格式化MYSQL的PHP​​输出以获取morris.js图表​​数据

Question

I have the following query for retrieving data in my MYSQL database:

SELECT a.PARNT_CIVSTATUS,count(a.PARNT_CIVSTATUS) from tbl_parnt a left join tbl_intrvw b 
on a.QN_NUMBR=b.QN_NUMBR left join tbl_barangay c on b.ZONE_NUM=c.BRGY_ZONE_NUM 
group by a.PARNT_CIVSTATUS

This is the output from phpmyadmin:

And this is the output using json_encode in PHP:

[
{"0":"L","PARNT_CIVSTATUS":"L","1":"14","count(a.PARNT_CIVSTATUS)":"14"},
{"0":"LSP","PARNT_CIVSTATUS":"LSP","1":"9","count(a.PARNT_CIVSTATUS)":"9"},
{"0":"M","PARNT_CIVSTATUS":"M","1":"4386","count(a.PARNT_CIVSTATUS)":"4386"},
{"0":"N","PARNT_CIVSTATUS":"N","1":"45","count(a.PARNT_CIVSTATUS)":"45"},
{"0":"NON","PARNT_CIVSTATUS":"NON","1":"1","count(a.PARNT_CIVSTATUS)":"1"},
{"0":"O","PARNT_CIVSTATUS":"O","1":"12","count(a.PARNT_CIVSTATUS)":"12"},
{"0":"S","PARNT_CIVSTATUS":"S","1":"681","count(a.PARNT_CIVSTATUS)":"681"},
{"0":"SGP","PARNT_CIVSTATUS":"SGP","1":"143","count(a.PARNT_CIVSTATUS)":"143"},
{"0":"SP","PARNT_CIVSTATUS":"SP","1":"148","count(a.PARNT_CIVSTATUS)":"148"},
{"0":"SPG","PARNT_CIVSTATUS":"SPG","1":"12","count(a.PARNT_CIVSTATUS)":"12"},
{"0":"W","PARNT_CIVSTATUS":"W","1":"239","count(a.PARNT_CIVSTATUS)":"239"},
{"0":"WGW","PARNT_CIVSTATUS":"WGW","1":"1","count(a.PARNT_CIVSTATUS)":"1"},
{"0":"WW","PARNT_CIVSTATUS":"WW","1":"2","count(a.PARNT_CIVSTATUS)":"2"}
]

I don't know how to format it like this:

{ y: 'L', a: 10 },
{ y: 'LSP', a: 75 },
{ y: 'M', a: 50 },
{ y: 'N', a: 75 },
{ y: 'SGP', a: 50 },
{ y: 'SP', a: 75 },
{ y: 'W', a: 57 }

It's for a morris.js bar chart data. How do I format it this way?

Answer 1

This will require modification in more than just your sql, but it is easily possible; firstly, using the AS keyword , you can create an alias for the columns which you reference:

SELECT a.PARNT_CIVSTATUS AS 'y', count(a.PARNT_CIVSTATUS) AS 'a' 
FROM tbl_parnt a
LEFT JOIN tbl_intrvw b ON a.QN_NUMBR=b.QN_NUMBR 
LEFT JOIN tbl_barangay c ON b.ZONE_NUM=c.BRGY_ZONE_NUM 
GROUP BY a.PARNT_CIVSTATUS

The above, assuming I did things correctly, should give you this output .

However, that still leaves the php side of things. Given you haven't posted your php, I'm going to address this in the two ways I know how; deprecated mysql_* functions, and PDO:

Firstly, the deprecated functions. If you're using these, I highly recommend against doing so, and changing to either PDO or mysqli; anyway, to the code!

$arr = array();
while ($row = mysql_fetch_assoc($result)) {
    $arr[] = $row;
}

The above uses mysql_fetch_assoc() , and the example enclosed within that documentation page. It fetches an associative array of the results, assuming they are like the above SQLFiddle demo I provided, and shoves them into an existing array; this array would of course then be used in json_encode to produce the desired result.

The second method I spoke, PDO, or php Data Objects, is one I stand by; I'm not going to get into a complete demo here, as that's out of the scope of this answer, but will replicate the above code in PDO:

$arr = $stmt->fetchAll(PDO::FETCH_ASSOC);

Here, I use fetchAll() to fetch every row, and use the constant PDO::FETCH_ASSOC to make sure it only gets the column names and their values; otherwise, you get what you had above.

There is of course a third option, or even a fourth. Mysqli, and some custom-made database abstraction layer. I'm not experienced in either, and thus I won't be providing any examples save php.net's mysqli quick start guide .