我有一个下拉输入选择“评估测试类型”，它根据选择显示某些数据，并在其下方显示一个提交按钮。现在我添加到:“评估测试类型”的默认值 <option selected='selected'></option>但是，如果选择了此选项并单击了提交 1，我想阻止提交按钮出现

$options = '';
$filter=mysql_query("select afnumber from employees WHERE Status='Employed'");
while($row = mysql_fetch_array($filter)) {
    $options .="<option >" . $row['afnumber'] . "</option>";
}
$menu="<form id='filter' name='filter' method='post' action=''>
AFNumber : <select name='SelectAF' id='filter' style='color:grey;'>" . $options . "</select>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Evaluation Test Type : <select name='Type' id='type' style='color:grey;'><option selected='selected'></option><option value='loyalty'>Loyalty</option><option value='performance'>Performance</option></select>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<input type='submit' name='submit1' value='Submit' style='width:80px; height:30px; text-align:center; padding:0px;'>
</form>
<br>
";
 echo $menu;

if(isset($_POST['submit1']))

{   
$type = $_POST['Type'];

$mysqli = new mysqli("localhost", "root", "Js", "jr");
/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

 if ( $result = $mysqli->query( "SELECT questiontext FROM questioninfo WHERE type='$type'" ) ) {


        $html=array();

        $html[]="
        <form action='' method='post' id='quiz'>
            <ol>";

        $counter=1;

        while( $row = $result->fetch_array() ) {


            $question=$row['questiontext'];
            $answerA=1;
            $answerB=2;
            $answerC=3;
            $answerD=4;
            $answerE=5;

            $html[]="
             <br/>
                <h3>Question {$counter}:&nbsp; {$question}</h3>

                <li>
                    <br/>
                    <input type='radio' name='question-{$counter}-answers' id='question-$counter-answersA' value='1' />
                    <label for='question-{$counter}-answers-A'> {$answerA} </label>
                    <br/>
                    <input type='radio' name='question-{$counter}-answers' id='question-{$counter}-answersB' value='2' />
                    <label for='question-{$counter}-answers-B'> {$answerB} </label>
                    <br/>
                    <input type='radio' name='question-{$counter}-answers' id='question-{$counter}-answersC' value='3' />
                    <label for='question-{$counter}-answers-C'> {$answerC} </label>
                    <br/>
                    <input type='radio' name='question-{$counter}-answers' id='question-{$counter}-answersD' value='4' />
                    <label for='question-{$counter}-answers-D'> {$answerD} </label>
                    <br/>
                    <input type='radio' name='question-{$counter}-answers' id='question-{$counter}-answersE' value='5' />
                    <label for='question-{$counter}-answers-E'> {$answerE} </label>

                </li>";

            $counter++;

        }

        $html[]="
            </ol>
        <input type='submit' name='submit' value='Submit' style='width:80px; height:30px; text-align:center; padding:0px;'>
        <input type='hidden' name='type' value='{$type}' />
        </form>";

        echo implode( PHP_EOL, $html );



    $result->close();

 }
}

if( isset( $_POST['submit'] ) ){ 

    $mysqli = new mysqli("localhost", "root", "Js", "jr");
    if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();}

if ($result = $mysqli->query("SELECT * FROM questioninfo WHERE Type='performance'")) {

    $row_cnt = $result->num_rows;
    $result->close();
}
if ($result = $mysqli->query("SELECT * FROM questioninfo WHERE Type='loyalty'")) {

    $row_cnt1 = $result->num_rows;
    $result->close();
} 

$numQuestions=$row_cnt;
$numQuestions1=$row_cnt1; 
    $type = $_POST['type']; 
if($type == 'performance')
{
for( $counter=1; $counter <= $numQuestions; $counter++ ){
$type = $_POST['type']; 
$answer = $_POST['question-'.$counter.'-answers']; 
$sql="insert into `question` (`Type`,`Value`) values ('".$type."','".$answer."')"; 
$mysqli->query($sql);
} 
}
    else if($type == 'loyalty')
    {
for( $counter=1; $counter <= $numQuestions1; $counter++ ){
$type = $_POST['type']; 
$answer = $_POST['question-'.$counter.'-answers']; 
$sql="insert into `question` (`Type`,`Value`) values ('".$type."','".$answer."')"; 
$mysqli->query($sql);
} 

}
    else
    {
    }

}

最佳答案

如果你只是想阻止用户选择空白选项，只需使用 disabled它的属性。然后使用 required选择元素的属性，以防止他们提交空白的“评估测试类型”值。不要忘记在空白选项上添加 value='' 以使必需的属性按照回答 here 工作。 .

Evaluation Test Type : 
<select name='Type' id='type' style='color:grey;' required>
    <option value='' selected disabled></option>
    <option value='loyalty'>Loyalty</option>
    <option value='performance'>Performance</option>
</select>

关于php - 操作下拉列表的默认值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31980238/