下拉列表可能与MYSQL PHP不一样

时间:2014-03-16 21:35:10

标签: javascript php html mysql drop-down-menu

我有两个下拉菜单,输入相同。所以" thuisteam1"可以和" uitteam1"相同。但这不是我想要的。我想要你选择" thuisteam1"你不能在" uitteam1"。

上选择同一个团队的团队

我尝试了一些javascript,但这对我没用。我希望它可以在PHP中完成。或者是一个可行的JavaScript。

希望你们能帮忙。

这是我的代码:

<form id="mySubmitForm" action="e2admin.php" method="post">
 <select name="thuisteam1">
                    <option value=""></option>
                    <?php
                        mysql_data_seek($result, 0);

                        if(mysql_num_rows($result) > 0){
                            while($row = mysql_fetch_array($result)) {
                               echo '<option value="' . $row['Team'] . '">' . $row['Team'] . '</option>';
                            }
                        }
                     ?>

                </select>
                <input onKeyPress="return alpha(event)" style="width:20px; text-align:center;" type="text" maxlength="2" class="form-control" name="scorethuis1" placeholder="0" />
                -
                <input onKeyPress="return alpha(event)" style="width:20px; text-align:center;" type="text" maxlength="2" class="form-control" name="scoreuit1" placeholder="0" />
                <select name="uitteam1">
                    <option value=""></option>
                    <?php
                        mysql_data_seek($result, 0);

                        if(mysql_num_rows($result) > 0){
                            while($row = mysql_fetch_array($result)) {
                               echo '<option value="' . $row['Team'] . '">' . $row['Team'] . '</option>';
                            }
                        }
                     ?>
                </select><br>
 <input style="margin-left:330px;" type="submit" class="form-control" value="Toevoegen" />

            </form>         

3 个答案:

答案 0 :(得分:2)

这个简单的jquery可以帮助您检测所选内容是否相同:

$( "select" ).change(function () {
    if ($( "select[name='thuisteam1'] option:selected" ).text() == $( "select[name='uitteam1'] option:selected" ).text()) {
        $("option:selected").removeAttr("selected");
        alert('can not select the same');
    };

});

我试过这段代码并且有效。

这里是所有示例代码:

<!doctype html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>test</title>

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>

</head>
<body>
<select name="thuisteam1" multiple="multiple">
    <option>team 1</option>
    <option>team 2</option>
    <option>team 3</option>
</select>

<select name="uitteam1" multiple="multiple">
    <option>team 1</option>
    <option>team 2</option>
    <option>team 3</option>
</select>

<script>
$( "select" )
    .change(function () {

    if ($( "select[name='thuisteam1'] option:selected" ).text() == $( "select[name='uitteam1'] option:selected" ).text()) {
        $("option:selected").removeAttr("selected");
        alert('same');
    };

});
</script>
</body>
</html>

答案 1 :(得分:0)

您可以发送表单并在Server PHP Way上验证

或者您在Browser JS Way中验证

在Jquery中

$('#<idselecttwo>').bind('change' function(){
 var valueOne = $('#<idselectone>').val();
 var valueTwo = $('#<idselecttwo>').val();
 if(valueOne == valueTwo)
 {
    alert('error');
 }

});

答案 2 :(得分:0)

根据要求,我从你的代码中做了一个例子。我没有测试过,但那是我的想法。此代码假定e2admin.php是完全相同的文件。

<?php

$send=$_POST['send'];
if($send==1)
{
    $team1=$_POST['thuisteam1'];
    $team2=$_POST['uitteam1'];
    if(isset($_POST['submitbutton']))
    {
        //Do your stuff
    }
}

echo '
    <form id="mySubmitForm" action="e2admin.php" method="post">
    <input type="hidden" name="send" value="1">
            <select name="thuisteam1" onchange="this.form.submit();">
                    <option value=""></option>
';
                        mysql_data_seek($result, 0);

                        if(mysql_num_rows($result) > 0)
                        {
                            while($row = mysql_fetch_array($result)) 
                            {
                                if($team1 != $row['Team'])
                                {
                                    echo '<option value="' . $row['Team'] . '">' . $row['Team'] . '</option>';
                                }
                            }
                        }
echo '
                </select>
                <select name="uitteam1" onchange="this.form.submit();">
                    <option value=""></option>
';
                        mysql_data_seek($result, 0);

                        if(mysql_num_rows($result) > 0)
                        {
                            while($row = mysql_fetch_array($result)) 
                            {
                                if($team2 != $row['Team'])
                                {
                                    echo '<option value="' . $row['Team'] . '">' . $row['Team'] . '</option>';
                                }
                            }
                        }
echo '
                </select><br>
                <input style="margin-left:330px;" type="submit" name="submitbutton" class="form-control" value="Toevoegen" />
            </form>
';

?>

编辑:在继续操作之前,您可能希望为提交按钮指定名称并设置chech。这样,只有在实际按下提交按钮时才会继续。