Question

im试图将jquery变量传递给php，这是代码

<label>Turnamen</label>
<div class="input-group">
   <div class="input-group-prepend">
      <span class="input-group-text">
         <input id="checkturnamen" type="checkbox">
      </span>
   </div>
   <select id="colturnamen" name="colturnamen" class="form-control" disabled="">
      @foreach ($nav as $navigasi)
         @if ($turnamen == $navigasi->waktu)
            <option value="terpilih" selected>{{$navigasi->waktu_text}}</option>
         @else
            <option>{{$navigasi->waktu_text}}</option>
         @endif
      @endforeach
   </select>
</div>

<label>Grup</label>
<div class="input-group">
    <div class="input-group-prepend">
        <span class="input-group-text">
           <input id="checkgrup" type="checkbox">
        </span>
    </div>
    <select id="colgrup" class="form-control" disabled="">
        <option value="terpilih" selected>NULL</option>
    </select>
</div>



<script>
   $(document).ready(function(){
      $('#colturnamen').change(function(){
          var pilihturnamen =  'February 2019';
          $('#colgrup').append('<?php $pilihturnamen ="'+pilihturnamen+'";echo($pilihturnamen); foreach($nav as $navigasi){if($pilihturnamen == $navigasi->waktu_text){if (is_null($navigasi->jumlah_grup)){}else{echo ("<option class="."autogrup".">A</option>");$alphabet = "A";for ($i = 1; $i < $navigasi->jumlah_grup; $i++){$alphabet++;echo ("<option class="."autogrup".">".$alphabet."</option>");}}}}?>');
          });
 </script>

如果我运行代码，它将显示

   <select id="colgrup" class="form-control">
       <option value="terpilih" selected="">NULL</option>
            February 2019
   </select>

然后我将脚本更改为此（我手动插入字符串）

<script>
       $(document).ready(function(){
          $('#colturnamen').change(function(){
              var pilihturnamen =  'February 2019';
              $('#colgrup').append('<?php $pilihturnamen ="February 2019";echo($pilihturnamen); foreach($nav as $navigasi){if($pilihturnamen == $navigasi->waktu_text){if (is_null($navigasi->jumlah_grup)){}else{echo ("<option class="."autogrup".">A</option>");$alphabet = "A";for ($i = 1; $i < $navigasi->jumlah_grup; $i++){$alphabet++;echo ("<option class="."autogrup".">".$alphabet."</option>");}}}}?>');
              });
 </script>

然后我再次运行它即可

<select id="colgrup" class="form-control">
   <option value="terpilih" selected="">NULL</option>
   February 2019
  <option class="autogrup">A</option>
  <option class="autogrup">B</option>
  <option class="autogrup">C</option>
  <option class="autogrup">D</option>
  <option class="autogrup">E</option>
</select>

问题是为什么我无法将jquery变量内的文本传递给该php代码？

Answer 1

页面请求：处理流程如下：

服务器端PHP代码呈现
已将响应发送到浏览器
浏览器开始呈现和执行JS

您想要进行AJAX调用，并在“成功”后执行JS代码。

以下是使用jQuery进行AJAX调用的示例：

$.ajax({
  type: "GET",
  url: url,
  success: success,
  ...
});

为什么我不能将Jquery变量传递给PHP

1 个答案: