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我先查询数据库以获取与特定用户ID相关的所有记录,然后我需要进入并修改数组,因为其中一个字段是一个ID,我需要与该ID相关联的名称。为什么PDOStatement-> columnCount不能返回正确的数字?
因此,我使用columnCount遍历id索引处的结果数组,并用正确的名称替换它,这对于前六个结果来说工作正常。 columnCount只返回6,但前六个按照它们应该重新命名。但是除此之外,它会从这个pdostatement获得结果并正常填充表格,并包含所有相关数据,现在有17行。
为什么它返回6,或者我在做什么来得到错误的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!男孩我觉得自己像一个tard。我试图得到返回结果的数量 – Ryan 2011-02-01 16:24:25