PHP how to insert select box value into database - php

Question

i have a question
i already make a form to view two select box value with database and one text filed value
i want to know how to insert those value back into database with different table
my script on view select box value is something like this
<?php
$query = "SELECT gejala FROM gejala where idatribut =110000";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>
<select name="gejala1">
<?php
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['gejala']."'>".$row['gejala']."</option>";
}
?>
</select>
<p>Subatribut2
<?php
$query = "SELECT gejala FROM gejala where idatribut =110000";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>
<select name="gejala2">
<?php
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['gejala']."'>".$row['gejala']."</option>";
}
?>
</select>
<p>
<label>value
<input type="text" name="textfield">
</label>
<p>
<input name="submit" type="button" value="submit">
<p>
how to know how to insert this value into another table(pucuk) with same field
thankyou
i already try insert script
<?php
if(isset($_POST['gejala1'])) {
$gejala1 = $_POST['gejala1'];
$sql = "INSERT INTO pucuk (gejala1) VALUES ({$gejala2})";
$dbLink = mysql_connect('localhost', 'root', 'root') or die(mysql_error());
mysql_select_db('cbrteh', $dbLink) or die(mysql_errno());
$result = mysql_query($sql);
if($result) {
echo "Record successfully inserted!";
}
else {
echo "Record not inserted! (". mysql_error() .")";
}
}
if(isset($_POST['gejala2'])) {
$gejala2 = $_POST['gejala2'];
$sql = "INSERT INTO pucuk (gejala2) VALUES ({$gejala2})";
$dbLink = mysql_connect('localhost', 'root', 'root') or die(mysql_error());
mysql_select_db('cbrteh', $dbLink) or die(mysql_errno());
$result = mysql_query($sql);
if($result) {
echo "Record successfully inserted!";
}
else {
echo "Record not inserted! (". mysql_error() .")";
}
}
}
?>
</form>
but when i hit submit button it still has no effect

Answer 1

In while loop itself u can write insert query and send the option box value into the field u want,try it !

Answer 2

after a submit, your select option value will be in the $_POST[] variable under the name of your select. Namely, here "gejala2"

Answer 3

put your code in a form with method=POST and after submitting your form you will get all your new values in _POST array
for example
if(isset($_POST['submit']))
{
$var1=$_POST['select1'];
$var2=$_POST['select2'];
$var3=$_POST['select3'];
//then write update/insert query as your requirement
}

Answer 4

Try this
<?php
if(isset($_POST['gejala1'])) {
$gejala1 = $_POST['gejala1'];
$sql = "INSERT INTO pucuk (gejala1) VALUES ({$gejala2})";
$dbLink = mysql_connect('localhost', 'root', 'root') or die(mysql_error());
mysql_select_db('cbrteh', $dbLink) or die(mysql_errno());
$result = mysql_query($sql);
if($result) {
echo "Record successfully inserted!";
}
else {
echo "Record not inserted! (". mysql_error() .")";
}
}
if(isset($_POST['gejala2'])) {
$gejala2 = $_POST['gejala2'];
$sql = "INSERT INTO pucuk (gejala2) VALUES ({$gejala2})";
$dbLink = mysql_connect('localhost', 'root', 'root') or die(mysql_error());
mysql_select_db('cbrteh', $dbLink) or die(mysql_errno());
$result = mysql_query($sql);
if($result) {
echo "Record successfully inserted!";
}
else {
echo "Record not inserted! (". mysql_error() .")";
}
}
?>
<form action="" method="POST"><!-- add this -->
<?php
$query = "SELECT gejala FROM gejala where idatribut =110000";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>
<select name="gejala1">
<?php
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['gejala']."'>".$row['gejala']."</option>";
}
?>
</select>
<p>Subatribut2
<?php
$query = "SELECT gejala FROM gejala where idatribut =110000";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>
<select name="gejala2">
<?php
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['gejala']."'>".$row['gejala']."</option>";
}
?>
</select>
<p>
<label>value
<input type="text" name="textfield">
</label>
<p>
<input name="submit" type="submit" value="submit"> <!-- changed from type="button" to type="submit"> -->
<p>
</form>