Question

我正在开发一个Android应用程序，我应该创建一个包含其图片的配方。我有一个名为create_recipe.php的php文件，在该文件中我创建了标题，成分，描述和类别 - 所有这些值都是在一个查询中创建的。另外，我有一个名为UploadImage.php的第二个php文件，用于从图库或相机中选择一张照片并将其上传到网络服务器，这也可以在一个查询中完成。在我的java代码中，我首先调用create_recipe.php，然后调用UploadImage.php。这样做会将信息保存在不同的行中。

有没有办法在数据库的单行中进行此查询？任何帮助将不胜感激！

以下是UploadImage.php

的代码

<?php

$target_path    =   "./images".basename( $_FILES['uploadedfile']['name']);
$pic            =   ($_FILES['photo']['name']); 
$file_path      =   $_FILES['tmp_name'];

// include db connect class
define('__ROOT__', dirname(dirname(__FILE__)));
require_once(__ROOT__.'/android_connect/db_connect.php');

// connecting to db
$db = new DB_CONNECT();

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name'])." has been uploaded";
    // Make your database insert only if successful and insert $target_path, not $file_path
    // Use mysqli_ or PDO with prepared statements

    // here I'm making the query to add the image
    $result =   mysql_query("INSERT INTO scb( name) VALUES('$target_path')");
} else
    echo "There was an error uploading the file, please try again!";

?>;

以下是create_recipe.php

的代码

<?php

/*
 * Following code will create a new recipe row
 * All recipe details are read from HTTP Post Request
 */

// array for JSON response
$response = array();

// check for required fields
if (isset($_POST['title']) 
    && isset($_POST['ingredients']) 
    && isset($_POST['description'])
    && isset($_POST['category']) ) {
//  && isset($_POST['image'])

    $title = $_POST['title'];
    $ingredients = $_POST['ingredients'];
    $description = $_POST['description'];
    $category = $_POST['category'];

    // include db connect class
    define('__ROOT__', dirname(dirname(__FILE__)));
    require_once(__ROOT__.'/android_connect/db_connect.php');

    // connecting to db
    $db = new DB_CONNECT();

    // mysql inserting a new row
    $result = mysql_query("INSERT INTO scb(title, ingredients, description, category, name) VALUES('$title', '$ingredients', '$description', '$category', '$target_path')");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Recipe successfully created.";

        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}

Answer 1

一个简单的更新查询就可以了！

$result = mysql_query("UPDATE scb SET name = '$target_path' ORDER BY id DESC LIMIT 1");

从2个php文件中一起添加2个查询

1 个答案: