douxunchen3498 2012-11-10 12:55
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在MySQL COUNT()创建的JSON对象上使用$ .each

I have been using JSON objects encoded in PHP from my MySQL database in many places with no problems. I am now trying to create one from a resultset using COUNT(column) in the syntax and I think it is causing a the problem but I have no idea how to fix it.

My JS looks like this...

createTeacherStatsTab = function(){
    $('#main').append('<div id="teacherStatsTab></div>"');
    $.getJSON("php/countMarkingPerTeacher.php", function(data) {
        $.each(data, function(key, val) {
            $('#teacherStatsTab').append(val.teacher + ' : ' + val.count(teacher));
        });
    });
}

And my PHP looks like this...

$result = mysql_query("
SELECT teacher, COUNT(teacher) 
FROM Assessment
GROUP BY teacher
ORDER BY COUNT(teacher) DESC
");

$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows[] = $r;
}
print json_encode($rows);

The data that come back looks like this...

[
{
teacher: "SDe",
COUNT(teacher): "413"
},
{
teacher: "OJe",
COUNT(teacher): "297"
},
{
teacher: "AMi",
COUNT(teacher): "257"
},
{
teacher: "ASt",
COUNT(teacher): "218"
},
{
teacher: "VJa",
COUNT(teacher): "194"
},
{
teacher: "SLa",
COUNT(teacher): "125"
},
{
teacher: "MCr",
COUNT(teacher): "99"
}
]

It feels like this is probably something to do with the GROUP BY stripping ID's or something?

Can someone please help?

Thanks!

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4条回答 默认 最新

  • dongxing1853 2012-11-10 13:15
    关注

    This is because JS interprets your property access as a function call because of the parenthesis:

    val.count(teacher)
    

    The most simple solution i guess is to give your COUNT() result an alias name:

    $result = mysql_query("
    SELECT 
      teacher, 
      COUNT(teacher) AS teacher_count
    FROM Assessment
    GROUP BY teacher
    ORDER BY COUNT(teacher) DESC
    ");
    

    And access it like:

    $('#teacherStatsTab').append(val.teacher + ' : ' + val.teacher_count);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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