我不知道它是否可能在mysql中，但任何建议将不胜感激。另外，我通过我的PHP脚本调用此查询并在谷歌图表中使用结果。
我的查询返回以下结果。
mysql> SELECT substring_index(xo.module_name, '/', 1) as Module, COUNT(date) AS Count, Year(date) as Year FROM xclude_run xr, join_run_object jro, xclude_object xo WHERE xr.syshost='darter' AND xo.module_name IS NOT NULL AND xr.run_id = jro.run_id AND jro.obj_id = xo.obj_id GROUP BY Module, Year ORDER BY Module, Year desc;
+--------------------------+-------+------+
| Module | Count | Year |
+--------------------------+-------+------+
| alps | 98 | 2015 |
| atp | 1220 | 2015 |
| atp | 279 | 2014 |
| boost | 28 | 2015 |
| cce | 10086 | 2015 |
| cp2k | 6 | 2014 |
| cray-hdf5 | 16 | 2014 |
| cray-hdf5-parallel | 114 | 2015 |
| cray-libsci | 721 | 2015 |
| cray-libsci | 514 | 2014 |
| cray-mpich | 1771 | 2015 |
| cray-mpich | 386 | 2014 |
| cray-netcdf-hdf5parallel | 50 | 2015 |
| cray-netcdf-hdf5parallel | 76 | 2014 |
| cray-shmem | 1832 | 2015 |
| cray-tpsl | 29 | 2015 |
| cray-trilinos | 104 | 2015 |
| craype-intel-knc | 2676 | 2015 |
| craype-intel-knc | 171 | 2014 |
| dmapp | 920 | 2015 |
| dmapp | 8 | 2014 |
| fftw | 288 | 2015 |
| fftw | 1791 | 2014 |
| gcc | 5371 | 2015 |
| gcc | 180 | 2014 |
| intel | 14 | 2015 |
| pmi | 2480 | 2015 |
| pmi | 416 | 2014 |
| rca | 920 | 2015 |
| udreg | 2479 | 2015 |
| udreg | 416 | 2014 |
| ugni | 2479 | 2015 |
| ugni | 416 | 2014 |
| wlm_detect | 49 | 2015 |
| xpmem | 2479 | 2015 |
| xpmem | 416 | 2014 |
+--------------------------+-------+------+
36 rows in set (0.00 sec)
我希望结果能够以一种简洁的方式逐年显示，例如 -
+--------------------------+-------+------+-------+-------+
| Module | Count | Year | Count | Year |
+--------------------------+-------+------+-------+-------+
| alps | 98 | 2015 | 0 | 2014 |
| atp | 1220 | 2015 | 279 | 2014 |
| boost | 28 | 2015 | 0 | 2014 |
| cce | 10086 | 2015 | 0 | 2014 |
| cp2k | 0 | 2015 | 6 | 2014 |
| cray-hdf5 | 0 | 2015 | 16 | 2014 |
| cray-hdf5-parallel | 114 | 2015 | 0 | 2014 |
| cray-libsci | 721 | 2015 | 514 | 2014 |
| cray-mpich | 1771 | 2015 | 386 | 2014 |
| cray-netcdf-hdf5parallel | 50 | 2015 | 76 | 2014 |
| cray-shmem | 1832 | 2015 | 0 | 2014 |
| cray-tpsl | 29 | 2015 | 0 | 2014 |
| cray-trilinos | 104 | 2015 | 0 | 2014 |
| craype-intel-knc | 2676 | 2015 | 171 | 2014 |
| dmapp | 920 | 2015 | 8 | 2014 |
| fftw | 288 | 2015 | 1791 | 2014 |
| gcc | 5371 | 2015 | 180 | 2014 |
| intel | 14 | 2015 | 0 | 2014 |
| pmi | 2480 | 2015 | 416 | 2014 |
| rca | 920 | 2015 | 0 | 2014 |
| udreg | 2479 | 2015 | 416 | 2014 |
| ugni | 2479 | 2015 | 416 | 2014 |
| wlm_detect | 49 | 2015 | 0 | 2014 |
| xpmem | 2479 | 2015 | 416 | 2014 |
+--------------------------+-------+------+-------+------+
有没有办法在mysql中实现这一点？ 谢谢你的建议。
这是最近的实施，因此我正在寻找2014年以后的数据。
答案 0 :(得分：1)
由于PHP已添加到标记中，因此一种简单的方法是使用连接构建查询，这些连接遍历多年：
// Your query for a particular year
function originalQuery($currYear) {
return "(SELECT substring_index(xo.module_name, '/', 1) as Module,
COUNT(date) AS Count, Year(date) as Year FROM xclude_run xr, join_run_object
jro, xclude_object xo WHERE xr.syshost='darter' AND xo.module_name IS NOT NULL
AND xr.run_id = jro.run_id AND jro.obj_id = xo.obj_id GROUP BY Module, Year
ORDER BY Module HAVING year = $currYear) as q$currYear ";
}
// The joined query can be generated using a for loop for multiple years.
$joinedQuery = "SELECT * " .
"FROM " . originalQuery(2015) . " JOIN " . originalQuery(2014) .
"ON q2015.Module = q2014.Module" .
"ORDER by q2015.Module";