Question

我创建了一个接受数据数组的函数我想通过创建一个已显示日期的表来返回此HMTL 但foreach不显示HTML而不显示任何错误。

我的问题是：如何让这个函数返回显示数组数据的表？

我的功能：

<?php

function GetTableResult($arrayData) {

  return "<table id='Grid' class='table table-striped table-bordered table-hover dataTable no-footer DTTT_selectable' role='grid' aria-describedby='dynamic-table_info'>
          <thead>
            <tr>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Date</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Name</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Procedure</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Provider Name</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Bravery</th>                                         
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Coparticipation</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Type</th>
              <th class='hidden-480' tabindex='0' aria-controls='dynamic-table' rowspan='1' colspan='1' aria-label='teste'>Date Reference`enter code here`</th>
            </tr>
          </thead>
          <tbody>";
            foreach($arrayData as $data){
             "<tr id='Grid_Linha_1' class='odd' role='row'>
              <td class = 'hidden-480'>" . $data['date']             . "</td>
              <td class = 'hidden-480'>" . $data['name']             . "</td>
              <td class = 'hidden-480'>" . $data['procedure']        . "</td>
              <td class = 'hidden-480'>" . $data['provider_name']   . "</td>
              <td class = 'hidden-480'>" . $data['bravery']          . "</td>
              <td class = 'hidden-480'>" . $data['coparticipation'  . "</td>
              <td class = 'hidden-480'>" . $data['type']             . "</td>
              <td class = 'hidden-480'>" . $data['date_reference']  . "</td>
            </tr>";
            }
          "</tbody>
        </table>";
}

Answer 1

这是一个简化的例子，但你应该得到要点：

<?php

function GetTableResult($arrayData) {
    $table = '<table>';

    foreach($arrayData as $data) {
        $table .= "<tr id='Grid_Linha_1' class='odd' role='row'>
              <td class = 'hidden-480'>${data['date']}</td>
              ...
            </tr>";
    }
    $table .= '</table>';

    return $table;
}

echo GetTableResult([
    ['date' => '2017-11-22'],
    ['date' => '2017-11-23'],
]);

关键是将生成的HTML存储到变量（$table）中，向其追加字符串（使用$table .= ...）并在字符串中使用${data['field_name']}来插入变量值。

我希望这有帮助！

一般建议：手动生成复杂的HTML是一个痛苦的过程。我建议您查看像Twig这样的模板引擎来缓解疼痛：）。

Answer 2

这是我在评论中所说的解决方案。你错过了一个括号，试着回应而不是返回。

window.onload

输出：

日期名称

1/1/10 peterparker

2/10/27克拉克肯特

我怎样才能使这个函数返回显示数组数据的表

2 个答案: