How To Update Data Using Onclick Even Checkbox Without Button Submit In Php And Mysql
How, can i update status when i click the check box?..this's my code... i already find and try hard but realy i cant finding anything, please to take out me from this problem,, thi
Solution 1:
You can try by this way
<pid="costumersdata">Print Sucess or Fail</p><inputclass="button blue small"name="btnBaca"type="checkbox"value="R"onClick="gotoupdate(this.value)"><inputclass="button red small"name="btnKirim"type="checkbox"value="D"onClick="gotoupdate(this.value)"/><inputclass="button orange small"name="btnPending"type="checkbox"value="P"onClick="gotoupdate(this.value)"/><scriptsrc="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script><script>functiongotoupdate(btnBaca){
$.post("ajax.php?btnBaca="+btnBaca,
function(data){
$("#costumersdata").html(data);
});
}
</script>
On your ajax.php page
<?phpif(isset($_POST["btnBaca"])) {
// Query Update status$Kode = $_POST["btnBaca"];
if($Kode == 'R'){
$mySql1 = "UPDATE pemesanan set status='DiBaca' WHERE kd_pesan='$Kode'";
}elseif($Kode == 'D'){
$mySql1 = "UPDATE pemesanan set status='DiKirim' WHERE kd_pesan='$Kode'";
}else{
$mySql1 = "UPDATE pemesanan set status='Pending' WHERE kd_pesan='$Kode'";
}
$myQry1 = mysql_query($mySql1, $koneksidb) ordie ("Gagal query".mysql_error());
if($myQry1){
echo"<meta http-equiv='refresh' content='0; url=?page=Pemesanan-Data'>";
}
exit;
}
?>
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