答案 0 :(得分:1)
试试这个
<?php
$date = date('Y-m-d'); //2017-03-14
$finaldate= date('Y-m-d', strtotime("+11 months", strtotime($date)));
$select_query = "SELECT *
FROM tablename
WHERE date > '".$finaldate."'";
?>
答案 1 :(得分:0)
这是你对mysql的SQL查询:
App app = new App();
app.onCreate();
app.getLanguageDetectionApi().getData("4978e60252ae102dfe1341146bb8cc3ec4bbbd78", textForRecognition).enqueue(new Callback<List<PostModel>>() {
@Override
public void onResponse(Call<List<PostModel>> call, Response<List<PostModel>> response) {
List<PostModel> posts = new ArrayList<>();
posts.addAll(response.body());
}
@Override
public void onFailure(Call<List<PostModel>> call, Throwable t) {
Toast.makeText(MainActivity.this, "An error occurred during networking", Toast.LENGTH_SHORT).show();
}
});
根据您的数据库更改您的表名和列名
您可以根据您的月,年或日进行更改
答案 2 :(得分:0)
您可以使用其他方式执行此操作,例如
$startdate = date('Y-m-d H:i:s',$your_date);
SELECT date,start_date,CONCAT('INTERVAL ',number_of_months,' ','MONTH') as comapre_date FROM table_name having date BETWEEN start_date AND (start_date - comapre_date)
尝试使用此查询希望这会有效..
答案 3 :(得分:0)
SELECT * FROM `extend_blog_details` as ExtendBlogDetail WHERE item_id='$item_id' and CURDATE() between start_date and DATE_ADD(start_date,INTERVAL TIMESTAMPDIFF(MONTH,NOW(),DATE_ADD(start_date, INTERVAL `number_of_months` month)) month) limit 1
我从此查询中获得了当前记录