Question

我一直在寻找如何将多个key-＆gt;值插入mysql-database。我可以插入单个对象，但不能在foreach循环中插入我必须做错的事情。考虑到我知道我的java代码有效，因此我能够将单个对象存储在数据库中的单个coloumn表中，它必须是我的php语法错误 - 即使我遵循教程。请帮我这是我的php代码，必须有错误

<?php

    require_once 'db.php';
  $json = file_get_contents('php://input');
  $obj = json_decode($json,true);





  $st = mysqli_prepare($con, 'INSERT INTO movies(title,year,genre,director) VALUES(?,?,?,?)');

  mysqli_stmt_bind_param($st, 'sdss', $title, $year, $genre, $director);


  foreach ($obj as $row){
  $title =  $obj['title'];
  $year =  $obj['year'];
  $genre =  $obj['genre'];
  $director =  $obj['director'];   

  mysqli_stmt_execute($st);

  }


  ?>

my require_once'db_php';连接到数据库，我从远程Java应用程序获取数据。这是我的javacode - 只是省略了我的数据库URL和方法创建

    HttpURLConnection connection= null;
    try {

        URL obj = new URL(url2);
         connection = (HttpURLConnection) obj.openConnection();
        connection.setDoOutput(true);
        connection.setDoInput(true);
        connection.setRequestMethod("POST");
        connection.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
    //  connection.setRequestProperty("Accept", "application/json");
        connection.connect();


        OutputStreamWriter writer = new OutputStreamWriter(connection.getOutputStream());   
        JSONObject request = new JSONObject();
        request.put("action","login");
        request.put("title","The 40 year old virgin");
        request.put("year","2004");
        request.put("genre","Horror");
        request.put("director","Takashi Shimizu");




        String output =request.toString();                                             
        writer.write(output);                                                           
        writer.flush();                                                                 
        writer.close();   


        InputStream input = connection.getInputStream();                                    
        BufferedReader reader = new BufferedReader(new InputStreamReader(input));       
        StringBuilder result = new StringBuilder();                                     
        String line;       

        int responseCode = connection.getResponseCode();
        System.out.println("ResponseCode: " + responseCode);


        while ((line = reader.readLine()) != null) {                                    
            result.append(line);                                                        
        }         

        System.out.println("result:" + result.toString());
    }


   catch (IOException e) {                                                           

    } finally {   

        connection.disconnect();

    }       
}

这是我在IDE中的控制台中出现的错误 - eclipse“：警告：mysqli_stmt_bind_param（）期望参数1为mysqli_stmt”但是，我不知道如何处理它

Answer 1

如果sql失败，$ st将不是mysqli语句对象，我通常使用mysqli作为对象，语句中的任何拼写错误通常都会显示为在非对象上调用bind param的错误，我怀疑你可能有一个类似的问题。您可以使用程序样式测试错误，如下所示：

if($st == false){
    error_log(mysqli_error($conn));
}

如果准备失败，mysqli_prepare返回false，那么你可以调用mysqli_error传递mysqli_erction以获取最后一个错误的详细信息。

将json关联数组插入mysql

1 个答案: