PHP PDO 动态查询构建

Question

$arr = array();
$from_date = '2015-01-01';
$to_date = '2015-01-31';
$order_no = '25215';
$sql = "SELECT * FROM test";
if(!empty($from_date)&&!empty($to_date))
{
    $sql.=" WHERE txn_date BETWEEN :from_date AND :to_date";
    $arr[] = ":from_date => $from_date";
    $arr[] = ":to_date => $to_date";
    $condition=true;
}
if(!empty($order_no))
{
    if($condition)
    {
       $sql.=" AND ref_number = :order_no";
       $arr[] = ":order_no => $order_no";
    }
    else
    {
       $sql.=" WHERE ref_number = :order_no";
       $arr[] = ":order_no => $order_no";
       $condition=true;
    }
}
$stmt = $db->prepare($sql);
$stmt->execute($arr);
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);

执行此查询时会显示类似

的警告

Warning: PDOStatement::execute(): SQLSTATE[HY093]: Invalid parameter number: parameter was not defined

有什么问题吗？

Answer 1

而不是

$arr[] = ":from_date => $from_date";

做

$arr['from_date'] = $from_date;

详细了解 php 个数组 here。

此外，我会做的是：

$sql = "SELECT * FROM test WHERE 1=1";

那么就不需要测试 $condition 了，只需简单地连接到 $sql.

Answer 2

在变量 $arr 中，你应该像这样替换所有关联索引

$arr[':from_date'] = $from_date;
$arr[':to_date'] = $to_date;
$arr[':order_no'] = $order_no;

此外，将公共代码移动到外部块而不是 if/else 梯形图

也是一种很好的做法

if(!empty($order_no))
{
    $arr[':order_no'] = $order_no;
    if($condition)
    {
       $sql.=" AND ref_number = :order_no";
    }
    else
    {
       $sql.=" WHERE ref_number LIKE :order_no";
       $condition=true;
    }
}

正如 op 在评论中所问的那样，这里是对此的解释

$sql.=" WHERE txn_date BETWEEN :from_date AND :to_date";

用这个替换这个

$sql.=" WHERE date(txn_date) BETWEEN date(:from_date) AND date(:to_date)";

ref_number = :order_no

在 if/else 语句

的两个 where 条件中，将其转换为无符号整数，而不是上面的一种类型

CONVERT(ref_number,UNSIGNED INTEGER) = :order_no

PHP PDO 动态查询构建

PHP PDO Dynamic Query Building

php

mysql

pdo