php - 将数据发布到 mysql DB

Question

我正在尝试将数据从 android 作为 JSON 发送到 PHP，以便解析它并保存在 MySQL DB 中

这是 PHP 代码的一部分

$JsonString = $_POST["DATA"];
$JsonData = json_decode($JsonString, TRUE);

$Add_First_Only = 0;
foreach ($JsonData['items'] as $item)
{
    $Order_ID = $item['Order_ID'];
    $Order_Row_Number = $item['Order_Row_Number'];
    $Order_Item_ID = $item['Order_Item_ID'];
    $Order_Course_ID = $item['Order_Course_ID'];
    $Order_Seat_No = $item['Order_Seat_No'];
    $Order_Row_Value_wo_Options = $item['Order_Row_Value_wo_Options'];
    $Order_Row_Value_with_options = $item['Order_Row_Value_with_options'];

if ($Add_First_Only == 0)
{
    $result = mysqli_query($con,
    "INSERT INTO order_items (Order_ID,Order_Row_Number,Order_Item_ID,Order_Course_ID,Order_Seat_No,Order_Row_Value_wo_Options, Order_Row_Value_with_options) 
    VALUES 
    (['$Order_ID'],['$Order_Row_Number'],['$Order_Item_ID'],['$Order_Course_ID'],
        ['$Order_Seat_No'],['$Order_Row_Value_wo_Options'],['$Order_Row_Value_with_options'])"
    );
    $Add_First_Only = 1;
}
}

这是我在 Eclipse LogCAT 上遇到的错误

12-16 02:00:01.800: V/TAG(1841): Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '['26'],['1'],['1'],['1'],['1'],['0'],['1'])' at line 4

正如您从错误中看到的那样，我有变量值，因此它们都不是空值

问题是我应该更改或添加什么到我的 sql 语法来修复此错误？

Answer 1

删除 ['$Order_ID'] 和其他内容周围的括号

在您的VALUES中使用'$Order_ID'而不是['$Order_ID']等

if ($Add_First_Only == 0)
{
    $result = mysqli_query($con,
    "INSERT INTO order_items (Order_ID,Order_Row_Number,Order_Item_ID,Order_Course_ID,Order_Seat_No,Order_Row_Value_wo_Options, Order_Row_Value_with_options) 
    VALUES 
    ('$Order_ID','$Order_Row_Number','$Order_Item_ID','$Order_Course_ID',
        '$Order_Seat_No','$Order_Row_Value_wo_Options','$Order_Row_Value_with_options')"
    );
    $Add_First_Only = 1;
}