我正在根据用户的搜索结果生成一个动态表,只使用那些搜索到的列。该表在AJAX页面中生成,并在另一个页面中作为响应返回。单靠分页是行不通的。请帮忙。我的代码:
<?php
require_once('../Connections/finalkms.php');
$data = json_decode(stripslashes($_POST['data']), true);
$qry = " SELECT AssetId,";
$qry .= $data;
$qry .= " from Completedetails";
mysql_select_db($database_finalkms, $finalkms);
$query_getcolumns = $qry;
$getcolumns = mysql_query($query_getcolumns, $finalkms) or die(mysql_error());
$row_getcolumns = mysql_fetch_assoc($getcolumns);
$totalRows_getcolumns = mysql_num_rows($getcolumns);
if (($getcolumns)||(mysql_errno == 0))
{
echo "<table width='50%' class='table table-striped table-bordered table-hover' align='center' id='sample_2'>
<thead><tr id='vstr'>";
if (mysql_num_rows($getcolumns)>0)
{
$i = 0;
while ($i < mysql_num_fields($getcolumns))
{
echo "<th align='center'>". mysql_field_name($getcolumns, $i) . "</th>";
$i++;
}
echo "</tr></thead>";
while ($rows = mysql_fetch_array($getcolumns,MYSQL_ASSOC))
{
echo "<tbody><tr>";
foreach ($rows as $data)
{
echo "<td align='center'>". $data . "</td>";
}
}
}else{
echo "<tr><td colspan='" . ($i+1) . "'>No Results found!</td></tr></tr>";
}
echo "</tbody></table>";
}else{
echo "Error in running query :". mysql_error();
}
?>
Here's My Java script code:
<script>
$("#sample_2").dataTable();
</script>
这就是解决方案:
<?php
require_once('../Connections/finalkms.php');
$data = json_decode(stripslashes($_POST['data']), true);
$qry = " SELECT AssetId,";
$qry .= $data;
$qry .= " from Completedetails";
mysql_select_db($database_finalkms, $finalkms);
$query_getcolumns = $qry;
$getcolumns = mysql_query($query_getcolumns, $finalkms) or die(mysql_error());
$row_getcolumns = mysql_fetch_assoc($getcolumns);
$totalRows_getcolumns = mysql_num_rows($getcolumns);
if (($getcolumns)||(mysql_errno == 0))
{
echo "<table width='50%' class='table table-striped table-bordered table-hover' align='center' id='sample_2'>
<thead><tr id='vstr'>";
if (mysql_num_rows($getcolumns)>0)
{
$i = 0;
while ($i < mysql_num_fields($getcolumns))
{
echo "<th align='center'>". mysql_field_name($getcolumns, $i) . "</th>";
$i++;
}
echo "</tr></thead>";
while ($rows = mysql_fetch_array($getcolumns,MYSQL_ASSOC))
{
echo "<tr>"; //removed body tag here
foreach ($rows as $data)
{
echo "<td align='center'>". $data . "</td>";
}
}
}else{
echo "<tr><td colspan='" . ($i+1) . "'>No Results found!</td></tr></tr>";
}
echo "</table>";
}else{
echo "Error in running query :". mysql_error();
}
?>
Here's My Java script code:
<script>
$("#sample_2").dataTable();
</script>