我想根据4列的任意两个组合在表格中搜索 1-INSID 2- CRN 3-TERM 4-CID。我认为它是六个组合* 2因为val1可能意味着vla2。有更好的解决方案吗?
<?php
// define variables and set to empty values
$val1= $val2="" ;
if (isset($_POST['submit'])){
$val1=test_input($_POST["Val1"]);
$val2=test_input($_POST["Val2"]);
}
function test_input($data) {
$data=trim($data);
$data=stripslashes($data);
$data=htmlspecialchars($data);
return $data;
}
?>
<div class="container">
<div class="row">
<form name="myForm" class="col s12" method="post" action="<?php echo htmlspecialchars($_SERVER[" PHP_SELF "]);?>">
<div class="row">
<div class="input-field col s2">
<p>Enter any combination of:<br>1)CRN<br>2)INSID<br>3)CID<br>4)TERM</p>
<input id="Val1" type="text" class="validate" name="Val1">
<label for="Val1">First key to search for</label>
</div>
<div class="input-field col s2">
<input id="Val2" type="text" class="validate" name="Val2">
<label for="Val2">Second key to search for</label>
</div>
</div>
<button class="waves-effect waves-light btn" type="submit" name="submit">Search
<i class="mdi-content-send right"></i>
</button>
</form>
</div>
</div>
<?php
if (isset($_POST[ 'submit'])) {
if ($val1 !="" and $val2 !="" ) {
$con=mysql_connect( 'localhost', 'root', '');
if (!$con) {
die( 'Could not connect: ' . mysqli_connect_error());
}
$DBName="ICS324" ;
//echo 'Connected successfully';
if (!mysql_select_db($DBName, $con)) {
die( 'Could not select database SWE363');
}
//echo ( '</br> DB students selected successfully');
mysql_query("SELECT * FROM `SECTION`
WHERE
(
`CRN` ='$val1'
AND `TERM` = '$val2'
)
||
(
`CRN` ='$val1'
AND `INSID` = '$val2'
)
||
(
`CRN` ='$val1'
AND `CID` = '$val2'
)
||
(
`INSID` ='$val1'
AND `CID` = '$val2'
)
||
(
`TERM` ='$val1'
AND `CID` = '$val2'
)
||
(
`TERM` ='$val1'
AND `INSID` = '$val2'
)
(
`CRN` = '$val2'
AND `TERM` = '$val1'
)
||
(
`CRN` ='$val2'
AND `INSID` = '$val1'
)
||
(
`CRN` ='$val2'
AND `CID` = '$val1'
)
||
(
`INSID` ='$val2'
AND `CID` = '$val1'
)
||
(
`TERM` ='$val2'
AND `CID` = '$val1'
)
||
(
`TERM` ='$val2'
AND `INSID` = '$val1'
)");
echo "The table has the following records:";
// query here
$result=mysql_query( "SELECT * FROM PERSON WHERE 1") or die ( "Error in query: $query. ".mysql_error());
echo "<table border='1'>
<tr>
<th>CRN</th>
<th>TERM</th>
<th>INSID</th>
<th>CID</th>
<th>ACTIV</th>
<th>DAYS</th>
<th>BLD</th>
<th>ROOM</th>
<th>STime</th>
<th>ETime</th>
<th>Enrolled</th>
<th>MaxEnrolled</th>
</tr>";
while($row=mysql_fetch_array($result,MYSQL_ASSOC)) {
print "<tr>
<td>".$row['CRN']. "</td>
<td>".$row['TERM']. "</td>
<td>".$row['INSID']. "</td>
<td>".$row['CID']. "</td>
<td>".$row['ACTIVITY']. "</td>
<td>".$row['DAYS']. "</td>
<td>".$row['BUILDING']. "</td>
<td>".$row['ROOM']. "</td>
<td>".$row['START_TIME']. "</td>
<td>".$row['END_TIME']. "</td>
<td>".$row['NUMBER_OF_ENROLLED_STUDENTS']. "</td>
<td>".$row['MAXIMUM_LIMIT_ON_ENROLLMENT']. "</td>
</tr>";
}
echo "</table>";
mysql_close($con);
} else {
echo "Please fill out all the fields !";
}
}
?>