Question

我有使用php / ajax / mysql传递2个参数的问题。

无法找出问题

这是Ajax代码

<script type="text/javascript"src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<script type="text/javascript">
$(document).ready(function(){
$('#myonoffswitch').click(function(){
var myonoffswitch=$('#myonoffswitch').val();
if ($("#myonoffswitch:checked").length == 0)
{
var a=myonoffswitch;
}
else
{
var a="off";
}

$.ajax({
type: "POST",
url: "process.php",
data: "ps="+a ,
success: function(html){
$("#display").html(html).show();
}
});

});
});
</script>

<script type="text/javascript">
$(document).ready( function(){
$(".cb-enable").click(function(){
var parent = $(this).parents('.switch');
$('.cb-disable',parent).removeClass('selected');
$(this).addClass('selected');
$('.checkbox',parent).attr('checked', true);
});
$(".cb-disable").click(function(){
var parent = $(this).parents('.switch');
$('.cb-enable',parent).removeClass('selected');
$(this).addClass('selected');
$('.checkbox',parent).attr('checked', false);
});
});
</script>

PHP：

$k=1; 
while($row = mysqli_fetch_array($result)) {
$id = $row['ID'];   
$ps = $row['Status'];

    echo "<div class=\"onoffswitch\">\n";
        echo "<input type=\"checkbox\" name=\"$k\" class=\"onoffswitch-checkbox\" id=\"$k\"\n";
        if($ps=="1") {
            echo "checked ";
        }
        echo ">";
        echo "<label class=\"onoffswitch-label\" for=\"$k\">\n";
        echo "  <div class=\"onoffswitch-inner\"></div>\n";
        echo "  <div class=\"onoffswitch-switch\"></div>\n";
        echo "</label>\n";
    echo "</div>\n";
    $k++;
}

进程.php类似于以下内容：

$ PS = $ _ POS [＆＃39; PS＆＃39;]; $ ID = $ _ POS [＆＃39; ID＆＃39;];

// mysql_query（＆＃34; UPDATE Tbl set Status =＆＃39; $ ps＆＃39;其中ID = $ id＆＃34;）; //不起作用

mysql_query（＆＃34; UPDATE Tbl set Status =＆＃39; $ ps＆＃39;其中ID = 1＆＃34;）;

代码在仅传递1个参数/参数时更新并正常工作。

POST或GET方法就可以了。有什么帮助吗？

Answer 1

PHP文件中有很多echo个语句。

将您的代码更改为此。告诉我它是否有效

$k = 1;

$content = null;

while ($row = mysqli_fetch_array($result) or die ("<p>DB Erroe!!!</p>")) {
    $id = $row['ID'];
    $ps = $row['Status'];

    $content .= "<div class=\"onoffswitch\">\n";
    $content .= "<input type=\"checkbox\" name=\"$k\" class=\"onoffswitch-checkbox\" id=\"$k\"\n";
    if ($ps == "1") {
        echo "checked ";
    }
    $content .= ">";
    $content .= "<label class=\"onoffswitch-label\" for=\"$k\">\n";
    $content .= "  <div class=\"onoffswitch-inner\"></div>\n";
    $content .= "  <div class=\"onoffswitch-switch\"></div>\n";
    $content .= "</label>\n";
    $content .= "</div>\n";
    $k++;
}

echo $content;

Answer 2

var send='{"ps":"'+a+'","id":"'+id+'"}';
var obj=JSON.parse(send);
$.ajax({
type: "GET", 
url: "process.php",
data: obj,
success: function(html){
$("#display").html(html).show();
}
});

Answer 3

不要，我再说一遍，不要使用mysql_ *接口。切换到mysqli或PDO。它已被弃用并在最新的PHP版本中消失。并且特别是不要混合它们！它不起作用！

PHP Ajax GET方法不会触发

3 个答案: