Question

我正在尝试将JS变量传递给PHP并让PHP回显JS变量。我一直得到一个空的空字符串。我究竟做错了什么？

function(u){

            if(u){
                var dt = {'ud':u};
                console.log(dt);

                $.post('xrege.php', dt, function(r){
                    console.log(r.responseText);
                    console.log(typeof(r.responseText));
                });
            }
        });


<?php

$ud = $_POST['ud'];
echo json_encode($ud);
?>

Answer 1

你应该将PHP代码分成不同的文件，它应该正常工作。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button class='showDiv' value="1">1</button>
<button class='showDiv' value="2">2</button>
<button class='showDiv' value="3">3</button>

<div class="layoutGroups" id="layoutGroup1">
  <h2>UK Map</h2>
  <div div style="width: 650px; height: 700px;"id = "MapDIV"></div>
  <div id="userUpdateDIV"></div>
  <div id = "BarChartDIV"></div>
  <div id="divPack1"></div>
</div>
<div class="layoutGroups" id="layoutGroup2">  
  <div id= "tree">Tree</div>
</div>
<div class="layoutGroups" id="layoutGroup3">
  <div id = "map">Map</div>
</div>

将上述代码命名为123.html，并将以下代码保留在xrege.php

中

if(u){
    var dt = {'ud':u};
    console.log(dt);

    $.post('xrege.php', dt, function(r){
        console.log(r);
        console.log(typeof r);
    },"json");
}

将JS变量传递给PHP

1 个答案: