如何使用onclick更新数据,即使没有按钮提交在PHP和MySQL中的CHECKBOX [英] How to update data using onclick even CHECKBOX without button submit in php and mysql
本文介绍了如何使用onclick更新数据,即使没有按钮提交在PHP和MySQL中的CHECKBOX的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
if(isset($ _ POST [btnBaca])){
//查询更新状态
$ Kode = isset($ _ GET ['Kode'])? $ _GET ['Kode']:$ _POST ['txtKode'];
$ mySql1 =UPDATE pemesanan set status ='DiBaca'WHERE kd_pesan ='$ Kode';
$ myQry1 = mysql_query($ mySql1,$ koneksidb)或死(Gagal查询.mysql_error());
if($ myQry1){
echo< meta http-equiv ='refresh'content ='0; url =?page = Pemesanan-Data'>;
}
exit;
elseif(!empty($ _ POST ['btnKirim'])){
$ Kode = isset($ _ GET ['Kode'])? $ _GET ['Kode']:$ _POST ['txtKode'];
$ mySql1 =更新pemesanan set status ='DiKirim'WHERE kd_pesan ='$ Kode';
$ myQry1 = mysql_query($ mySql1,$ koneksidb)或死(Gagal查询.mysql_error());
if($ myQry1){
echo< meta http-equiv ='refresh'content ='0; url =?page = Pemesanan-Data'>;
}
exit;
elseif(!empty($ _ POST ['btnPending'])){
$ Kode = isset($ _ GET ['Kode'])? $ _GET ['Kode']:$ _POST ['txtKode'];
$ mySql1 =UPDATE pemesanan set status ='Pending'WHERE kd_pesan ='$ Kode';
$ myQry1 = mysql_query($ mySql1,$ koneksidb)或死(Gagal查询.mysql_error());
if($ myQry1){
echo< meta http-equiv ='refresh'content ='0; url =?page = Pemesanan-Data'>;
}
exit;
}
=========
< td align =center>
< input class =button blue smallname =btnBacatype =checkboxvalue =R>
< input class =button red smallname =btnKirimtype =checkboxvalue =D/>
< input class =button orange smallname =btnPendingtype =checkboxvalue =P/>
<?php echo $ myData ['status']; ?>
< / td>
解决方案
您可以通过这种方式尝试
< p id =costumersdata>打印成功或失败< / p>
< input class =button blue smallname =btnBacatype =checkboxvalue =RonClick =gotoupdate(this.value)>
< input class =button red smallname =btnKirimtype =checkboxvalue =DonClick =gotoupdate(this.value)/>
< input class =button orange smallname =btnPendingtype =checkboxvalue =PonClick =gotoupdate(this.value)/>
< script src =https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js>< / script>
< script>
函数gotoupdate(btnBaca){
$ .post(ajax.php?btnBaca =+ btnBaca,
函数(数据){
$(#costumersdata)。 html(data);
});
}
< / script>
在您的ajax.php页面上
<$ p如果(isset($ _POST [btnBaca])){
//查询更新状态
$ Kode
<?php
= $ _POST [btnBaca];
if($ Kode =='R'){
$ mySql1 =UPDATE pemesanan set status ='DiBaca'WHERE kd_pesan ='$ Kode';
} else if($ Kode =='D'){
$ mySql1 =UPDATE pemesanan set status ='DiKirim'WHERE kd_pesan ='$ Kode';
} else {
$ mySql1 =UPDATE pemesanan set status ='Pending'WHERE kd_pesan ='$ Kode';
$ b $ myQry1 = mysql_query($ mySql1,$ koneksidb)或die(Gagal query.mysql_error());
if($ myQry1){
echo< meta http-equiv ='refresh'content ='0; url =?page = Pemesanan-Data'> ;
}
exit;
}
?>
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,, this's my some script...thank you
if(isset($_POST["btnBaca"])) {
// Query Update status
$Kode = isset($_GET['Kode']) ? $_GET['Kode'] : $_POST['txtKode'];
$mySql1 = "UPDATE pemesanan set status='DiBaca' WHERE kd_pesan='$Kode'";
$myQry1 = mysql_query($mySql1, $koneksidb) or die ("Gagal query".mysql_error());
if($myQry1){
echo "<meta http-equiv='refresh' content='0; url=?page=Pemesanan-Data'>";
}
exit;
}
elseif(!empty($_POST['btnKirim'])){
$Kode = isset($_GET['Kode']) ? $_GET['Kode'] : $_POST['txtKode'];
$mySql1 = "UPDATE pemesanan set status='DiKirim' WHERE kd_pesan='$Kode'";
$myQry1 = mysql_query($mySql1, $koneksidb) or die ("Gagal query".mysql_error());
if($myQry1){
echo "<meta http-equiv='refresh' content='0; url=?page=Pemesanan-Data'>";
}
exit;
}
elseif(!empty($_POST['btnPending'])){
$Kode = isset($_GET['Kode']) ? $_GET['Kode'] : $_POST['txtKode'];
$mySql1 = "UPDATE pemesanan set status='Pending' WHERE kd_pesan='$Kode'";
$myQry1 = mysql_query($mySql1, $koneksidb) or die ("Gagal query".mysql_error());
if($myQry1){
echo "<meta http-equiv='refresh' content='0; url=?page=Pemesanan-Data'>";
}
exit;
}
=========
<td align="center">
<input class="button blue small" name="btnBaca" type="checkbox" value="R" >
<input class="button red small" name="btnKirim" type="checkbox" value="D" />
<input class="button orange small" name="btnPending" type="checkbox" value="P" />
<?php echo $myData['status']; ?>
</td>
解决方案
You can try by this way
<p id="costumersdata">Print Sucess or Fail</p>
<input class="button blue small" name="btnBaca" type="checkbox" value="R" onClick="gotoupdate(this.value)">
<input class="button red small" name="btnKirim" type="checkbox" value="D" onClick="gotoupdate(this.value)"/>
<input class="button orange small" name="btnPending" type="checkbox" value="P" onClick="gotoupdate(this.value)"/>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
function gotoupdate(btnBaca){
$.post("ajax.php?btnBaca="+btnBaca,
function(data){
$("#costumersdata").html(data);
});
}
</script>
On your ajax.php page
<?php
if(isset($_POST["btnBaca"])) {
// Query Update status
$Kode = $_POST["btnBaca"];
if($Kode == 'R'){
$mySql1 = "UPDATE pemesanan set status='DiBaca' WHERE kd_pesan='$Kode'";
}else if($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, $koneksidb) or die ("Gagal query".mysql_error());
if($myQry1){
echo "<meta http-equiv='refresh' content='0; url=?page=Pemesanan-Data'>";
}
exit;
}
?>
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