Question

Wanted result Result - error- trying to use two statements

我已经尝试了好几天才能成功建立一个可以给我两个下拉框的表单，在那里我可以从同一个数据库中的不同表中选择记录。我的目标是能够将酒店房间从一个参与者转移到另一个参与者。为此，我需要先选择房间捐赠者，然后再选择收件人。我是在cms内做这个，所以我必须能够在同一个PHP脚本中进行选择。

一旦我加入第二个选项，我就会遇到问题。我不明白如何在同一表单上实现两个选择选项。有没有人有我的解决方案？这是我的代码：

    if (!isset($_POST['selected']))
    {   
        $DonorSelect = "Select * from rog_jomres_guests where property_uid in ($RallyHotels) order by surname";
        $DonorResult = mysql_query($DonorSelect) or die("Donor lookup Query failed" . mysql_error());
        $numrecords=mysql_num_rows($DonorResult); $rows = 0;
        ?>
        <html>
        <body>
        <form method="post" action="<?php echo $PHP_SELF;?>">
        <select name="donor">
        <?php
        while ($JosRow = mysql_fetch_array($DonorResult))
        {
            $Name = $JosRow['guests_uid'] . " " . $JosRow['surname'] . "," . $JosRow['firstname'];
            $id = $JosRow['mos_userid'];
            $gid=$JosRow['guests_uid'];
            $IDName = $Name; 

            $AttendSelect = "select * from rog_sembookings where userid = $id AND semid = $ThisRally";
            $AttendResult = mysql_query($AttendSelect) or die("$ThisRally query failed" . mysql_error());
            $AttendRow = mysql_fetch_array($AttendResult);

            if ($Name == "Administrator"){ }
            elseif (!$AttendRow) { }
            else
                {
                    ?>          
                    <option value="<?php echo $gid; ?>"><?php echo $IDName;?></option>
                <?php
                }
        }
        ?>
        </select>
        <?php /**  

        <select name="transferto">
        <?php
        $RecipientSelect = "Select * from rog_users order by name";
        $RecipientResult = mysql_query($RecipientSelect) or die("Recipient lookup Query failed" . mysql_error());
        $Recipientrecords=mysql_num_rows($RecipientResult); 
        while ($RecipientRow = mysql_fetch_array($RecipientResult))
        {
            $RecipientName = $RecipientRow['name'];
            $Recipientid = $RecipientRow['id'];
            $RIDName = $RecipientName; 
            $PartSelect = "select * from rog_users where userid = $Recipientid";
            $PartResult = mysql_query($PartSelect) or die("$ThisRally PartSelect query failed" . mysql_error());
            $PartRow = mysql_fetch_array($PartResult);

            if ($RecipientName == "Administrator"){ }
            elseif (!$PartRow) { }
            else
            {
                    ?>          
                    <option value="<?php echo $Recipientid; ?>"><?php echo $RIDName;?></option>
                <?php  
            }
        }
        ?>
        </select>
        <?php **/ ?>

        <input type="submit" value="<?php echo $Valg; ?>" name="selected">
        </input>
        </form>
        </body>
        </html>
        <?php

    }       // end of clause if not set     $_POST['selected']
    else
    {       // beginning clause if set  $_POST['selected'] let's do this thing......

        if ($AdminUser==1)
            { $CurrGuest    = $_POST["donor"];}

非常感谢任何帮助！感谢

Answer 1

试试这段代码：

if (!isset($_POST['selected']))
{   
    $DonorSelect = "Select * from rog_jomres_guests where property_uid in ($RallyHotels) order by surname";
    $DonorResult = mysql_query($DonorSelect) or die("Donor lookup Query failed" . mysql_error());
    $numrecords=mysql_numrows($DonorResult); $rows = 0;
    ?>
    <html>
    <body>
    <form method="post" action="<?php echo $PHP_SELF;?>">
    <select name="donor">
    <?php
    while ($JosRow = mysql_fetch_array($DonorResult))
    {
        $Name = $JosRow['guests_uid'] . " " . $JosRow['surname'] . "," . $JosRow['firstname'];
        $id = $JosRow['mos_userid'];
        $gid=$JosRow['guests_uid'];
        $IDName = $Name; 

        $AttendSelect = "select * from rog_sembookings where userid = $id AND semid = $ThisRally";
        $AttendResult = mysql_query($AttendSelect) or die("$ThisRally query failed" . mysql_error());
        $AttendRow = mysql_fetch_array($AttendResult);

        if ($Name == "Administrator"){ }
        elseif (!$AttendRow) { }
        else
            {
                ?>          
                <option value="<?php echo $gid; ?>"><?php echo $IDName;?></option>
            <?php
            }
    }
    ?>
    </select>

    <select name="transferto">
    <?php
    $RecipientSelect = "Select * from rog_users order by name";
    $RecipientResult = mysql_query($RecipientSelect) or die("Recipient lookup Query failed" . mysql_error());
    $Recipientrecords=mysql_numrows($RecipientResult); 
    while ($RecipientRow = mysql_fetch_array($RecipientResult))
    {
        $RecipientName = $RecipientRow['name'];
        $Recipientid = $RecipientRow['id'];
        $RIDName = $RecipientName; 
        $PartSelect = "select * from rog_users where userid = $Recipientid";
        $PartResult = mysql_query($PartSelect) or die("$ThisRally PartSelect query failed" . mysql_error());
        $PartRow = mysql_fetch_array($PartResult);

        if ($RecipientName == "Administrator"){ }
        elseif (!$PartRow) { }
        else
        {
                ?>          
                <option value="<?php echo $Recipientid; ?>"><?php echo $RIDName;?></option>
            <?php  
        }
    }
    ?>
    </select>

    <input type="submit" value="<?php echo $Valg; ?>" name="selected">
    </input>
    </form>
    </body>
    </html>
    <?php

}       // end of clause if not set     $_POST['selected']
else
{       // beginning clause if set  $_POST['selected'] let's do this thing......

    if ($AdminUser==1)
        { $CurrGuest    = $_POST["donor"];}

Answer 2

我自己解决了这个挑战。一些结构性逻辑挑战。案件结案

在PHP

2 个答案: