如果我想列出一个地址类似于用户输入的人的语法是什么?这是我的代码,请帮忙。表单输入将是地址。 我想做这样的事情:http://dev.mysql.com/doc/refman/5.0/en/pattern-matching.html
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("Hospital", $con);
$result = mysql_query("SELECT * FROM nais WHERE ADDRESS='{$_POST["address"]}'");
echo "<table border='1'>
<tr>
<th>HospNum</th>
<th>RoomNum</th>
<th>LastName</th>
<th>FirstName</th>
<th>MidName</th>
<th>Address</th>
<th>TelNum</th>
<th>Nurse</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['HOSPNUM'] . "</td>";
echo "<td>" . $row['ROOMNUM'] . "</td>";
echo "<td>" . $row['LASTNAME'] . "</td>";
echo "<td>" . $row['FIRSTNAME'] . "</td>";
echo "<td>" . $row['MIDNAME'] . "</td>";
echo "<td>" . $row['ADDRESS'] . "</td>";
echo "<td>" . $row['TELNUM'] . "</td>";
echo "<td>" . $row['NURSE'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
答案 0 :(得分:4)
试试这个:
$result = mysql_query("SELECT * FROM nais WHERE ADDRESS LIKE '%" . $_POST['address'] . "%';");
您还应该使用预备语句或mysql_real_escape_string()
,请参阅SQL Injections。