php - PHP を使用して動的 SQL ピボット テーブルを提供する際の構文エラー

Question

この動的なピボット SQL は MS Studio で完全に実行されますが、サーバー上の php で実行することはできません。「、」の近くで不正な構文が表示されます。また、「p」付近の構文が正しくありません。私が知っているあらゆる種類の「実行」と、あらゆる種類の句読点を試しました。8時間で十分です。専門家を呼んでください。

SQL:

declare @col varchar(8000)
declare @sql varchar(8000)

if OBJECT_ID('tempdb..#q1') is not null
drop table #q1

if OBJECT_ID('tempdb..#q2') is not null
drop table #q2

select top 5 row_number() over (
    order by PTC_ASSESSMENT.ASSESSMENT_DATE desc
    ) as row, PTC_ASSESSMENT.ASSESSMENT_ID
into #q1
from PTC_ASSESSMENT
inner join PT_BASIC
on PTC_ASSESSMENT.PATIENT_ID = PT_BASIC.PATIENT_ID
inner join SYS_AS_DDF
on PTC_ASSESSMENT.AS_DDF_ID = SYS_AS_DDF.AS_DDF_ID
    and PTC_ASSESSMENT.DISCIPLINE_CODE = SYS_AS_DDF.DISCIPLINE_CODE
where (PT_BASIC.PATIENT_CODE = '60085')
and (PTC_ASSESSMENT.DISCIPLINE_CODE = 'MD')
group by PTC_ASSESSMENT.ASSESSMENT_DATE, PTC_ASSESSMENT.ASSESSMENT_ID

select convert(varchar, ASSESSMENT_DATE, 101) as dates, QUESTION_TEXT, 
case when ENTRY_TEXT is not null then ENTRY_TEXT else ANSWER_TEXT end as   ANSWER_TEXT
into #q2
from VW_ASSESSMENT_DATA
inner join PT_BASIC
on VW_ASSESSMENT_DATA.PATIENT_ID = PT_BASIC.PATIENT_ID
inner join SYS_AS_DDF
on VW_ASSESSMENT_DATA.AS_DDF_ID = SYS_AS_DDF.AS_DDF_ID
inner join #q1
on VW_ASSESSMENT_DATA.ASSESSMENT_ID = #q1.ASSESSMENT_ID
where (PT_BASIC.PATIENT_CODE = '60085')
and (SYS_AS_DDF.DISCIPLINE_CODE = 'MD')
and VW_ASSESSMENT_DATA.QUESTION_ID in ( 22222, 22223, 22237, 22238, 22239, 22241, 22242 )

select @col = COALESCE(@col + ',', '') + QUOTENAME(dates)
from #q2
group by dates

set @sql = 'select QUESTION_TEXT, ' + @col + '
                from (
                    select distinct *
                    from #q2
                    ) p
                pivot
                (
                max(ANSWER_TEXT) 
                for dates 
                in (' + @col + ' )) as pvt'

execute (@sql)

PHP: これを使用して取得します。

$result = mssql_query($sql) or exit ("select query failed");

PHP スクリプト全体:

$sql = "above sql";

$result = mssql_query($sql) or exit ("select query failed");
$numrows = mssql_num_rows($result);

echo "<table width='60%' cellpadding='5%'>
  <tr>";
if ($numrows > 0) { 
 //loop thru the field names to print the correct headers 
   $i = 0; 
   while ($i < mssql_num_fields($result)) 
 { 
  echo "<th>". mssql_field_name($result, $i) . "</th>"; 
$i++; 
 } 
  echo "</tr>"; 
  //display the data 
  while ($rows = mssql_fetch_array($result)) 
{ 
  echo "<tr>"; 
  foreach ($rows as $data) 
    { 
  echo "<td align='left'>". $data . "</td>"; 
} 
    } 
    }else{ 
echo "<tr><td colspan='" . ($i+1) . "'>No patient medical history found</td></tr>"; 
} 
echo "</table>";