='test'>

  <option value="test1">test1</option>

  <option value="test2">test2</option>

  <option value="test3">test3</option>

  <option value="test4">test4</option>

</select>

[trq]

Then all you need to do is make php produce the same.

[amg182]

Try:

 

<form name=form action="" method=POST>
<select name=test>
<?php
$query = "SELECT Column_name FROM table_name group by column_name";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo "<option value='". $row['column_name'] ."'>". $row['column_name'] ."</option>";
}
?>
</select>
</form>