Question

我报告的问题更简洁一些。（删除其他）

我有一个我建立的时间报告系统，我正在尝试创建一个遵守类型报告。

基本上我正在尝试执行以下操作：

根据where子句和字段practice = ccv＆amp;，收集我数据库中的所有唯一名称（$ fn，$ ln）。年= 2016年（例如）
然后我还想收集另一个名单（$ fn，$ ln），基于practice = ccv＆amp; year = 2016和其他where子句= 8
这将为我提供2个可以比较的DISTINCT列表，差异将是第8周的优秀时间卡。

从那里，我需要显示提交的所有名称（来自上面的＃3）以及未完成的名单（通过比较上面的＃1和＃2）

到目前为止，我可以获得名单。但是，我需要使用数据库中的每个人（Region，Timestamp等）显示其他数据。

期望输出将是这样的：
的提交
的 EMEA
约翰J Wally R
北美
戴维斯D
鲍勃C
约翰Y
的杰出
的 EMEA
凯尔D
北美
Cory T
休R 约翰J 的 LATAM
等....

到目前为止，这就是我所拥有的......我确信我会以错误的方式解决这个问题。

<?php
$con = mysqli_connect("localhost","xxx","xxx!","xxx");                  

//Get distinct names from practice
if ($con) {
$SQL = "SELECT DISTINCT fn,ln FROM $table WHERE year='$year' AND practice='$practice' AND archived!='yes' ORDER BY region,fn,ln";   
}

$result = mysqli_query($con,$SQL);

if (!$result) die('Couldn\'t fetch records'); 

while ( $db_field = mysqli_fetch_assoc($result) ) {
$name[] = trim($db_field['fn']. " " .$db_field['ln']);
}


//Get distinct names for current week practice
if ($con) {
$SQL2 = "SELECT DISTINCT fn,ln FROM $table WHERE year='$year' AND week_num='$week' AND practice='$practice' AND archived!='yes' ORDER BY region,fn,ln"; 
}

$result2 = mysqli_query($con,$SQL2);

if (!$result2) die('Couldn\'t fetch records Again'); 

while ( $db_field = mysqli_fetch_assoc($result2) ) {

$name2[] = trim($db_field['fn']. " " .$db_field['ln']);
}

//$SQL3 = "SELECT fn,ln,week_start_date,region FROM $table WHERE fn IN ( '".implode("', '", $differences)."' )");

mysqli_close($con);
?>

<TABLE><TR><TD>
<B>All Names</B><BR>
<?php
foreach ( $name as $item ) {
    echo $item . "<br/>";
}
?>
</TD>
<TD>
<B>Names for Week 8</B><BR>
<?php
foreach ( $name2 as $item2 ) {
    echo $item2 . "<br/>";
}
?>
</TD>
<TD>
<B>Outstanding</B><BR>
<?php
$results = array_diff($name, $name2);

foreach($results as $val) {
    echo $val ." - ".$val2."<BR>";          
}
?>

</TD>
</TR>
</TABLE>

Answer 1

我认为你应该在MySQL级别解决这个问题。

// All names with all data
$SQL = "SELECT DISTINCT *, CONCAT(fn,' ',ln) AS `name` FROM $table WHERE year='$year' AND practice='$practice' AND archived!='yes' ORDER BY region,fn,ln";

// Names for specific week with all data
$SQL = "SELECT DISTINCT *, CONCAT(fn,' ',ln) AS `name` FROM $table WHERE year='$year' AND week_num='$week' AND practice='$practice' AND archived!='yes' ORDER BY region,fn,ln";

// Outstanding rows with all data
$SQL = "SELECT * FROM (SELECT DISTINCT *, CONCAT(fn,' ',ln) AS `name` FROM $table WHERE year='$year' AND practice='$practice' AND archived!='yes') AS `subquery` WHERE `name` NOT IN (SELECT DISTINCT CONCAT(fn,' ',ln) AS `name` FROM $table WHERE year='$year' AND week_num='$week' AND practice='$practice' AND archived!='yes') ORDER BY region,fn,ln";

这样您就不需要在PHP级别对数据进行排序和处理。无法测试这个，因为我没有数据库环境，但它应该工作，可能需要一些小的改动。

显示MySQL数据库中的唯一值

1 个答案: