使用 sqlsrv 在 PHP 中参数化 IN 子句

Question

我有一种方法可以根据特定输入的整数数组来选择记录。但是，当我尝试用从方法参数 sqlsrv_query() returns false 传入的数组替换该数组时。我确信这很简单，但是 'obvious' 可能性对我不起作用。

工作版本

public function FindLocationRecords($locationIds)
{
    require("./Location.php");
    $resource = sqlsrv_connect($this->Server,$this->ConnectionInfo);
    $tsql = "select Date, PlaceName from rde_613949.dbo.Locations where id  in (14, 15, 16, 17); SELECT SCOPE_IDENTITY() as id";
    $results = sqlsrv_query($resource, $tsql);
    $locations = array();
    while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
    {
        $location = new Location($row['Date'], $row['PlaceName']);
        array_push($locations, $location);
    }
    return $locations;
}

版本 1 不工作

public function FindLocationRecords($locationIds)
{
    require("./Location.php");
    $resource = sqlsrv_connect($this->Server,$this->ConnectionInfo);
    $tsql = "select Date, PlaceName from rde_613949.dbo.Locations where id  in ?; SELECT SCOPE_IDENTITY() as id";
    $results = sqlsrv_query($resource, $tsql, array($locationIds));
    $locations = array();
    while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
    {
        $location = new Location($row['Date'], $row['PlaceName']);
        array_push($locations, $location);
    }
    return $locations;
}

版本 2 不工作

public function FindLocationRecords($locationIds)
{
    require("./Location.php");
    $resource = sqlsrv_connect($this->Server,$this->ConnectionInfo);
    $tsql = "select Date, PlaceName from rde_613949.dbo.Locations where id  in (?); SELECT SCOPE_IDENTITY() as id";
    $results = sqlsrv_query($resource, $tsql, array($locationIds));
    $locations = array();
    while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
    {
        $location = new Location($row['Date'], $row['PlaceName']);
        array_push($locations, $location);
    }
    return $locations;
}

版本 3 不工作

public function FindLocationRecords($locationIds)
{
    require("./Location.php");
    $resource = sqlsrv_connect($this->Server,$this->ConnectionInfo);
    $tsql = "select Date, PlaceName from rde_613949.dbo.Locations where id  in ?; SELECT SCOPE_IDENTITY() as id";
    $results = sqlsrv_query($resource, $tsql, $locationIds);
    $locations = array();
    while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
    {
        $location = new Location($row['Date'], $row['PlaceName']);
        array_push($locations, $location);
    }
    return $locations;
}

Answer 1

您需要生成占位符的动态列表 (?) 并将其包含在 SQL 语句中。此外，请始终检查 sqlsrv_connect() 和 sqlsrv_query() 执行的结果。

此示例基于问题中的代码，是您问题的可能解决方案：

<?php
public function FindLocationRecords($locationIds) {
    require("./Location.php");

    // Connection   
    $resource = sqlsrv_connect($this->Server, $this->ConnectionInfo);
    if ($resource === false) {
        echo "Error (sqlsrv_connect): ".print_r(sqlsrv_errors(), true); 
        return false;
    }   
    
    // Statement
    $tsql = "
        SELECT [Date], PlaceName 
        FROM rde_613949.dbo.Locations 
        WHERE id IN (".substr(str_repeat(',?', count($locationIds)), 1).")"
    ;
    $results = sqlsrv_query($resource, $tsql, $locationIds);
    if ($results === false) {
        echo "Error (sqlsrv_query): ".print_r(sqlsrv_errors(), true);   
        return false;
    }   

    // Data
    $locations = array();
    while ($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC)) {
        $location = new Location($row['Date'], $row['PlaceName']);
        array_push($locations, $location);
    }
    return $locations;
}
?>

使用 sqlsrv 在 PHP 中参数化 IN 子句

Parameterising an IN clause in PHP with sqlsrv

php

sql-server

sqlsrv