php - 将 NULL 传递给旨在在 PHP 中构建 MySQL 查询的函数

Question

语言:PHP 和 MySQL

数据库中有一些行的列值为 NULL。我有一个函数，可以在收到 $params 数组时构建查询。以下是该函数中的相关代码(用于构建 WHERE 子句):

    if (isset($params['where'])) {
        $query .= "WHERE ";
        $count = count($params['where']);
        foreach ($params['where'] as $key => $value) {
            $count--;
            if ($count) {
                $query .= "`$key` ".$value['sign']." '".$value['text']."' AND ";
            } else {
               $query .= "`$key` ".$value['sign']." '".$value['text']."' ";
            }
        }
    }

问题是，当我将 $value['text'] 设置为 null 时，结果如下:

SELECT * FROM `db_profiles` WHERE `committed` IS '' LIMIT 100

如果我将其设置为 $value['text'] = 'NULL'，则会得到以下结果:

SELECT * FROM `db_profiles` WHERE `committed` IS 'NULL' LIMIT 100

这当然不会产生任何结果。有什么想法吗？

答案

    if (isset($params['where'])) {
        $where_segment = array ();
        foreach ($params['where'] as $key => $value) {
            if (is_null($value['text'])) {
                $where_segment[] = "`".$key."`"." IS NULL";
            } else {
                $where_segment[] = "`$key` ".$value['sign']." '".$value['text']."' ";
            }
        }
        $query .= " WHERE ".implode('AND', $where_segment);
    }

Answer 1

  $string_array = array();

    foreach($params['where'] as $field => $value) {

        if(is_null($value)) {
            $string_array[] = "`$field` IS NULL";
        }
        else {
            $string_array[] = "`$field`='$value'";
        }

    }

 $where_string = "WHERE ".implode(" AND ", $string_array);


 $sql = "SELECT * FROM `db_profiles` $where_string";

请注意，如果 $value 为 NULL，我会省略 NULL 短语周围的 '' 标记。