tyngtyng Posted October 2, 2006 Share Posted October 2, 2006 hi...i'm new to the php and now i'm getting an error to insert my data inside the form..and also undefined index in latest_updated......Now,i'm having trouble catching what i'm doing wrong.... Here's my code to the query entry...<?phpinclude 'config.php';$pili_id=$_POST['pili_id'];$branch=$_POST['branch'];$zone=$_POST['zone'];$location=$_POST['location'];$type=$_POST['type'];$subtype=$_POST['subtype'];$alamat=$_POST['alamat'];$latest_updated=$_POST['latest_updated'];$query = "INSECT INTO pili VALUES (pili_id, branch, zone, location, type, subtype, alamat, latest_updated)('$pili_id', '$branch', '$zone', '$location', '$type', '$subtype', '$alamat', '$latest_updated')";mysql_query($query) or die('Error, insert query failed');echo " '$pili_id' <b> '$branch' </b> ,'$zone', '$location', '$type', '$subtype', '$alamat','$latest_updated'";?>Can anyone give me a clue..?thanks...!!!! Link to comment Share on other sites More sharing options...
www.mihalism.com Posted October 2, 2006 Share Posted October 2, 2006 Try this <?phpinclude 'config.php';$pili_id = $_POST['pili_id'];$branch = $_POST['branch'];$zone = $_POST['zone'];$location = $_POST['location'];$type = $_POST['type'];$subtype = $_POST['subtype'];$alamat = $_POST['alamat'];$latest_updated = $_POST['latest_updated'];$con = mysql_connect("localhost","user","pass");if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("my_db", $con); $query = "INSERT INTO pili VALUE ('$pili_id', '$branch', '$zone', '$location', '$type', '$subtype', '$alamat', '$latest_updated')";mysql_query($query) or die('Error, insert query failed');echo " '$pili_id' <b> '$branch' </b> ,'$zone', '$location', '$type', '$subtype', '$alamat','$latest_updated'";?> Link to comment Share on other sites More sharing options...
eguru Posted October 2, 2006 Share Posted October 2, 2006 <?phpinclude 'config.php';$pili_id=$_POST['pili_id'];$branch=$_POST['branch'];$zone=$_POST['zone'];$location=$_POST['location'];$type=$_POST['type'];$subtype=$_POST['subtype'];$alamat=$_POST['alamat'];$latest_updated=$_POST['latest_updated'];$connection = mysql_connect("localhost","username","password");if (!$connection) { die('Could not connect: ' . mysql_error()); }mysql_select_db("databasename" , $connection) or die ('unable to select the database');$query = "INSERT INTO pili VALUES (pili_id, branch, zone, location, type, subtype, alamat, latest_updated)('$pili_id', '$branch', '$zone', '$location', '$type', '$subtype', '$alamat', '$latest_updated')";mysql_query($query) or die('Error, insert query failed');echo "<table cellpadding=8 border=1>"; echo "<tr>";echo "<td>$pili_id</td>";echo "<td>$branch</td>";echo "<td>$zon</td>";echo "<td>$location</td>";// and similarly add for othersecho "</tr>";echo "</table>";// OR echo " '$pili_id' <b> '$branch' </b> ,'$zone', '$location', '$type', '$subtype', '$alamat','$latest_updated'"; // this could be useful if u want to display your results in a table, ?>Well if u still get erros please do tell us, there would be someone to help u all the time. Link to comment Share on other sites More sharing options...
www.mihalism.com Posted October 2, 2006 Share Posted October 2, 2006 Hey eguru you have a error in your code look at this line echo "</table">; but it should say echo "</table>"; also the sql query says "INSECT INTO" but shouldent it be "INSERT INTO" Link to comment Share on other sites More sharing options...
eguru Posted October 2, 2006 Share Posted October 2, 2006 Thanx for notifying me, edited the lines, i did not notice. Link to comment Share on other sites More sharing options...
tyngtyng Posted October 2, 2006 Author Share Posted October 2, 2006 i still cant insert and save my data into my DB, even after i made a little modification...how was tat..?here's my codes again.. <html><head><title>Add New Pili Bomba</title></head><h1>Add New Pili Bomba Information</h1><hr/><?phpif(isset($_POST['submit'])){$con = mysql_connect("localhost","root","");if (!$con) { die('Could not connect:'.mysql_error()); }mysql_select_db("pili_bomba",$con)or die ('unable to select the database');include 'config.php';$pili_id=$_POST['pili_id'];$branch=$_POST['branch'];$zone=$_POST['zone'];$location=$_POST['location'];$type=$_POST['type'];$subtype=$_POST['subtype'];$alamat=$_POST['alamat'];$latest_updated=$_POST['latest_updated']; $query = "INSERT INTO pili(pili_id, branch, zone, location, type, subtype, alamat, latest_updated) VALUES ('$pili_id', '$branch', '$zone', '$location', '$type', '$subtype', '$alamat', '$latest_updated')";mysql_query($query) or die('Error, insert query failed');echo "<table cellpadding=8 border=1>"; echo "<tr>";echo "<td>$pili_id</td>";echo "<td>$branch</td>";echo "<td>$zon</td>";echo "<td>$location</td>";echo "<td>$type</td>";echo "<td>$subtype</td>";echo "<td>$alamat</td>";echo "<td>$latest_updated</td>";echo "</tr>";echo "</table>";echo "New pili Bomba added";}else{?><form action="" method=post> <TABLE cellSpacing=2 cellPadding=6 align=center border=1> <TR> <TD colSpan=4> <H3 align=center>Add Pili Bomba </H3></TD></TR> <TR> <TD> Branch</TD> <TD><SELECT name=branch> <OPTION value=selected>select branch <OPTION value=branch>Tun Mustapha <OPTION value=branch>Kg Jawa <OPTION value=branch>Layang-layangan </OPTION> </SELECT></TD> <TD> Location</TD> <TD><INPUT name= location></TD> <TR> <TD> Pili ID</TD> <TD><INPUT name=pili_id></TD> <TD> Zone</TD> <TD><INPUT name=zone></TD> <TR> <TD> Type</TD> <TD><SELECT name=type> <OPTION value=0 selected>select type <OPTION value=type>PH <OPTION value=type>GH </OPTION> </SELECT></TD> <TD> Subtype</TD> <TD><SELECT name=subtype> <OPTION value=0 selected>select subtype <OPTION value=subtype>A <OPTION value=subtype>V </OPTION> </SELECT></TD> <TR> <TD>Alamat</TD> <TD><INPUT name=alamat></TD> <TR> <TD> Latest updated</TD> <TD><INPUT name=latest_updated</TD> <TD><INPUT type='submit' value='Save'></TD> <TD><INPUT type='reset' value='Reset'></TD> <TR> </TABLE> </form> <?php } ?> </body> </html>what would be the solution for this? thankx Link to comment Share on other sites More sharing options...
eguru Posted October 2, 2006 Share Posted October 2, 2006 check the connection to your database, username and password Link to comment Share on other sites More sharing options...
tyngtyng Posted October 2, 2006 Author Share Posted October 2, 2006 wow...its not a problem from my connection,its a problem from my $query,!!! everythin hv been solve now..!!! anyway,thx alot for ur kindly help ya..!! Link to comment Share on other sites More sharing options...
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