我最近完成了我的搜索引擎,但现在我有一个新的挑战。
下面的代码我正在使用它从我的数据库中的呼叫流表中读取值并将它们显示在一个表中,让您知道呼叫是否被回答是或否。
if(isset($res))
{
//creating table
echo '<table style="width:1500px; cell-padding:4px; cell-spacing:0; margin:auto;">';
echo'<th>Time</th><th>Answered Y/N</th></th><th>Naam</th><th>Caller ID</th>';
while($result = mysql_fetch_assoc($res))
{
echo '<tr>';
echo '<td>'.$result['statusCalling'].'</td>';
if ($result['statusAnswered'] =="NULL"||$result['statusAnswered'] =="Null" || $result['statusAnswered'] =="null" || $result['statusAnswered'] =="")
{
echo "<td>Not Answered!</td>";
}
else
{
echo "<td>Answered!</td>";
}
echo '<td>'.$result['calleridname'].'</td>'.'<td>'.$result['calleridnum'].'</td>' ;
echo '</tr>';
}
echo '</table>';
}
我现在需要在搜索引擎结果中显示这些结果!我试过这个,但我不工作!不知道该怎么做!请帮忙!
$output = '';
//collect
if(isset($_POST['asd'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysql_query('SELECT * FROM callflow WHERE statusCalling LIKE "%'.$searchq.'%" OR calleridname LIKE "%'.$searchq.'%" OR calleridnum LIKE "%'.$searchq.'%" OR $results LIKE "%'.$searchq'%"');
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'There was no search results!';
}else{
while($row = mysql_fetch_array($query)) {
$statusCalling = $row['statusCalling'];
$calleridname = $row['calleridname'];
$calleridnum = $row['calleridnum'];
$results = $row['statusAnswered'];
$id = $row['ID'];
$output .= '<div>'.$statusCalling.' '.$calleridname.' '.$calleridnum.' '.$results.'</div>';
}
}
}
我知道 mysql 已被弃用,我仍在学习编程,我想如果我不知道 mysql,我就无法学习 pdo,因为我不明白什么是什么。请帮忙!