php复选框在编辑页面上选中

Question

我有 3 张桌子：

people:
------------------------
peopleID,
firstname

peopletype:
------------------------
peopletypeID,
type

peoplepeopletype (junction table):
------------------------
peopleID,
peopletypeID

在添加表单上一切都很好，但我无法显示为分配的人员类型选中的复选框

这是我的代码。

从 peoplepeopletype 表中检索数据：

/*PEOPLE TYPE ************* */

    $stmt = $conn->prepare("SELECT * FROM peoplepeopletype WHERE peopleID=?");


    // set parameters and execute
    if ( !$stmt ) { echo "error"; }
    else if ( !$stmt->bind_param('i', $_GET['peopleID']) ) { echo "error";}
    else if ( !$stmt->execute() ) { echo "error"; }
  else {
      $result = $stmt->get_result();

      while($row = $result->fetch_assoc()) {
          $peopletypeIDfromdb = $row ['peopletypeID'];
          echo $peopletypeIDfromdb; /*echo only for test purposes, but I don't know how to use this in the form */
      }  
  } /* end else */

显示复选框：

 <?php /*retrieve peopletype from db */     
 $sql = "SELECT * from peopletype";
$result = $conn->query($sql);

while($row = $result->fetch_assoc()) { 
    echo '<input required="required" type="checkbox" name="peopletypeID[]" value="' . $row["peopletypeID"] . '"'; 
    if  ($peopletypeIDfromdb = $row["peopletypeID"]) { /*problem is probably here*/
               echo 'checked';
            }
    echo '>' . $row["type"];

}
              ?>

感谢您的帮助！

Answer 1

如果您可以将多个类型附加到一个人，则需要更改第一个循环以将所有类型存储在一个数组中。我将它们存储为键，以便将来在分配给人员的类型中检查某些类型是否存在的复杂性最小。

$result = $stmt->get_result();
$types = array();

while($row = $result->fetch_assoc()) {
    $types[$row ['peopletypeID']] = 1;
}

然后在第二个循环中使用它：

while($row = $result->fetch_assoc()) { 
    echo '<input required="required" type="checkbox" name="peopletypeID[]" value="' . 
         $row["peopletypeID"] . '" '; 
    if(isset($types[$row["peopletypeID"]]))
    {
         echo 'checked';
    }
    echo '>' . $row["type"];
}