Insert Mysql problem

<script type="text/javascript">
function populateHarga() {
var x=document.getElementById("barang").value;
document.getElementById("harga").value=x;  
var y=document.getElementById('barang')[document.getElementById('barang').selectedIndex].innerHTML;
document.getElementById('nama_brg').value=y;
}
function populateTotal()
{
var y=document.getElementById("jumlah").value;
var x=document.getElementById("harga").value;
document.getElementById("totalnya").value=x*y;
}
</script>

	<body>
		&nbsp;<p>

		

		

		<?php
			mysql_connect("localhost","root","");
			mysql_select_db("stageco");
	
		?>
		<table border="1" cellpadding="0" cellspacing="0">
			<?php
				$nama_b=$_GET['nama_brg'];
				$harga_b=$_GET['harga'];
				$jumlah_b=$_GET['jumlah'];
				$total_b=$_GET['totalnya'];
				if($nama_b!="")
				{
					mysql_query("INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')");
				}
				$result=mysql_query("SELECT * FROM temp");
				$i = 0;
				while ($row = mysql_fetch_array($result)) {
					if ($i > 0) {
						echo "<tr valign='bottom'>";
						echo "<td bgcolor='#ffffff' height='1'></td>";
						echo "</tr>";
					}
					echo "<tr valign='middle'>";
					echo "<td class='tabval'><width='10' height='20'></td>";
					echo "<td class='tabval'><b>".$row['t_nama']."</b></td>";
					echo "<td class='tabval'>".$row['t_jumlah']."&nbsp;</td>";
					echo "<td class='tabval'>".$row['t_harga']."&nbsp;</td>";
					echo "<td class='tabval'>".$row['t_total']."&nbsp;</td>";
					echo "<td class='tabval'></td>";
					echo "</tr>";
					$i++;
					}
				echo "<tr valign='bottom'>";
				echo "<td bgcolor='#fb7922' colspan='6'></td>";
				echo "</tr>";
							?>

		</table>

<h2>add Entry</h2>
<?
echo "<form action=phonebook.php method=get>";
$get = mysql_query("select * from inventori");

echo"<table width=649 border=0 cellspacing=3 cellpadding=3>";

echo"<tr>";
echo"<td width=99>Nama Produk</td>";
echo"<td width=198><select id=barang onchange=populateHarga()>";
echo "<option></option>";
while($list=mysql_fetch_array($get))
{
echo "<option value='$list[harga]'>$list[nama_barang]</option>";
}
echo"</select></td>";
echo"</tr>";
echo"<tr>";
echo"<td>Jumlah</td>";
echo"<td><input id=jumlah name=jumlah type=text size=12 onblur=populateTotal() /></td>";
echo"</tr>";
echo"<tr>";
echo"<td>Harga</td>";
echo"<td>";
echo"<input id=harga  name=harga type=text />";
echo"</td>";
echo"</tr><tr><td>";
echo"Total :";
echo"</td><td>";
echo"<input type=text name=totalnya id=totalnya /> </td></tr>";
echo"<input type=text name=nama_brg id=nama_brg />";
echo"<tr>";
echo"<td colspan=2><div align=right><input type=submit  value=Next />
  <input type=reset id=batal value=Reset />
</div></td></tr>";
echo"</table></form>";
?>

	</body>
</html>

My problem is i want to insert the DB from a form i make in the same page.. but it doesnt work at all..thanks b4

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

10 Months Ago

Is there any error?

MooGeek

Posting Whiz

355 posts since Mar 2009

Reputation Points: 26

Solved Threads: 33

10 Months Ago

Post your errors.
Then it is easy to find out solution.

divyakrishnan

Posting Whiz in Training

201 posts since May 2010

Reputation Points: 30

Solved Threads: 24

10 Months Ago

Post your errors. Then it is easy to find out solution.

the problem is i cant insert the data i filled to the database..no error detected..it confuse me

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

10 Months Ago

is this phonebook.php file?

Karthik_pranas

Posting Pro

564 posts since Feb 2011

Reputation Points: 96

Solved Threads: 97

10 Months Ago

is this phonebook.php file?

yes.. i just testing some part of the main source code..and this little part has a problem.. so the name i use is not really important.

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

10 Months Ago

after 34th line echo out your query

echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')"

copy the query from output and execute it in the SQL section of your phpmyadmin and see what happens

Karthik_pranas

Posting Pro

564 posts since Feb 2011

Reputation Points: 96

Solved Threads: 97

10 Months Ago

after 34th line echo out your query
echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')"
copy the query from output and execute it in the SQL section of your phpmyadmin and see what happens

SQL query:

echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')"

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b'' at line 1

I dont get it

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

10 Months Ago

dont execute this

echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')"

just put this line in your code after 35th line.
Run this page in browser.
copy the printed query from your browser and execute this in phpmyadmin

Karthik_pranas

Posting Pro

564 posts since Feb 2011

Reputation Points: 96

Solved Threads: 97

10 Months Ago

dont execute this
echo "INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')"
just put this line in your code after 35th line. Run this page in browser. copy the printed query from your browser and execute this in phpmyadmin

Yeah done it and gives me error :
INSERT INTO temp( 't_nama', 't_harga', 't_jumlah', 't_total' )
VALUES (
'Kabel merah', '3000', '9', '27000'
)

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''t_nama','t_harga','t_jumlah','t_total') VALUES('Kabel merah','3000','9','27000'' at line 1

just to mention my type of data:
t_nama=varchar
t_harga=int
t_jumlah=int
t_total=int

thx b4..

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

10 Months Ago

replace your line 36 with

mysql_query("INSERT INTO temp(`t_nama`,`t_harga`,`t_jumlah`,`t_total`) VALUES('$nama_b','$harga_b','$jumlah_b','$total_b')");

just replace the "'" with"`" in your query.

Karthik_pranas

Posting Pro

564 posts since Feb 2011

Reputation Points: 96

Solved Threads: 97

10 Months Ago

Your mistakes are uniformly distributed from line 23-38. Replace those lines with this;

<?php		$conn = mysql_connect("localhost","root","");
			         mysql_select_db("stageco"); 		?>
<table border="1" cellpadding="0" cellspacing="0">			
<?php				
                  $nama_b=$_GET['nama_brg'];
		$harga_b=$_GET['harga'];
		$jumlah_b=$_GET['jumlah'];
		$total_b=$_GET['totalnya'];	
			[INDENT]if($nama_b!="")	{					
 mysql_query("INSERT INTO temp('t_nama','t_harga','t_jumlah','t_total') VALUES     ('$nama_b','$harga_b','$jumlah_b','$total_b')", $conn);

  }[/INDENT]				

                  $result=mysql_query("SELECT * FROM temp", $conn);

When ever you are sending a query you need your connection identifier; stored in the variable "$conn".

faroukmuhammad

Junior Poster

140 posts since Sep 2008

Reputation Points: 20

Solved Threads: 16

10 Months Ago

OK.. thx for your help

Lius

Newbie Poster

10 posts since Sep 2009

Reputation Points: 10

Solved Threads: 0

This question has already been solved

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