怎么实现多个查找框内只需填写至少一个查找内容即可进行查找
日期：2014-05-17 浏览次数：20522 次
如何实现多个查找框内只需填写至少一个查找内容即可进行查找
apply.html
<html>
<head>
<title>A new Application for a classroom</title>
</head>
<body>
<h1>CRMS - New classroom Entry</h1>
<form action="2.php" method="post">
<table border="0">
<tr>
<td>Classroom ID</td>
<td><input type="text" name="Cno" maxlength="13" size="13"></td>
</tr>
<tr>
<td>Course ID</td>
<td> <input type="text" name="CID" maxlength="30" size="30"></td>
</tr>
<tr>
<td colspan="2"><input type="submit" value="Register"></td>
</tr>
</table>
</form>
</body>
</html>
2.php
<html>
<head>
<title>A new Application for a classroom</title>
</head>
<body>
<h1>CRMS - New classroom Entry Results</h1>
<?php
// create short variable names
$Cno=$_POST['Cno'];
$CID=$_POST['CID'];
if (!$Cno) {$Cno = "%";}
if (!$CID) {$CID = "%";}
if (!get_magic_quotes_gpc()) {
$Cno = addslashes($Cno);
$CID = addslashes($CID);
}
$con = mysql_connect("localhost","root","");
//通过服务器locahost建立连接，用户名为root，无密码
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("crms", $con);
$search_classroom = "SELECT * FROM use2
WHERE Cno LIKE '{$Cno}' AND CID LIKE $CID";
$result = mysql_query($search_classroom, $con);
while($row = mysql_fetch_array($result))
{
echo $row['Cno'];
echo $row['CID'];
echo "<br />";
}
mysql_close($con)
?>
</body>
</html>
为什么这样不能实现当Classroom ID 或者Course ID没有填写时，任然能够成功搜索。我已经写了 if (!$Cno) {$Cno = "%";}
if (!$CID) {$CID = "%";}如果没有填就用通配符，而且查询时也用了LIKE（如果查询时LIKE全部换为=则成功）。为什么？在线等！
------解决方案--------------------
因为你用like了。去查下mysql手册。
apply.html
<html>
<head>
<title>A new Application for a classroom</title>
</head>
<body>
<h1>CRMS - New classroom Entry</h1>
<form action="2.php" method="post">
<table border="0">
<tr>
<td>Classroom ID</td>
<td><input type="text" name="Cno" maxlength="13" size="13"></td>
</tr>
<tr>
<td>Course ID</td>
<td> <input type="text" name="CID" maxlength="30" size="30"></td>
</tr>
<tr>
<td colspan="2"><input type="submit" value="Register"></td>
</tr>
</table>
</form>
</body>
</html>
2.php
<html>
<head>
<title>A new Application for a classroom</title>
</head>
<body>
<h1>CRMS - New classroom Entry Results</h1>
<?php
// create short variable names
$Cno=$_POST['Cno'];
$CID=$_POST['CID'];
if (!$Cno) {$Cno = "%";}
if (!$CID) {$CID = "%";}
if (!get_magic_quotes_gpc()) {
$Cno = addslashes($Cno);
$CID = addslashes($CID);
}
$con = mysql_connect("localhost","root","");
//通过服务器locahost建立连接，用户名为root，无密码
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("crms", $con);
$search_classroom = "SELECT * FROM use2
WHERE Cno LIKE '{$Cno}' AND CID LIKE $CID";
$result = mysql_query($search_classroom, $con);
while($row = mysql_fetch_array($result))
{
echo $row['Cno'];
echo $row['CID'];
echo "<br />";
}
mysql_close($con)
?>
</body>
</html>
为什么这样不能实现当Classroom ID 或者Course ID没有填写时，任然能够成功搜索。我已经写了 if (!$Cno) {$Cno = "%";}
if (!$CID) {$CID = "%";}如果没有填就用通配符，而且查询时也用了LIKE（如果查询时LIKE全部换为=则成功）。为什么？在线等！
------解决方案--------------------
因为你用like了。去查下mysql手册。
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