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

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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

当我点击复选框时,如何更新状态?..这是我的代码...我已经找到并努力了,但是我真的不能找到任何东西,请从这个问题中拿出我,这是我的一些问题脚本...谢谢

  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;
}
?>

这篇关于如何使用onclick更新数据,即使没有按钮提交在PHP和MySQL中的CHECKBOX的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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