PHP：显示NULL但实际包含数据

Question

产品table结构如下

    CREATE TABLE `products` (
  `id` int(6) unsigned NOT NULL AUTO_INCREMENT,
  `category_id` int(6) unsigned NOT NULL,
  `name` varchar(30) NOT NULL,
  `unit_id` int(6) unsigned NOT NULL,
  `brand_id` int(6) unsigned NOT NULL,
  `orignalCost` int(30) NOT NULL,
  `saleprice` int(30) NOT NULL,
  `deleted` tinyint(1) NOT NULL DEFAULT 0,
  PRIMARY KEY (`id`),
  KEY `fk-to-uom` (`unit_id`),
  KEY `fk-to-brand` (`brand_id`),
  KEY `category_id` (`category_id`),
  CONSTRAINT `fk-to-brand` FOREIGN KEY (`brand_id`) REFERENCES `brands` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  CONSTRAINT `products_ibfk_1` FOREIGN KEY (`category_id`) REFERENCES `categories` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=230 DEFAULT CHARSET=utf8mb4

谁能告诉我我犯了什么错误，因为 var_dump() 显示 NULl 值……mysqli_fetch_array 有什么错误吗？它显示为 "array(1) { [0]=> NULL } "

$selected_items_values = $_POST['product_id']; 
   $prices = [];
   foreach($selected_items_values as $prud)
   {
     if(isset($prud))
     {
      $priceSql = "SELECT saleprice from products where id = ' $prud ' ";
      $price=mysqli_query($db,$priceSql);
      $price = mysqli_fetch_array($price);
      array_push($prices , $price);
      var_dump($prices);

     } 
  }

Answer 1

编辑后的答案

这是我创建的示例 table。（我删除了外键）

CREATE TABLE `products` (
 `id` int(6) unsigned NOT NULL AUTO_INCREMENT,
 `category_id` int(6) unsigned NOT NULL,
 `name` varchar(30) COLLATE utf8_turkish_ci NOT NULL,
 `unit_id` int(6) unsigned NOT NULL,
 `brand_id` int(6) unsigned NOT NULL,
 `orignalCost` int(30) NOT NULL,
 `saleprice` int(30) NOT NULL,
 `deleted` tinyint(1) NOT NULL DEFAULT 0,
 PRIMARY KEY (`id`)
)

这是示例插入语句

INSERT INTO `products`VALUES (NULL, '1', 'name1', '1', '1', '100', '200', '0');
INSERT INTO `products`VALUES (NULL, '2', 'name2', '2', '2', '100', '200', '0');
INSERT INTO `products`VALUES (NULL, '3', 'name3', '3', '3', '100', '200', '0');
INSERT INTO `products`VALUES (NULL, '4', 'name4', '4', '4', '100', '200', '0');
INSERT INTO `products`VALUES (NULL, '5', 'name5', '5', '5', '100', '200', '0');
INSERT INTO `products`VALUES (NULL, '6', 'name6', '6', '6', '100', '200', '0');

这是我输入的示例数据。

+----+-------------+-------+---------+----------+-------------+-----------+---------+
| id | category_id |  name | unit_id | brand_id | orignalCost | saleprice | deleted |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  1 |      1      | name1 |    1    |     1    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  2 |      2      | name2 |    2    |     2    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  3 |      3      | name3 |    3    |     3    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  4 |      4      | name4 |    4    |     4    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  5 |      5      | name5 |    5    |     5    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+
|  6 |      6      | name6 |    6    |     6    |     100     |    200    |    0    |
+----+-------------+-------+---------+----------+-------------+-----------+---------+

这里是 db connect 通过 PHP 获取所有数据。

$conn = mysqli_connect("$db_host","$db_username","$db_pass","$db_name");
$query = $conn->query("SELECT * FROM `products`;");
var_dump($query);

这是上面的结果。你可以看到我有 6 行。

mysqli_result Object
(
    [current_field] => 0
    [field_count] => 8
    [lengths] => 
    [num_rows] => 6
    [type] => 0
)

为了遍历 php 对象使用 foreach

foreach ($query as $key => $value) {
  var_dump($value);
}

Foreach 结果如下

Array
(
    [id] => 1
    [category_id] => 1
    [name] => name1
    [unit_id] => 1
    [brand_id] => 1
    [orignalCost] => 100
    [saleprice] => 200
    [deleted] => 0
)

Array
(
    [id] => 2
    [category_id] => 2
    [name] => name2
    [unit_id] => 2
    [brand_id] => 2
    [orignalCost] => 100
    [saleprice] => 200
    [deleted] => 0
)

It continues like that...

为了只得到 saleprice 像下面这样使用它。

foreach ($query as $key => $value) {
  var_dump($value['saleprice']);
}

这就是您的获取方式。剩下的就看你怎么用了。

Answer 2

除了特定于您的实施的错误之外，此代码还有两个重要问题：

您是运行一个循环中的查询，使您正在执行的查询数量成倍增加并产生额外的数据库流量。
您正在将不受信任的数据连接到您的查询中，让自己容易受到 SQL 注入

第一个问题的答案是使用SELECT...WHERE...IN...，从而在一个查询中选择所有需要的数据第二个问题是通过使用准备好的语句来解决的。对于可变长度数据，它们的设置可能很繁琐，但提高安全性和性能是值得的。

对于您的示例，我们需要获取表单的查询

SELECT `saleprice` FROM `products` WHERE `id` IN (?,?,?);  // could be any number of placeholders here.

    $mysqli = new mysqli('localhost','user','password','schema');
    // create the placeholders
    $paramList = str_repeat("?,", count($_POST['product_id']));
    $query = "select saleprice from products where id in (".trim($paramList, ',').")";
    $stmt = $mysqli->prepare($query);
    // create a string of parameter types
    $types = str_repeat('s', count($_POST['product_id']));  // All $_POST values are strings.
    // Now bind the data to the query
    $stmt->bind_param($types, ...$_POST['product_id']); // Use the ... spread operator to unpack the array
    $stmt->execute();
    $stmt->bind_result($resultData);
    while ($row = $stmt->fetch()) {
        echo $resultData;
        echo "<br>\n";
    }

来源：代码取自 this blog

PHP：显示NULL但实际包含数据

PHP: displaying NULL but actually contains data

php

mysqli

var-dump

编辑后的答案