PHP Error : Parse error: syntax error, unexpected '<' in filename.php

Question

我有三个按钮，编辑删除和查看。在视图部分，我试图回显行 ID 以获取详细信息。但是我得到了这个解析错误。

这是代码：

 while( $row=mysqli_fetch_array($query) ) { 
 $active = '';
 if($row['active'] == 1) {
    $active = '<label class="label label-success">Complete</label>';
 } else {
    $active = '<label class="label label-danger">Incomplete</label>'; 
 } 


 $actionButton = '
<div class="btn-group">
  <button type="button" class="btn btn-default dropdown-toggle" data-     toggle="dropdown" aria-haspopup="true" aria-expanded="false">
    Action <span class="caret"></span>
  </button>
  <ul class="dropdown-menu">
    <li><a type="button" data-toggle="modal" data-backdrop="static" data-keyboard="false"  data-target="#editMemberModal"  onclick="editMember('.$row['id'].')"> <span class="glyphicon glyphicon-edit"></span> Edit</a></li>
    <li><a type="button" data-toggle="modal" data-target="#removeMemberModal" onclick="removeMember('.$row['id'].')"> <span class="glyphicon glyphicon-trash"></span> Remove</a></li>
    <li><a class="view_data"  type="button" data-toggle="modal"   data-target="#dataModal" id="'.<?php echo $row['id']; ?>.'"><span class="glyphicon glyphicon-trash"></span> View</a></li> 
  </ul>
</div>
    ';

我在下拉菜单第三个列表中遇到此错误.. class 是 view_data

Answer 1

检查你的最后一个 li 按如下更改：

<li><a class="view_data"  type="button" data-toggle="modal"   data-target="#dataModal" id="' . $row['id'] . '"><span class="glyphicon glyphicon-trash"></span> View</a></li>

同样在上面的代码中结尾 } 最后丢失了

PHP Error : Parse error: syntax error, unexpected '<' in filename.php

PHP Error : Parse error: syntax error, unexpected '<' in filename.php

php

html

php-parse-error