使用php和基于用户输入值的表单从mysql检索所有数据

我有一个这样的数据库

  IDNo             Column1           Column2         .............    
  12341
  12342
  12343  
    .......

我向用户n提供了一个表单，告诉用户在表单中输入IDNO，以便用户可以检索相应的IDNO信息

如果用户在表单中输入12341并提交，则输出将是12341的信息 即 IDNO COUNN1第2列。。。。。。。。。 12341213 1231

这将是输出，为此我使用下面的代码

<form id="form" action="" method="post">
<td><p align="center"> Hallticket Number : <input type="text" name="id" id="id" maxlength="10"></p></td>
<input type="submit" id="submit" class='btnExample' value="Click here to get your Result">
</form>
<?PHP

$user_name = "admin";
$password = "123456";
$database = "exam";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);

if ($db_found) {
$id = mysql_real_escape_string($_POST['id']);
$SQL = "SELECT * FROM jbit WHERE htno='$id'";
$result = mysql_query($SQL);
echo "
<center><table id='mytable' cellspacing='0'  border=3 align=center>
<tr>
<TH scope='col'>SUBJECT CODE</TH>
<TH scope='col'>SUBJECT</TH>
<TH scope='col'>INTERNAL MARKS</TH>
<TH scope='col'>EXTERNAL MARKS</TH>
<TH scope='col'>TOTAL MARKS</TH>
<TH scope='col'>CREDITS</TH>
<TH scope='col'>RESULT</TH>
</tr><center>";          

while ($db_field = mysql_fetch_assoc($result)) {


echo "<tr>";
echo "<td align=center>" . $db_field['subjectcode'] . "</td>";
if ($db_field['credits']=="0")
{
echo "<td align=center><font color='red'>" . $db_field['subject'] . "</font></td>";
} else {
echo "<td align=center>" . $db_field['subject'] . "</td>";
}
echo "<td align=center>" . $db_field['im'] . "</td>";
echo "<td align=center>" . $db_field['em'] . "</td>";
echo "<td align=center>" . $db_field['tm'] . "</td>";
echo "<td align=center>" . $db_field['credits'] . "</td>";
if ($db_field['credits']>"0")
{
echo "<td align=center> <font color='green'><b> PASS </b></font></td>";
} else {
echo "<td align=center> <font color='red'><b> FAIL </b></font> </td>";
}
echo "</tr>";

}

如果用户单击12341，则输出应为

 IDNO          COLUMN1            Column2            ...
12341            321                323             ....

---

应该做什么？？？

您的sql应该使用

类似的
运算符：

那应该给你

$SQL = "SELECT * FROM jbit WHERE htno LIKE '1234%'";

$SQL = "SELECT * FROM jbit WHERE htno LIKE '$id%'";

$SQL = "SELECT * FROM jbit WHERE htno LIKE '1234%'";