Insert and Update data in database at the same time - php

Question

What i want- i want to insert data into database if no existing data, and update if the data already exist in database.
Problem - If i click submit button without enter new data, the data still insert into database.
This is my code so far:
<script type="text/javascript" src="js/jquery-1.7.1.js"></script>
<?php
mysql_connect("localhost","root","");
mysql_select_db("cuba");
if(isset($_POST['hantar']))
{
for($a=1; $a<=count($_POST['name']); $a++)
{
/*$sqlQuery = "REPLACE INTO a (id,name,ic) VALUES ('".$_POST['id'][$a]."','".$_POST['name'][$a]."', '".$_POST['ic'][$a]."')"; */
$sqlQuery = "INSERT INTO a (id,name,ic) VALUES ('".$_POST['id'][$a]."','".$_POST['name'][$a]."', '".$_POST['ic'][$a]."' ) on DUPLICATE KEY UPDATE name='".$_POST['name'][$a]."', ic ='".$_POST['ic'][$a]."'";
mysql_query($sqlQuery)or die(mysql_error());
}
}
?>
<form action="" method="post">
<div class="b">
<?php
$a = 1;
$share = mysql_query("select *from a")or die(mysql_error());
while($share_papar = mysql_fetch_array($share))
{
?>
<table class="bb" style="border:1px solid #003;">
<tr>
<td width="200">Name</td>
<td> : <input type="text" class="input_teks" name="name[]" value="<? php echo $share_papar['name'];?>" />
</td>
</tr>
<tr>
<td>ic</td>
<td> : <input type="text" class="input_teks" name="ic[]" value="<?php echo $share_papar['ic'];?>" /></td>
</tr>
<tr>
<td>id</td>
<td> : <input type="text" class="input_teks" name="id[]" value="<?php echo $share_papar['id'];?>" id="akhir"/></td>
</tr>
<tr>
<td colspan="2" align="right" class="b_buang">Remove</td>
</tr>
</table>
<?php
}?>
</div>
<input type="submit" value="ss" name="hantar">
</form>
<span class="b_tambah">Add new group</span>
<script type="text/javascript">
$(function()
{
//nk clone
$('.b_tambah').on('click',function(e)
{
$(".b").find(".bb").last().clone(true).appendTo(".b");
alert( $(".bb:last").find("input[id=akhir]").last().val());
var m = $(".bb:last").find("input[id=akhir]").last();
m.val("");
});
//nk buang
$(".b_buang").click(function () {
if($('.bb').length < 2)
{
alert("Remove operation can be done if GROUP more than one");
}
else
{
$(this).parents(".bb").remove();
}
});
});
</script>
<!-- close clone Shareholders -->
</td>
I'm using jquery to clone form same with update form, then reset value to blank for input type name (id), so i can have non match id with existing data. But if i submit form without cloning those form, the data will insert into database. What i want is, no need to insert data into database if i dont cloning the new data and just update the existing. How i can achieve that..

Answer 1

Use this code
if(isset($_POST['hantar']))
{
for($a=1; $a<=count($_POST['name']); $a++)
{
$name = $_POST['name'][$a];
$ic = $_POST['ic'][$a];
$name = $_POST['name'][$a];
$sel = mysql_query("select * from a where name = '$name' and ic = '$ic' ");
if(mysql_num_rows($sel) > 0){
//write update query
$fet =mysql_fetch_array($sel);
$auto_increment_id = $fet['id'];
$sqlQuery = " update a set name = '$name' , ic = '$ic' where id = '$auto_increment_id' ";
}
else {
// write insert query
$sqlQuery = "INSERT INTO a (id,name,ic) VALUES ('".$_POST['id'][$a]."','".$_POST['name'][$a]."', '".$_POST['ic'][$a]."' ) ";
}
mysql_query($sqlQuery)or die(mysql_error());
}
}

Answer 2

first check the $_POST['name'] in SELECT query if it is exsits update in the query if not then insert into database

Answer 3

try replacing
for($a=1; $a<=count($_POST['name']); $a++)
to
for($a=0; $a < count($_POST['name']); $a++)
And also check validation
e.g.
if(isset($_POST['name'][$a]) && trim($_POST['name'][$a]) != "")

Answer 4

Finally.. This code save me..
if(isset($_POST['hantar']))
{
for($a=0; $a < count($_POST['name']); $a++)
{
if($_POST['name'][$a]=="")
{
}else
{
if($_POST['id'][$a]=="")
{
$sqlQuery = "INSERT INTO a (name,ic) VALUES ('".$_POST['name'][$a]."', '".$_POST['ic'][$a]."' )";
mysql_query($sqlQuery)or die(mysql_error());
}
else
{
$sqlQuery = " update a set name = '{$_POST['name'][$a]}' , ic = '{$_POST['ic'][$a]}' where id = '{$_POST['id'][$a]}' ";
mysql_query($sqlQuery)or die(mysql_error());
}
}
}
}