Question

我有4个输入字段，它让我抓狂，因为我无法弄清楚何时动态添加AND条件

现在我的原始查询如下所示：

SELECT *
FROM `staff`
" . $where . $location . "
" . $and_1 . $expertise . "
" . $and_2 . $education . "
" . $and_3 . $salary . "
ORDER BY 
`created_on` DESC LIMIT " . $limit;

因为我想允许空白字段为查询添加较少的条件我一直试图找出将WHERE和AND添加到每个条件的逻辑，具体取决于条件的数量被要求。

我的尝试：

($location == '' ? $location = '' : $location = '`location` = "'.$location.'"');
($expertise == '' ? $expertise = '' : $expertise = '`expertise` = "' . $expertise . '"');
($education == '' ? $education = '' : $education = '`education` = "' . $education . '"');
($location != '' && ($expertise != '' || $education != '') ? $and_1 = 'AND ' : $and_1 = '');
($location != '' && ($and_1 != '') ? $and_2 = 'AND ' : $and_2 = '');
($and_1 != '' || $and_2 != '' && $salary != 'no preference' ? $salary : '');
($and_1 == '' || $and_2 == '' && $salary != 'no preference' ? '' : $salary);

我认为我只是迷失在自己的心理逻辑中，因为我编码我甚至不理解我自己的代码逻辑lol;（

Answer 1

$location = ($location)?'`location` = "'.$location.'"':null;
....
....

更确切地说

$vars = array(
    ($location)?'`location` = "'.$location.'"'): null,
    /**
    ...
    ...
    **/
 );

function myfilterarray($var)
{
    return !empty($var)?$var: null;
}

$newvars = array_filter($vars, 'myfilterarray');

$where = join(" AND ", $newvars);

Answer 2

这将有助于您的HTML输入，但这是一个解决方案，甚至不需要弄乱查询的部分内容。

//do for each input
//or similar check to create empty string when there's not a valid input.
$input = is_empty($input) ? '' : $input;

...

SELECT *
FROM staff
WHERE location LIKE '%$location%'
AND expertise LIKE '%$expertise%'
AND education LIKE '%$education%'
AND salary LIKE '%$salary%'

对于一系列工资，您需要进行检查，类似于您在问题中的想法。

首先，将$salary设置为您需要的内容，$salary = 'no preference';或$salary = array($num1, $num2);。这应该在检查其他输入时完成。

使用可以使用LIKE：

的内容构建基本查询

$query = "SELECT *
          FROM staff
          WHERE location LIKE '%$location%'
          AND expertise LIKE '%$expertise%'
          AND education LIKE '%$education%'";

完成后，根据$ salary：

的内容构建薪水条款

if( is_array($salary) )
{
  $query .= " AND salary BETWEEN $salary[0] AND $salary[1]";
}
else
{
  $query .= " AND salary LIKE '%no preference%'";
  //or "AND salary = 'no preference'" if that field is exactly that
};

然后你可以执行$query。请注意，如果用户尝试在您的查询中执行恶意操作，则需要清理输入。

帮助我从输入构建查询的逻辑

2 个答案: