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php - 如何在多个查询中进行多个字符串替换?

转载 作者:行者123 更新时间:2023-11-29 01:52:42 25 4
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我有一个名为 $queries 的数组,如下所示:

Array ( 
[0] => SELECT * FROM StudentRecord WHERE Year = '%YEAR%' AND Session = '%SESSION%' AND StudentID = '%SID%';
[1] => SELECT * FROM Student_ApplicationRecord WHERE ApplicationYear = '%YEAR%' AND Session%SESSION% = 1;
)

我还有第二个名为 $variables 的数组,如下所示:

$variables = array('YEAR' => 2016, 'SESSION' => 1, 'SID' => $_SESSION['sid']);

我想用 $variables 中的正确值替换 $queries 中的变量。

我已经尝试了很多解决方案,到目前为止我想出的最好的是:

foreach ($queries as $query)
foreach($variables as $key=>$value)
$newqueries[] = str_replace("%".$key."%", $value, $query);

然而,这只是在每个查询中一次替换一个变量。

即当我这样做时:

foreach ($newqueries as $query)
print ($query);

每次查询的结果是:

SELECT * FROM StudentRecord 
WHERE Year = '%YEAR%' AND Session = '%SESSION%' AND SID = '%SID%';

SELECT * FROM StudentRecord
WHERE Year = '%YEAR%' AND Session = '1' AND SID = '%SID%';

SELECT * FROM StudentRecord
WHERE Year = '%YEAR%' AND Session = '%SESSION%' AND SID = '1234';

我想要的是一个最终数组$newqueries,如下所示:

Array ( 
[0] => SELECT * FROM StudentRecord WHERE Year = '2016' AND Session = '1' AND StudentID = '1234';
[1] => SELECT * FROM Student_ApplicationRecord WHERE ApplicationYear = '2016' AND Session1 = 1;
)

有人能帮帮我吗?如果之前有人问过这个问题,我深表歉意,但我找不到任何类似的问题。

最佳答案

str_replace()接受 $search$replace 参数的数组,您可以使用它。

$search = ['%YEAR%', '%SESSION%', '%SID%'];
$replace = ['2016', '1', $_SESSION['id']];
$newQueries = [];
foreach ($queries as $query) {
$newQueries[] = str_replace($search, $replace, $query);
}

关于php - 如何在多个查询中进行多个字符串替换?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37064969/

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