Question

我会保持简短快捷。指出我正确的方向或告诉我如何做。

我有一个数组（如下所示） - 我希望能够替换值？与第二个数组中的元素。那么第一个？被替换为数组2中的第一个值，依此类推。

我没有推出自己的PDO，代码库很老，目前还没有使用PDO。只需要知道我如何在下面的数组中进行准备和绑定，使其成为一个快速的sql语句：

    Dim Current_Day As String = Format(Now, "dddd")
    Dim Day_Name_Val As String = "0"

    Select Case Current_Day
        Case "Saturday"
            Day_Name_Val = "1"
        Case "Monday"
            Day_Name_Val = "2"
        Case "Tuesday"
            Day_Name_Val = "3"
        Case "Wednesday"
            Day_Name_Val = "4"
        Case "Thursday"
            Day_Name_Val = "5"
        Case "Friday"
            Day_Name_Val = "6"
        Case "Sunday"
            Day_Name_Val = "7"
    End Select

Var Dump Output

Update tblA set name = 'SRTest12', agency = 'AUS', feature_tag = 'SRTest1', sart_date = '2000-01-01 00:00:00', end_date = '2050-12-31 00:00:00', ongoing = '0', hide = '1' where id = 165

Print_r输出

>     array(2) {
>       [0]=>
>       string(126) "UPDATE tblA SET name = ?,agency = ?,feature_Tag = ?,start_Date = ?,end_Date = ?,hide_text = ?,ongoing =
> ? WHERE id = ?"
>       [1]=>
>       array(8) {
>         [0]=>
>         string(8) "SRtest12"
>         [1]=>
>         string(5) "AUS"
>         [2]=>
>         string(7) "SRTest1"
>         [3]=>
>         string(19) "2000-01-01 00:00:00"
>         [4]=>
>         string(19) "2050-12-31 00:00:00"
>         [5]=>
>         string(1) "0"
>         [6]=>
>         string(1) "1"
>         [7]=>
>         string(3) "165"
>       }
>     }

Answer 1

最直接的方法可能是preg_replace_callback，带有一个计数器：

$query = 'UPDATE...';
$data = ['foo', 'bar', ...];
$counter = 0;

$result = preg_replace_callback('/\?/', function () use (&$counter, $data) {
    return $data[$counter++];
}, $query);

您还希望添加代码以在回调中转义/引用SQL语法的值。为了完整起见，我可能还会检查$data[$counter]是否存在，如果不存在则抛出异常。

来自Array的PHP构建语句

1 个答案: