Question

我试图从我的页面1上的表单中获取一个变量到我在第2页上进行的查询，这样当查询工作时它将使用第1页上输入的变量

Page 1表格

<form method="post" action="testformQ.php"> <input type="text" name="testform"> <input type="submit">enter

第2页查询

'$software = mysql_query("SELECT software.SoftwareID, rooms.RoomID 
    FROM softwareroom 
    INNER JOIN software 
    ON software.SoftwareID=softwareroom.SoftwareID 
    INNER JOIN rooms ON rooms.RoomID=softwareroom.RoomID 
    WHERE rooms.RoomNum=$_GET[testform]");              (the where clause is where i want the variable)
        while($rec = mysql_fetch_array($software))
        {
            echo $rec['SoftwareID'] . " " . $rec['RoomID'];
        }'

所以GET testform在查询中的位置是我想要表单变量的地方。任何帮助将不胜感激

Answer 1

此代码适用于您

$data = $_REQUEST['testform'];

$software = mysql_query("SELECT software.SoftwareID, rooms.RoomID 
    FROM softwareroom 
    INNER JOIN software 
    ON software.SoftwareID=softwareroom.SoftwareID 
    INNER JOIN rooms ON rooms.RoomID=softwareroom.RoomID 
    WHERE rooms.RoomNum='$data'");        
    while($rec = mysql_fetch_array($software))
        {
            echo $rec['SoftwareID'] . " " . $rec['RoomID'];
        }'

Answer 2

对此进行Page02查询。

    $servername = "servername";
    $username = "username";
    $password = "password";
    $db = "db_name";

    $query = "SELECT software.SoftwareID, rooms.RoomID FROM softwareroom INNER JOIN software ON software.SoftwareID = softwareroom.SoftwareID"
            + "INNER JOIN rooms ON rooms.RoomID = softwareroom.RoomID WHERE rooms.RoomNum = '$_POST["testform"]'";

    //Create db connection
    $connection = mysqli_connect($servername, $username, $password, $db);

    //Check connection
    if(!$connection){
        echo "Failed to connect to database ".mysqli_connect_error();
    }
    $software = mysqli_query($connection, $query);
    mysqli_close($connection);

将表单变量传递给Php查询

2 个答案: