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 您的PHP代码似乎混淆了列名称date （显示在您的WHERE子句中）和post_date （它显示为您的fetch_assoc()子句产生的列名。您实际上同时拥有date和post_date列吗？ 
 
 此外，看起来你的WHERE子句中的不等式可能是一个接一个。 
 
 如果你想从今天起最近一年，那就是 
 
WHERE date >= CURDATE() - INTERVAL 1 YEAR
 
 在那之前的那一年，它是 
 
WHERE date >= CURDATE() - INTERVAL 2 YEAR
  AND date <  CURDATE() - INTERVAL 1 YEAR
 
 如果它是您想要的当前日历年 ，请使用此日历年 。 它有点复杂，因为它允许在date列上使用索引。 DATE_FORMAT()表达式将当前日期（例如2015-03-10）更改为2015-01-01。 
 
WHERE date >= DATE_FORMAT(CURDATE(), '%Y-01-01')
 
 类似地，在此之前的日历年是这样完成的。 
 
WHERE date >= DATE_FORMAT(CURDATE(), '%Y-01-01') - INTERVAL 1 YEAR
  AND date <  DATE_FORMAT(CURDATE(), '%Y-01-01')

Your php code seems to be confusing the column names date (which appears in your WHERE clause) and post_date (which appears as a column name resulting from your fetch_assoc() clause. Do you actually have both date and post_date columns?
 
Also, it looks like the inequalities in your WHERE clauses might be off-by-one. 
 
If you want to get the most recent year from today, it's
 
WHERE date >= CURDATE() - INTERVAL 1 YEAR
 
For the year prior to that, it's
 
WHERE date >= CURDATE() - INTERVAL 2 YEAR
  AND date <  CURDATE() - INTERVAL 1 YEAR
 
If it's the present calendar year you want, use this. It's a little complicated because it allows the use of an index on the date column. The DATE_FORMAT() expression turns the present date, e.g. 2015-03-10, into 2015-01-01.
 
WHERE date >= DATE_FORMAT(CURDATE(), '%Y-01-01')
 
Similarly, the calendar year prior to the present one is done like this.
 
WHERE date >= DATE_FORMAT(CURDATE(), '%Y-01-01') - INTERVAL 1 YEAR
  AND date <  DATE_FORMAT(CURDATE(), '%Y-01-01')