1
我先查詢數據庫以獲取與特定用戶ID相關的所有記錄,然後我需要進入並修改數組,因爲其中一個字段是一個ID,我需要與該ID相關聯的名稱。爲什麼PDOStatement-> columnCount不能返回正確的數字?
因此,我使用columnCount遍歷id索引處的結果數組,並用正確的名稱替換它,這對於前六個結果來說工作正常。 columnCount只返回6,但前六個按照它們應該重新命名。但是除此之外,它會從這個pdostatement獲得結果並正常填充表格,幷包含所有相關數據,現在有17行。
爲什麼它返回6,或者我在做什麼來得到錯誤的columncount?
global $__CMS_CONN__;
$timeqry = 'SELECT facility_id, program, date, visit_length, mileage, served FROM timesheet_db WHERE volunteer_id = '.$_SESSION['user_id'];
$stmt = $__CMS_CONN__->prepare($timeqry);
$stmt->execute();
$columns = $stmt->columnCount();
print $columns;
if($stmt)
{
$arrValues = $stmt->fetchAll(PDO::FETCH_ASSOC);
for($x=0;$x<$stmt->columnCount();$x++)
{
global $__CMS_CONN__;
$qry = 'SELECT facility FROM facility_db WHERE id = '.$arrValues[$x]['facility_id'];
$stmt1 = $__CMS_CONN__->prepare($qry);
$stmt1->execute();
if($stmt1)
{
$facilityName = $stmt1->fetchAll(PDO::FETCH_ASSOC);
foreach ($facilityName as $item)
{
foreach ($item as $key => $val)
{
$arrValues[$x]['facility_id'] = $val;
}
}
}
}
print "<table style=\"font-size:90%\">\n";
print "<tr>\n";
print "<th style=\"width:100%\">Facility</th>";
print "<th>Program</th>";
print "<th>Date</th>";
print "<th>Visit Length</th>";
print "<th>Mileage</th>";
print "<th>Served</th>";
print "</tr>";
foreach ($arrValues as $row)
{
print "<tr>";
foreach ($row as $key => $val)
{
print "<td>$val</td>";
}
print "</tr>\n";
}
print "</table>\n";
}
LOL!男孩我覺得自己像一個tard。我試圖得到返回結果的數量 – Ryan 2011-02-01 16:24:25