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
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2022-06-15 12:27:21
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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
A new Application for a classroom
2.php
A new Application for a classroom
// 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 $row['CID'];
echo "
";
}
mysql_close($con)
?>
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
CRMS - New classroom Entry
2.php
CRMS - New classroom Entry Results
// 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 $row['CID'];
echo "
";
}
mysql_close($con)
?>
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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