我正在使用基于用户搜索的PHP从MySQL检索结果
我想在结果
中每2秒后自动添加一个<?php
include_once("config.php");
isset( $_REQUEST['name'] ) ? $name=$_REQUEST['name'] : $name='';
$name = mysql_real_escape_string( $name );
if (strlen($name) >= 4) {
$sql = "select * from places where speciality like '%$name%'";
$rs = mysql_query( $sql ) or die('Database Error: ' . mysql_error
$num = mysql_num_rows( $rs );
if($num >= 1 ){
echo "<table id='result-table'><tr>";
while($row = mysql_fetch_array( $rs )){
echo "<td>row[cname]</td>"; //here i want to add </tr><tr> after 2 <td>s
}
}else{
echo "no records found</table>";
}
}
?>
答案 0 :(得分:2)
echo "<table id='result-table'><tr>";
$currentCount = -1;
while($row = mysql_fetch_array($rs))
{
echo "<td>row[cname]</td>";
$currentCount = ($currentCount + 1) % 2;
if($currentCount == 1)
{
echo '</tr><tr>';
}
}
答案 1 :(得分:2)
已使用$i = 1;
if ($i % 2 == 0)
和$i++;
<?php
include_once ("config.php");
isset ($_REQUEST['name']) ? $name = $_REQUEST['name'] : $name = '';
$name = mysql_real_escape_string($name);
if (strlen($name) >= 4) {
$sql = "select * from places where speciality like '%$name%'";
$rs = mysql_query($sql) or die('Database Error: '.mysql_error $num = mysql_num_rows($rs);
if ($num >= 1) {
$i = 1;
echo "<table id='result-table'><tr>";
while ($row = mysql_fetch_array($rs)) {
echo "<td>row[cname]</td>"; //here i want to add </tr><tr> after 2 <td>s
if ($i % 2 == 0)
echo "</tr><tr>";
$i++;
}
}
else {
echo "no records found</table>";
}
}
?>
答案 2 :(得分:0)
我建议不要把它们放在桌子上;而是简单地将它们放入<div>
个元素中,使用float:left
和width:50%
个样式。
如果您必须将它们放入表中,并且您希望按照您的要求进行操作,则可以使用PHP的模运算符(%
)来划分当前记录号2并得到余数。如果为1,则添加<tr>
标记:
if(++$rowNum % 2) {print "</tr><tr>";}