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，但是我有问题显示已为指定的人员类型检查的复选框

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，则您需要更改第一个循环以将所有类型存储在一个数组中。我将它们存储为密钥，以便在分配给该人的类型中检查将来某种类型存在的最低复杂性。

$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"];
}

在编辑页面上选中了php复选框

1 个答案: