在更新时预先填写php表单

时间:2014-01-09 17:57:23

标签: php oop sql-update crud

我目前只使用一种形式进行更新和插入,效果很好。但我想在更新时预先填写。我不确定如何做到这一点。如果我使用php变量设置值,我会在日志中收到错误,指出它未定义。任何帮助表示赞赏。谢谢。

以下是表格:

<?php 
include_once('Crud_class.php');
$obj = new Crud("loocalhost","root","password","mydb");
$id = $_GET['id'];
if (isset($_GET['id']) && $_GET['id'] > 0) {
  echo "Update for record id#:" . $_GET['id'];
} else {
  echo "Insert new book";
}
 ?>

 <html>
 <head>
 <title>Add New Product</title>
 </head>

 <body>
 <form method="post" action="actions.php">
 <ul class="form">

   <li><label for"title">Title:</label>
      <input type="text" name="title" value="<?php echo $title?>" /></li>

    <li><label for="author">Author:</label>
    <input type="text" name="author"/></li>

     <li><label for="category">Category:</label>
     <select name="category">
       <option value="General">General</option>
       <option value="HTML/CSS">HTML/CSS</option>
       <option value="Javascript">Javascript</option>
       <option value="PHP">PHP</option>
       <option value="Other">Other</option>
      </select></li>

   <li><label for"description">Description:</label>
   <textarea name="description"></textarea></li>

   <li><label for="img_path">Image Path:</label>
    <input type="text" name="img_path"/></li>

      <input type="hidden" name="id" value="<?php if ($id > 0) { echo $id;} else {echo 0;} ?>"/>
     <li><input class="submit" type="submit" name="submit" value="Submit"/></li>
 </ul>
 </form>

  </body>
  </html>

这是我的行动档案:

<?php
include('Crud_class.php');
$obj=new Crud("localhost","root","password","mydb");
if (isset($_POST['id']) && $_POST['id'] > 0) {
//update
extract($_REQUEST);
$obj->update($id,$title,$author,$category,$description,$img_path);

} else {
// insert
extract($_REQUEST); 
$obj->insert($title,$author,$category,$description,$img_path);
 }

?>

和我的crud文件

       <?php
class Crud{
public $mysqli;
public $data;
public function __construct($host,$username,$password,$db_name){
    $this->mysqli = new mysqli('localhost', 'root', 'password', 'mydb');            
}

// BOOKS Table
//READ
public function readAll(){
    $query="SELECT * FROM books";
    $result= $this->mysqli->query($query);
    $num_result=$result->num_rows;      

    if($num_result>0){
        while($rows=$result->fetch_assoc()){        
            $this->data[]=$rows;
            //print_r($rows);
        }       
        return $this->data;
    }
}
//INSERT
public function insert($title,$author,$category,$description,$img_path){
    $query="INSERT INTO books SET title='$title', author='$author', category='$category', description='$description', img_path='$img_path'";
    $result= $this->mysqli->query($query) or die(mysqli_connect_errno()."Product Failed to Insert");

    if($result){
        header('location:read.php?insert_status=success');  
    }
}
//UPDATE
    public function update($id,$title,$author,$category,$description,$img_path){
        $query="UPDATE books SET title='$title', author='$author', category='$category', description='$description', img_path='$img_path' WHERE id='$id'";
        $result= $this->mysqli->query($query) or die(mysqli_connect_errno()."Cannot update");

        if($result){
            header('location:read.php?update_status=success');  
        }
    } 
//Delete
public function delete($id){
    $query="DELETE FROM books WHERE id='$id'";      
    $result= $this->mysqli->query($query) or die(mysqli_connect_errno()."Failed to Delete");

    if($result){
        header('location:read.php?delete_status=success');  
    }               
}


}




 ?>

4 个答案:

答案 0 :(得分:2)

您需要从表单页面上正在编辑的记录中获取数据。

if (isset($_GET['id']) && $_GET['id'] > 0) {
  echo "Update for record id#:" . $_GET['id'];
  // GET row from database, something like $obj->read($_GET['id']);
}

为此你需要在你的CRUD类中添加一些读取函数,它选择一行id = some input integer。

如果您有数据,请将其提取到对象或数组,并使用以下格式的值:

<input type="text" value="<?php if (isset($data)) echo $data['author']; ?>">

(当然选择框需要更多工作)

答案 1 :(得分:1)

您已经设置了自己的ID和条件。现在您需要初始化您需要的变量:

if (isset($_GET['id']) && $_GET['id'] > 0) {
  echo "Update for record id#:" . $_GET['id'];
  $formValues = readRecordByID($_GET['id']);  // you will need to create this crud select function
} else {
  echo "Insert new book";
  //initialize blank values:
  $formValues = array(
        'author'=>'',
        'category'=>'',
        'description'=>'',
        'img_path'=>'');
}

现在,您可以安全地使用:

<input type="text" name="author" value="$formValues['author']"/>

答案 2 :(得分:0)

预填表单取决于在HTML中嵌入预填充内容,其中包含文本框的value=属性或selected="selected" option中的<select>Title客户端javascript基于通过AJAX检索的数据或者再次嵌入HTML中的数据来实现。

php中的value="<?php echo $title?>" 字段就是我所说的一个例子 - 你有这个:

selected="selected"

这就是所有其他字段所需的内容(以及<select>字段{{1}})。

另请参阅this question

答案 3 :(得分:0)

将php更改为

<?php 
include_once('Crud_class.php');
$obj = new Crud("loocalhost","root","password","mydb");
$id = $_GET['id'];

if (isset($_GET['id']) && $_GET['id'] > 0) {
echo "Update for record id#:" . $_GET['id'];
}

if(isset($_REQUEST['id'])){ // this is added
$id=$_REQUEST['id'];
$result=$obj->mysqli->query("SELECT * FROM books WHERE id='$id'");

$rows=$result->fetch_assoc();

extract($rows);
} else {
  echo "Insert new book";
}
?>

并输入值

     value="<?php echo $title; ?>"

作品!

感谢大家的帮助和意见! :)